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n! = n · (n − 1) · (n − 2) · · · 2 · 1 with 0! = 1 as a special case. (We will study factorials in greater detail in Section 9.4.) The world famous Fibonacci Numbers are deﬁned recursively and are explored in the exercises. While none of the sequences worked out to be the sequence in (1), they do give us some insight into what kinds of patterns to look for. Two patterns in particular are given in the next deﬁnition. Deﬁnition 9.2. Arithmetic and Geometric Sequences: Suppose {an}∞ n=k is a sequencea If there is a number d so that an+1 = an + d for all n ≥ k, then {an}∞ arithmetic sequence. The number d is called the common diﬀerence. n=k is called an If there is a number r so that an+1 = ran for all n ≥ k, then {an}∞ n=k is called a geometric sequence. The number r is called the common ratio. aNote that we have adjusted for the fact that not all ‘sequences’ begin at n = 1. Both arithmetic and geometric sequences are deﬁned in terms of recursion equations. In English, an arithmetic sequence is one in which we proceed from one term to the next by always adding the ﬁxed number d. The name ‘common diﬀerence’ comes from a slight rewrite of the recursion equation from an+1 = an + d to an+1 − an = d. Analogously, a geometric sequence is one in which we proceed from one term to the next by always multiplying by the same ﬁxed number r. If r = 0, we can rearrange the recursion equation to get an+1 = r, hence the name ‘common ratio.’ Some an sequences are arithmetic, some are geometric and some are neither as the next example illustrates.3 Example 9.1.2. Determine if the following sequences are arithmetic, geometric or neither. arithmetic, ﬁnd the common diﬀerence d; if geometric, ﬁnd the common ratio r. If 1. an = 5n−1 3n, n ≥ 1 3. {2n − 1}∞ n=1 2
. bk = (−1)k 2k + 1, k ≥ 0 4, − 27 16,... 2We’re basically talking about the ‘countably inﬁnite’ subsets of the real number line when we do this. 3Sequences which are both arithmetic and geometric are discussed in the Exercises. 9.1 Sequences 655 Solution. A good rule of thumb to keep in mind when working with sequences is “When in doubt, write it out!” Writing out the ﬁrst several terms can help you identify the pattern of the sequence should one exist. 1. From Example 9.1.1, we know that the ﬁrst four terms of this sequence are 1 27 and 125 81. To see if this is an arithmetic sequence, we look at the successive diﬀerences of terms. We ﬁnd that a2 − a1 = 5 27. Since we get diﬀerent numbers, there is no ‘common diﬀerence’ and we have established that the sequence is not arithmetic. To investigate whether or not it is geometric, we compute the ratios of successive terms. The ﬁrst three ratios 9 and a3 − a2 = 25 9 = 10 27 − 5 3 = 2 9 − 1 9, 25 3, 5 a2 a1 = 5 9 1 3 = 5 3, a3 a2 = = 5 3 and a4 a3 = 25 27 5 9 125 81 25 27 = 5 3 suggest that the sequence is geometric. To prove it, we must show that an+1 an = r for all n. an+1 an = 5(n+1)−1 3n+1 5n−1 3n = 5n 3n+1 · 3n 5n−1 = 5 3 This sequence is geometric with common ratio r = 5 3. 3, 1 2. Again, we have Example 9.1.1 to thank for providing the ﬁrst four terms of this sequence: 15. Hence, the sequence is not 5. Since there is no 1, − 1 3 and b2 − b1 = 8 3 and b2 arithmetic. To see if it is geometric, we compute b1 b1 b0 ‘common ratio,’ we conclude the sequence is not geometric, either. 7. We ﬁnd b1 − b0 = −
4 5 and − 1 = − 1 = − 3 3. As we saw in Example 9.1.1, the sequence {2n − 1}∞ n=1 generates the odd numbers: 1, 3, 5, 7,.... Computing the ﬁrst few diﬀerences, we ﬁnd a2 − a1 = 2, a3 − a2 = 2, and a4 − a3 = 2. This suggests that the sequence is arithmetic. To verify this, we ﬁnd an+1 − an = (2(n + 1) − 1) − (2n − 1) = 2n + 2 − 1 − 2n + 1 = 2 This establishes that the sequence is arithmetic with common diﬀerence d = 2. To see if it is geometric, we compute a2 3. Since these ratios are diﬀerent, we conclude the a1 sequence is not geometric. = 3 and a3 a2 = 5 4. We met our last sequence at the beginning of the section. Given that a2 − a1 = − 5 4 and a3 − a2 = 15 8, the sequence is not arithmetic. Computing the ﬁrst few ratios, however, gives us 2, a3 a2 = − 3 2. Since these are the only terms given to us, we assume that a2 a1 the pattern of ratios continue in this fashion and conclude that the sequence is geometric. 2 and a4 a3 = − 3 = − 3 We are now one step away from determining an explicit formula for the sequence given in (1). We know that it is a geometric sequence and our next result gives us the explicit formula we require. 656 Sequences and the Binomial Theorem Equation 9.1. Formulas for Arithmetic and Geometric Sequences: An arithmetic sequence with ﬁrst term a and common diﬀerence d is given by an = a + (n − 1)d, n ≥ 1 A geometric sequence with ﬁrst term a and common ratio r = 0 is given by an = arn−1, n ≥ 1 While the formal proofs of the formulas in Equation 9.1 require the techniques set forth in Section 9.3, we attempt to motivate them here. According to Deﬁnition 9.2, given an arithmetic sequence with ﬁrst term a and common
diﬀerence d, the way we get from one term to the next is by adding d. Hence, the terms of the sequence are: a, a + d, a + 2d, a + 3d,.... We see that to reach the nth term, we add d to a exactly (n − 1) times, which is what the formula says. The derivation of the formula for geometric series follows similarly. Here, we start with a and go from one term to the next by multiplying by r. We get a, ar, ar2, ar3 and so forth. The nth term results from multiplying a by r exactly (n − 1) times. We note here that the reason r = 0 is excluded from Equation 9.1 is to avoid an instance of 00 which is an indeterminant form.4 With Equation 9.1 in place, we ﬁnally have the tools required to ﬁnd an explicit formula for the nth term of the sequence given in (1). We know from Example 9.1.2 that it is geometric with common ratio r = − 3 2. The ﬁrst term is a = 1 for n ≥ 1. After a touch of simplifying, we get an = (−3)n−1 for n ≥ 1. Note that we can easily check our answer by substituting in values of n and seeing that the formula generates the sequence given in (1). We leave this to the reader. Our next example gives us more practice ﬁnding patterns. 2 so by Equation 9.1 we get an = arn−1 = 1 − 3 2 n−1 2n 2 Example 9.1.3. Find an explicit formula for the nth term of the following sequences. 1. 0.9, 0.09, 0.009, 0.0009,... 2. 2 5, 2. 1, − 2 7, 4 13, − 8 19,... Solution. 1. Although this sequence may seem strange, the reader can verify it is actually a geometric for 10n, n ≥ 1. There is more to this sequence than meets the sequence with common ratio r = 0.1 = 1 n ≥ 0. Simplifying, we get an = 9 eye and we shall return to this example in the next section. 10. With a = 0.9 = 9 10, we get an = 9 1
10 n−1 10 2. As the reader can verify, this sequence is neither arithmetic nor geometric. In an attempt to ﬁnd a pattern, we rewrite the second term with a denominator to make all the terms appear as fractions. We have 2 7,.... If we associate the negative ‘−’ of the last two −3, 2 1, 2 terms with the denominators we get 2 −7,.... This tells us that we can tentatively sketch out the formula for the sequence as an = 2 where dn is the sequence of denominators. dn 4See the footnotes on page 237 in Section 3.1 and page 418 of Section 6.1. 9.1 Sequences 657 Looking at the denominators 5, 1, −3, −7,..., we ﬁnd that they go from one term to the next by subtracting 4 which is the same as adding −4. This means we have an arithmetic sequence on our hands. Using Equation 9.1 with a = 5 and d = −4, we get the nth denominator by the formula dn = 5 + (n − 1)(−4) = 9 − 4n for n ≥ 1. Our ﬁnal answer is an = 2 9−4n, n ≥ 1. 3. The sequence as given is neither arithmetic nor geometric, so we proceed as in the last problem to try to get patterns individually for the numerator and denominator. Letting cn and dn denote the sequence of numerators and denominators, respectively, we have an = cn. After dn some experimentation,5 we choose to write the ﬁrst term as a fraction and associate the negatives ‘−’ with the numerators. This yields 1 19,.... The numerators form the sequence 1, −2, 4, −8,... which is geometric with a = 1 and r = −2, so we get cn = (−2)n−1, for n ≥ 1. The denominators 1, 7, 13, 19,... form an arithmetic sequence with a = 1 and d = 6. Hence, we get dn = 1 + 6(n − 1) = 6n − 5, for n ≥ 1. We obtain our formula for an = cn dn 6n−5, for n ≥ 1. We leave it to the
reader to show that this checks out. = (−2)n−1 13, −8 1, −2 7, 4 While the last problem in Example 9.1.3 was neither geometric nor arithmetic, it did resolve into a combination of these two kinds of sequences. If handed the sequence 2, 5, 10, 17,..., we would be hard-pressed to ﬁnd a formula for an if we restrict our attention to these two archetypes. We said before that there is no general algorithm for ﬁnding the explicit formula for the nth term of a given sequence, and it is only through experience gained from evaluating sequences from explicit formulas that we learn to begin to recognize number patterns. The pattern 1, 4, 9, 16,... is rather recognizable as the squares, so the formula an = n2, n ≥ 1 may not be too hard to determine. With this in mind, it’s possible to see 2, 5, 10, 17,... as the sequence 1 + 1, 4 + 1, 9 + 1, 16 + 1,..., so that an = n2 + 1, n ≥ 1. Of course, since we are given only a small sample of the sequence, we shouldn’t be too disappointed to ﬁnd out this isn’t the only formula which generates this sequence. For example, consider the sequence deﬁned by bn = − 1 2 n − 5, n ≥ 1. The reader is encouraged to verify that it also produces the terms 2, 5, 10, 17. In fact, it can be shown that given any ﬁnite sample of a sequence, there are inﬁnitely many explicit formulas all of which generate those same ﬁnite points. This means that there will be inﬁnitely many correct answers to some of the exercises in this section.6 Just because your answer doesn’t match ours doesn’t mean it’s wrong. As always, when in doubt, write your answer out. As long as it produces the same terms in the same order as what the problem wants, your answer is correct. Sequences play a major role in the Mathematics of Finance, as we have already seen with Equation 6.2 in Section 6.5. Recall that if we invest P dollars at an annual percentage rate r and compound the interest n times per year,
the formula for Ak, the amount in the account after k compounding periods, is Ak = P 1 + r, k ≥ 1. We now spot this as a geometric sequence with ﬁrst term P 1 + r. In retirement planning, it is seldom n the case that an investor deposits a set amount of money into an account and waits for it to grow. Usually, additional payments of principal are made at regular intervals and the value of the investment grows accordingly. This kind of investment is called an annuity and will be discussed in the next section once we have developed more mathematical machinery. 1 + r n and common ratio n2 + 25 2 n3 − 31 4 n4 + 5 k−1 k n 5Here we take ‘experimentation’ to mean a frustrating guess-and-check session. 6For more on this, see When Every Answer is Correct: Why Sequences and Number Patterns Fail the Test. 658 Sequences and the Binomial Theorem 9.1.1 Exercises In Exercises 1 - 13, write out the ﬁrst four terms of the given sequence. 1. an = 2n − 1, n ≥ 0 3. {5k − 2}∞ k=1 5. xn n2 ∞ n=1 2. dj = (−1) j(j+1) 2, j ≥ 1 4. 6. ∞ n=0 ∞ n2 + 1 n + 1 ln(n) n n=1 7. a1 = 3, an+1 = an − 1, n ≥ 1 8. d0 = 12, dm = dm-1 100, m ≥ 1 9. b1 = 2, bk+1 = 3bk + 1, k ≥ 1 11. a1 = 117, an+1 = 1 an, n ≥ 1 10. c0 = −2, cj = cj-1 (j + 1)(j + 2), j ≥ 1 12. s0 = 1, sn+1 = xn+1 + sn, n ≥ 0 13. F0 = 1, F1 = 1, Fn = Fn-1 + Fn-2, n ≥ 2 (This is the famous Fibonacci Sequence ) In Exercises 14 - 21 determine if the given sequence is arithmetic, geometric or neither. If it is arithmetic, ﬁnd the common diﬀerence
d; if it is geometric, ﬁnd the common ratio r. 14. {3n − 5}∞ n=1 16. 1 3, 1 6, 1 12, 1 24,... 15. an = n2 + 3n + 2, n ≥ 1 3 1 5 17. n−1∞ n=1 18. 17, 5, −7, −19,... 19. 2, 22, 222, 2222,... 20. 0.9, 9, 90, 900,... 21. an = n! 2, n ≥ 0. In Exercises 22 - 30, ﬁnd an explicit formula for the nth term of the given sequence. Use the formulas in Equation 9.1 as needed. 22. 3, 5, 7, 9,... 25. 1, 2 3, 1 3, 4 27,... 23. 1,... 26. 1, 1 4, 1 9, 1 16,...,... 24. 1, 2 3, 27. x, − 4 5 x3 3,, 8 7 x5 5, − x7 7,... 9.1 Sequences 659 28. 0.9, 0.99, 0.999, 0.9999,... 29. 27, 64, 125, 216,... 30. 1, 0, 1, 0,... 31. Find a sequence which is both arithmetic and geometric. (Hint: Start with an = c for all n.) 32. Show that a geometric sequence can be transformed into an arithmetic sequence by taking the natural logarithm of the terms. 33. Thomas Robert Malthus is credited with saying, “The power of population is indeﬁnitely greater than the power in the earth to produce subsistence for man. Population, when unchecked, increases in a geometrical ratio. Subsistence increases only in an arithmetical ratio. A slight acquaintance with numbers will show the immensity of the ﬁrst power in comparison with the second.” (See this webpage for more information.) Discuss this quote with your classmates from a sequences point of view. 34. This classic problem involving sequences shows the power of geometric sequences. Suppose that a wealthy benefactor agrees to give you one penny today and then double the amount she gives you each day for
30 days. So, for example, you get two pennies on the second day and four pennies on the third day. How many pennies do you get on the 30th day? What is the total dollar value of the gift you have received? 35. Research the terms ‘arithmetic mean’ and ‘geometric mean.’ With the help of your classmates, show that a given term of a arithmetic sequence ak, k ≥ 2 is the arithmetic mean of the term immediately preceding, ak−1 it and immediately following it, ak+1. State and prove an analogous result for geometric sequences. 36. Discuss with your classmates how the results of this section might change if we were to examine sequences of other mathematical things like complex numbers or matrices. Find an explicit formula for the nth term of the sequence i, −1, −i, 1, i,.... List out the ﬁrst four terms of the matrix sequences we discussed in Exercise 8.3.1 in Section 8.3. 660 Sequences and the Binomial Theorem 9.1.2 Answers 1. 0, 1, 3, 7 3. 3, 8, 13, 18 5. x, x2 4, x3 9, x4 16 7. 3, 2, 1, 0 9. 2, 7, 22, 67 11. 117, 1 117, 117, 1 117 13. 1, 1, 2, 3 14. arithmetic, d = 3 16. geometric, r = 1 2 18. arithmetic, d = −12 20. geometric, r = 10 2. −1, −1, 1, 1 4. 1, 1, 5 3, 5 2 6. 0, ln(2) 2, ln(3) 3, ln(4) 4 8. 12, 0.12, 0.0012, 0.000012 10. −2, − 1 3, − 1 36, − 1 720 12. 1, x + 1, x2 + x + 1, x3 + x2 + x + 1 15. neither 17. geometric, r = 1 5 19. neither 21. neither 22. an = 1 + 2n, n ≥ 1 23. an = − 1 2 n−1, n ≥ 1 24. an = 2n−1 2n−1, n ≥ 1 25. an = n 3n−1, n ≥ 1 26. an = 1 n
2, n ≥ 1 27. (−1)n−1x2n−1 2n−1, n ≥ 1 28. an = 10n−1 10n, n ≥ 1 29. an = (n + 2)3, n ≥ 1 30. an = 1+(−1)n−1 2, n ≥ 1 9.2 Summation Notation 661 9.2 Summation Notation In the previous section, we introduced sequences and now we shall present notation and theorems concerning the sum of terms of a sequence. We begin with a deﬁnition, which, while intimidating, is meant to make our lives easier. Deﬁnition 9.3. Summation Notation: Given a sequence {an}∞ satisfying k ≤ m ≤ p, the summation from m to p of the sequence {an} is written n=k and numbers m and p p n=m an = am + am+1 +... + ap The variable n is called the index of summation. The number m is called the lower limit of summation while the number p is called the upper limit of summation. In English, Deﬁnition 9.3 is simply deﬁning a short-hand notation for adding up the terms of the sequence {an}∞ n=k from am through ap. The symbol Σ is the capital Greek letter sigma and is shorthand for ‘sum’. The lower and upper limits of the summation tells us which term to start with and which term to end with, respectively. For example, using the sequence an = 2n − 1 for n ≥ 1, we can write the sum a3 + a4 + a5 + a6 as 6 (2n − 1) = (2(3) − 1) + (2(4) − 1) + (2(5) − 1) + (2(6) − 1) n=3 = 5 + 7 + 9 + 11 = 32 The index variable is considered a ‘dummy variable’ in the sense that it may be changed to any letter without aﬀecting the value of the summation. For instance, 6 (2n − 1) = n=3 6 k=3 (2k − 1) = 6 j=3 (2j − 1) One place you may encounter summation notation is in mathematical
deﬁnitions. For example, summation notation allows us to deﬁne polynomials as functions of the form f (x) = n k=0 akxk for real numbers ak, k = 0, 1,... n. The reader is invited to compare this with what is given in Deﬁnition 3.1. Summation notation is particularly useful when talking about matrix operations. For example, we can write the product of the ith row Ri of a matrix A = [aij]m×n and the jth column Cj of a matrix B = [bij]n×r as Ri · Cj = n k=1 aikbkj 662 Sequences and the Binomial Theorem Again, the reader is encouraged to write out the sum and compare it to Deﬁnition 8.9. Our next example gives us practice with this new notation. Example 9.2.1. 1. Find the following sums. (a) 4 k=1 13 100k (b) 4 n=0 n! 2 (c) 5 n=1 (−1)n+1 n (x − 1)n 2. Write the following sums using summation notation. (b) 1 − (a) 1 + 3 + 5 +... + 117 1 3 1 4 (c) 0.9 + 0.09 + 0.009 +... 0 zeros + −... + 1 117 1 2 − + Solution. 1. (a) We substitute k = 1 into the formula 13 100k and add successive terms until we reach k = 4. 4 k=1 13 100k = 13 1002 + 13 1001 + 13 1003 + = 0.13 + 0.0013 + 0.000013 + 0.00000013 = 0.13131313 13 1004 (b) Proceeding as in (a), we replace every occurrence of n with the values 0 through 4. We recall the factorials, n! as deﬁned in number Example 9.1.1, number 6 and get: 4 n=0 n! 2 + + + 2! 2 2 · 1 2 = 4! 3 + 12 1! 2 1 2 1 2 = = 0! 2 1 2 1 2 = 17 = + 4 · 3 · 2 · 1 2 (c) We proceed as before, replacing the index
n, but not the variable x, with the values 1 through 5 and adding the resulting terms. 9.2 Summation Notation 663 5 n=1 (−1)n+1 n (x − 1)n = (−1)1+1 1 (x − 1)1 + (−1)2+1 2 (x − 1)2 + (−1)3+1 3 (x − 1)3 + (−1)1+4 4 = (x − 1) − (x − 1)4 + (x − 1)2 2 + (−1)1+5 5 (x − 1)3 3 (x − 1)5 − (x − 1)4 4 + (x − 1)5 5 2. The key to writing these sums with summation notation is to ﬁnd the pattern of the terms. To that end, we make good use of the techniques presented in Section 9.1. (a) The terms of the sum 1, 3, 5, etc., form an arithmetic sequence with ﬁrst term a = 1 and common diﬀerence d = 2. We get a formula for the nth term of the sequence using Equation 9.1 to get an = 1 + (n − 1)2 = 2n − 1, n ≥ 1. At this stage, we have the formula for the terms, namely 2n − 1, and the lower limit of the summation, n = 1. To ﬁnish the problem, we need to determine the upper limit of the summation. In other words, we need to determine which value of n produces the term 117. Setting an = 117, we get 2n − 1 = 117 or n = 59. Our ﬁnal answer is 1 + 3 + 5 +... + 117 = 59 (2n − 1) n=1 (b) We rewrite all of the terms as fractions, the subtraction as addition, and associate the negatives ‘−’ with the numerators to get 1 1 + −1 2 + 1 3 + −1 4 +... + 1 117 The numerators, 1, −1, etc. can be described by the geometric sequence1 cn = (−1)n−1 for n ≥ 1, while the denominators are given by the arithmetic sequence2 dn = n for n ≥ 1. Hence, we get the formula an = (−1
)n−1 for our terms, and we ﬁnd the lower and upper limits of summation to be n = 1 and n = 117, respectively. Thus 117 = 117 n=1 (−1)n−1 n (c) Thanks to Example 9.1.3, we know that one formula for the nth term is an = 9 10n for n ≥ 1. This gives us a formula for the summation as well as a lower limit of summation. To determine the upper limit of summation, we note that to produce the n − 1 zeros to the right of the decimal point before the 9, we need a denominator of 10n. Hence, n is 1This is indeed a geometric sequence with ﬁrst term a = 1 and common ratio r = −1. 2It is an arithmetic sequence with ﬁrst term a = 1 and common diﬀerence d = 1. 664 Sequences and the Binomial Theorem the upper limit of summation. Since n is used in the limits of the summation, we need to choose a diﬀerent letter for the index of summation.3 We choose k and get 0.9 + 0.09 + 0.009 +... 0. 0 · · · 0 n − 1 zeros 9 = n k=1 9 10k The following theorem presents some general properties of summation notation. While we shall not have much need of these properties in Algebra, they do play a great role in Calculus. Moreover, there is much to be learned by thinking about why the properties hold. We invite the reader to prove these results. To get started, remember, “When in doubt, write it out!” Theorem 9.1. Properties of Summation Notation: Suppose {an} and {bn} are sequences so that the following sums are deﬁned. p n=m p n=m p (an ± bn) = p p bn an ± n=m n=m c an = c p n=m an, for any real number c. j an + p an = an, for any natural number m ≤ j < j + 1 ≤ p. n=m n=m n=j+1 p an = p+r n=m n=m+r an−r, for any whole number r. We now turn our attention to the sums involving arithmetic
and geometric sequences. Given an arithmetic sequence ak = a + (k − 1)d for k ≥ 1, we let S denote the sum of the ﬁrst n terms. To derive a formula for S, we write it out in two diﬀerent ways S = S = (a + (n − 1)d) + (a + (n − 2)d) +... + (a + d) + a +... + (a + (n − 2)d) + (a + (n − 1)d) (a + d) + a If we add these two equations and combine the terms which are aligned vertically, we get 2S = (2a + (n − 1)d) + (2a + (n − 1)d) +... + (2a + (n − 1)d) + (2a + (n − 1)d) The right hand side of this equation contains n terms, all of which are equal to (2a + (n − 1)d) so we get 2S = n(2a + (n − 1)d). Dividing both sides of this equation by 2, we obtain the formula 3To see why, try writing the summation using ‘n’ as the index. 9.2 Summation Notation 665 S = n 2 (2a + (n − 1)d) If we rewrite the quantity 2a + (n − 1)d as a + (a + (n − 1)d) = a1 + an, we get the formula S = n a1 + an 2 A helpful way to remember this last formula is to recognize that we have expressed the sum as the product of the number of terms n and the average of the ﬁrst and nth terms. To derive the formula for the geometric sum, we start with a geometric sequence ak = ark−1, k ≥ 1, and let S once again denote the sum of the ﬁrst n terms. Comparing S and rS, we get S = a + ar + ar2 +... + arn−2 + arn−1 rS = ar + ar2 +... + arn−2 + arn−1 + arn Subtracting the second equation from the ﬁrst forces all of the terms except
a and arn to cancel out and we get S − rS = a − arn. Factoring, we get S(1 − r) = a (1 − rn). Assuming r = 1, we can divide both sides by the quantity (1 − r) to obtain S = a 1 − rn 1 − r If we distribute a through the numerator, we get a − arn = a1 − an+1 which yields the formula S = a1 − an+1 1 − r In the case when r = 1, we get the formula times = n a Our results are summarized below. 666 Sequences and the Binomial Theorem Equation 9.2. Sums of Arithmetic and Geometric Sequences: The sum S of the ﬁrst n terms of an arithmetic sequence ak = a + (k − 1)d for k ≥ 1 is S = n k=1 ak = n a1 + an 2 = n 2 (2a + (n − 1)d) The sum S of the ﬁrst n terms of a geometric sequence ak = ark−1 for k ≥ 1 is 1. S = 2. S = n k=1 n k=1 ak = a1 − an+1 1 − r = a 1 − rn 1 − r, if r = 1. ak = n k=1 a = na, if r = 1. While we have made an honest eﬀort to derive the formulas in Equation 9.2, formal proofs require the machinery in Section 9.3. An application of the arithmetic sum formula which proves useful in Calculus results in formula for the sum of the ﬁrst n natural numbers. The natural numbers themselves are a sequence4 1, 2, 3,... which is arithmetic with a = d = 1. Applying Equation 9.2(n + 1) 2 2 = 5050. So, for example, the sum of the ﬁrst 100 natural numbers5 is 100(101) An important application of the geometric sum formula is the investment plan called an annuity. Annuities diﬀer from the kind of investments we studied in Section 6.5 in that payments are deposited into the account on an on-going basis, and this complicates the mathematics a little.6 Suppose you have an account with annual interest rate r which is compounded n times per year. We let i
= r n denote the interest rate per period. Suppose we wish to make ongoing deposits of P dollars at the end of each compounding period. Let Ak denote the amount in the account after k compounding periods. Then A1 = P, because we have made our ﬁrst deposit at the end of the ﬁrst compounding period and no interest has been earned. During the second compounding period, we earn interest on A1 so that our initial investment has grown to A1(1 + i) = P (1 + i) in accordance with Equation 6.1. When we add our second payment at the end of the second period, we get A2 = A1(1 + i) + P = P (1 + i) + P = P (1 + i) 1 + 1 1 + i The reason for factoring out the P (1 + i) will become apparent in short order. During the third compounding period, we earn interest on A2 which then grows to A2(1 + i). We add our third 4This is the identity function on the natural numbers! 5There is an interesting anecdote which says that the famous mathematician Carl Friedrich Gauss was given this problem in primary school and devised a very clever solution. 6The reader may wish to re-read the discussion on compound interest in Section 6.5 before proceeding. 9.2 Summation Notation 667 payment at the end of the third compounding period to obtain A3 = A2(1 + i) + P = P (1 + i) 1 + 1 1 + i (1 + i) + P = P (1 + i)1 + i)2 During the fourth compounding period, A3 grows to A3(1+i), and when we add the fourth payment, we factor out P (1 + i)3 to get A4 = P (1 + i)1 + i)2 + 1 (1 + i)3 This pattern continues so that at the end of the kth compounding, we get Ak = P (1 + i)k−1 + i)2 +... + 1 (1 + i)k−1 The sum in the parentheses above is the sum of the ﬁrst k terms of a geometric sequence with a = 1 and r = 1 1+i. Using Equation 9.2, we get 1 + 1 1 + i + 1 (1 +
i)2 +... + 1 (1 + i)k−1 = 1     Hence, we get 1 − 1 (1 + i)1 + i) 1 − (1 + i)−k i Ak = P (1 + i)k−1 (1 + i) 1 − (1 + i)−k i P (1 + i)k − 1 i = If we let t be the number of years this investment strategy is followed, then k = nt, and we get the formula for the future value of an ordinary annuity. Equation 9.3. Future Value of an Ordinary Annuity: Suppose an annuity oﬀers an annual interest rate r compounded n times per year. Let i = r n be the interest rate per compounding period. If a deposit P is made at the end of each compounding period, the amount A in the account after t years is given by A = P (1 + i)nt − 1 i The reader is encouraged to substitute i = r n into Equation 9.3 and simplify. Some familiar equations arise which are cause for pause and meditation. One last note: if the deposit P is made a the beginning of the compounding period instead of at the end, the annuity is called an annuitydue. We leave the derivation of the formula for the future value of an annuity-due as an exercise for the reader. 668 Sequences and the Binomial Theorem Example 9.2.2. An ordinary annuity oﬀers a 6% annual interest rate, compounded monthly. 1. If monthly payments of $50 are made, ﬁnd the value of the annuity in 30 years. 2. How many years will it take for the annuity to grow to $100,000? Solution. 1. We have r = 0.06 and n = 12 so that i = r n = 0.06 12 = 0.005. With P = 50 and t = 30, A = 50 (1 + 0.005)(12)(30) − 1 0.005 ≈ 50225.75 Our ﬁnal answer is $50,225.75. 2. To ﬁnd how long it will take for the annuity to grow to $100,000, we set A = 100000 and solve for t. We isolate the exponential and take natural
logs of both sides of the equation. 100000 = 50 (1 + 0.005)12t − 1 0.005 10 = (1.005)12t − 1 (1.005)12t = 11 ln (1.005)12t = ln(11) 12t ln(1.005) = ln(11) ln(11) t = 12 ln(1.005) ≈ 40.06 This means that it takes just over 40 years for the investment to grow to $100,000. Comparing this with our answer to part 1, we see that in just 10 additional years, the value of the annuity nearly doubles. This is a lesson worth remembering. We close this section with a peek into Calculus by considering inﬁnite sums, called series. Consider the number 0.9. We can write this number as 0.9 = 0.9999... = 0.9 + 0.09 + 0.009 + 0.0009 +... From Example 9.2.1, we know we can write the sum of the ﬁrst n of these terms as 0. 9 · · · 9 n nines =.9 + 0.09 + 0.009 +... 0. 0 · · · 0 n − 1 zeros 9 = n k=1 9 10k Using Equation 9.2, we have 9.2 Summation Notation 669   k=1   1 − n 9 10 9 10k = 1 10n+1 1 10 It stands to reason that 0.9 is the same value of 1 − 1 10n+1 as n → ∞. Our knowledge of exponential 10n+1 → 1. We have expressions from Section 6.1 tells us that just argued that 0.9 = 1, which may cause some distress for some readers.7 Any non-terminating decimal can be thought of as an inﬁnite sum whose denominators are the powers of 10, so the phenomenon of adding up inﬁnitely many terms and arriving at a ﬁnite number is not as foreign of a concept as it may appear. We end this section with a theorem concerning geometric series. 10n+1 → 0 as n → ∞, so 1 − 1   = 1 − 1 10n+1
1 − 1 Theorem 9.2. Geometric Series: Given the sequence ak = ark−1 for k ≥ 1, where \|r\| < 1, a + ar + ar2 +... = ∞ ark−1 = k=1 If \|r\| ≥ 1, the sum a + ar + ar2 +... is not deﬁned. a 1 − r The justiﬁcation of the result in Theorem 9.2 comes from taking the formula in Equation 9.2 for the sum of the ﬁrst n terms of a geometric sequence and examining the formula as n → ∞. Assuming \|r\| < 1 means −1 < r < 1, so rn → 0 as n → ∞. Hence as n → ∞, n k=1 ark−1 = a 1 − rn 1 − r → a 1 − r As to what goes wrong when \|r\| ≥ 1, we leave that to Calculus as well, but will explore some cases in the exercises. 7To make this more palatable, it is usually accepted that 0.3 = 1 3 so that 0.9 = 3 0.3 = 3 1 3 = 1. Feel better? 670 Sequences and the Binomial Theorem 9.2.1 Exercises In Exercises 1 - 8, ﬁnd the value of each sum using Deﬁnition 9.3. 1. 5. 9 (5g + 3) g=4 4 i=1 1 4 (i2 + 1) 2. 6. 8 k=3 1 k 100 (−1)n n=1 3. 7. 5 j=0 5 n=1 2j (n + 1)! n! 4. 8. 2 (3k − 5)xk k=0 3 j=1 5! j! (5 − j)! In Exercises 9 - 16, rewrite the sum using summation notation. 9. 8 + 11 + 14 + 17 + 20 10 11. x − x3 3 + x5 5 − x7 7 13 15 16 + 1 25 − 1 36 12. 1 + 2 + 4 + · · · + 229 14. − ln(3) + ln(4) − ln(5) + · · · + ln(20) 16. 1 2 (x − 5) + 1 4 (x −
5)2 + 1 6 (x − 5)3 + 1 8 (x − 5)4 In Exercises 17 - 28, use the formulas in Equation 9.2 to ﬁnd the sum. 17. 20. 10 n=1 5n + 3 n 10 n=1 1 2 18. 21. 20 n=1 2n − 1 n 5 n=1 3 2 19. 22. 15 k=0 3 − k k 5 k=0 1 4 2 23. 1 + 4 + 7 +... + 295 24. 4 + 2 + 0 − 2 −... − 146 25. 1 + 3 + 9 +... + 2187 26 256 27 256 28. 10 n=1 −2n + n 5 3 In Exercises 29 - 32, use Theorem 9.2 to express each repeating decimal as a fraction of integers. 29. 0.7 30. 0.13 31. 10.159 32. −5.867 9.2 Summation Notation 671 In Exercises 33 - 38, use Equation 9.3 to compute the future value of the annuity with the given terms. In all cases, assume the payment is made monthly, the interest rate given is the annual rate, and interest is compounded monthly. 33. payments are $300, interest rate is 2.5%, term is 17 years. 34. payments are $50, interest rate is 1.0%, term is 30 years. 35. payments are $100, interest rate is 2.0%, term is 20 years 36. payments are $100, interest rate is 2.0%, term is 25 years 37. payments are $100, interest rate is 2.0%, term is 30 years 38. payments are $100, interest rate is 2.0%, term is 35 years 39. Suppose an ordinary annuity oﬀers an annual interest rate of 2%, compounded monthly, for 30 years. What should the monthly payment be to have $100,000 at the end of the term? 40. Prove the properties listed in Theorem 9.1. 41. Show that the formula for the future value of an annuity due is A = P (1 + i) (1 + i)nt − 1 i 42. Discuss with your classmates what goes wrong when trying to ﬁnd the following sums.8 (a) ∞ k=1 2k−1 ∞
(1.0001)k−1 (b) k=1 ∞ (−1)k−1 (c) k=1 8When in doubt, write them out! 672 Sequences and the Binomial Theorem 9.2.2 Answers 1. 213 5. 17 2 9. 13. 5 (3k + 5) k=1 5 k=1 k + 1 k 17. 305 21. 633 32 25. 3280 29. 7 9 2. 341 280 6. 0 10. 14. 8 k=1 20 k=3 (−1)k−1k (−1)k ln(k) 18. 400 22. 26. 30. 1365 512 255 256 13 99 3. 63 7. 20 11. 15. 4 k=1 6 k=1 (−1)k−1 x2k−1 2k − 1 (−1)k−1 k2 19. −72 4. −5 − 2x + x2 8. 25 12. 16. 20. 30 k=1 4 k=1 2k−1 1 2k (x − 5)k 1023 1024 23. 14652 24. −5396 27. 31. 513 256 3383 333 28. 17771050 59049 32. − 5809 990 33. $76,163.67 34. $20,981.40 35. $29,479.69 36. $38,882.12 37. 49,272.55 38. 60,754.80 39. For $100,000, the monthly payment is ≈ $202.95. 9.3 Mathematical Induction 673 9.3 Mathematical Induction The Chinese philosopher Confucius is credited with the saying, “A journey of a thousand miles begins with a single step.” In many ways, this is the central theme of this section. Here we introduce a method of proof, Mathematical Induction, which allows us to prove many of the formulas we have merely motivated in Sections 9.1 and 9.2 by starting with just a single step. A good example is the formula for arithmetic sequences we touted in Equation 9.1. Arithmetic sequences are deﬁned recursively, starting with a1 = a and then an+1 = an + d for n ≥ 1. This tells us that we start the sequence with a and we go from one term to the next by successively
adding d. In symbols, a, a + d, a + 2d, a + 3d, a + 4d +... The pattern suggested here is that to reach the nth term, we start with a and add d to it exactly n − 1 times, which lead us to our formula an = a + (n − 1)d for n ≥ 1. But how do we prove this to be the case? We have the following. The Principle of Mathematical Induction (PMI): Suppose P (n) is a sentence involving the natural number n. IF 1. P (1) is true and 2. whenever P (k) is true, it follows that P (k + 1) is also true THEN the sentence P (n) is true for all natural numbers n. The Principle of Mathematical Induction, or PMI for short, is exactly that - a principle.1 It is a property of the natural numbers we either choose to accept or reject. In English, it says that if we want to prove that a formula works for all natural numbers n, we start by showing it is true for n = 1 (the ‘base step’) and then show that if it is true for a generic natural number k, it must be true for the next natural number, k + 1 (the ‘inductive step’). The notation P (n) acts just like function notation. For example, if P (n) is the sentence (formula) ‘n2 + 1 = 3’, then P (1) would be ‘12 + 1 = 3’, which is false. The construction P (k + 1) would be ‘(k + 1)2 + 1 = 3’. As usual, this new concept is best illustrated with an example. Returning to our quest to prove the formula for an arithmetic sequence, we ﬁrst identify P (n) as the formula an = a + (n − 1)d. To prove this formula is valid for all natural numbers n, we need to do two things. First, we need to establish that P (1) is true. In other words, is it true that a1 = a + (1 − 1)d? The answer is yes, since this simpliﬁes to a1 = a, which is part of the deﬁnition of the arithmetic sequence. The second thing we
need to show is that whenever P (k) is true, it follows that P (k + 1) is true. In other words, we assume P (k) is true (this is called the ‘induction hypothesis’) and deduce that P (k + 1) is also true. Assuming P (k) to be true seems to invite disaster - after all, isn’t this essentially what we’re trying to prove in the ﬁrst place? To help explain this step a little better, we show how this works for speciﬁc values of n. We’ve already established P (1) is true, and we now want to show that P (2) 1Another word for this you may have seen is ‘axiom.’ 674 Sequences and the Binomial Theorem is true. Thus we need to show that a2 = a + (2 − 1)d. Since P (1) is true, we have a1 = a, and by the deﬁnition of an arithmetic sequence, a2 = a1 +d = a+d = a+(2−1)d. So P (2) is true. We now use the fact that P (2) is true to show that P (3) is true. Using the fact that a2 = a + (2 − 1)d, we show a3 = a +(3 − 1)d. Since a3 = a2 + d, we get a3 = (a +(2 − 1)d)+ d = a +2d = a +(3 − 1)d, so we have shown P (3) is true. Similarly, we can use the fact that P (3) is true to show that P (4) is true, and so forth. In general, if P (k) is true (i.e., ak = a+(k −1)d) we set out to show that P (k +1) is true (i.e., ak+1 = a + ((k + 1) − 1)d). Assuming ak = a + (k − 1)d, we have by the deﬁnition of an arithmetic sequence that ak+1 = ak + d so we get ak+1 = (a + (k − 1)d) + d = a + kd = a + ((k
+ 1) − 1)d. Hence, P (k + 1) is true. In essence, by showing that P (k + 1) must always be true when P (k) is true, we are showing that the formula P (1) can be used to get the formula P (2), which in turn can be used to derive the formula P (3), which in turn can be used to establish the formula P (4), and so on. Thus as long as P (k) is true for some natural number k, P (n) is true for all of the natural numbers n which follow k. Coupling this with the fact P (1) is true, we have established P (k) is true for all natural numbers which follow n = 1, in other words, all natural numbers n. One might liken Mathematical Induction to a repetitive process like climbing stairs.2 If you are sure that (1) you can get on the stairs (the base case) and (2) you can climb from any one step to the next step (the inductive step), then presumably you can climb the entire staircase.3 We get some more practice with induction in the following example. Example 9.3.1. Prove the following assertions using the Principle of Mathematical Induction. 1. The sum formula for arithmetic sequences: n (a + (j − 1)d) = j=1 2. For a complex number z, (z)n = zn for n ≥ 1. n 2 (2a + (n − 1)d). 3. 3n > 100n for n > 5. 4. Let A be an n × n matrix and let A be the matrix obtained by replacing a row R of A with cR for some real number c. Use the deﬁnition of determinant to show det(A) = c det(A). Solution. 1. We set P (n) to be the equation we are asked to prove. For n = 1, we compare both sides of the equation given in P (n) 1 j=1 (a + (j − 1)d) a + (1 − 1)2a + (1 − 1)d) (2a) a = a 2Falling dominoes is the most widely used metaphor in the mainstream College Algebra books. 3This is how Carl climbed the stairs in the Cologne Cathedral. Well, that, and encouragement from
Kai. 9.3 Mathematical Induction 675 This shows the base case P (1) is true. Next we assume P (k) is true, that is, we assume k (a + (j − 1)d) = j=1 k 2 (2a + (k − 1)d) and attempt to use this to show P (k + 1) is true. Namely, we must show k+1 (a + (j − 1)d) = j=1 k + 1 2 (2a + (k + 1 − 1)d) To see how we can use P (k) in this case to prove P (k + 1), we note that the sum in P (k + 1) is the sum of the ﬁrst k + 1 terms of the sequence ak = a + (k − 1)d for k ≥ 1 while the sum in P (k) is the sum of the ﬁrst k terms. We compare both side of the equation in P (k + 1). k+1 (a + (j − 1)d) j=1 summing the ﬁrst k + 1 terms k (a + (j − 1)d) + (a + (k + 1 − 1)d) j=1 summing the ﬁrst k terms adding the (k + 1)st term? = k + 1 2 (2a + (k + 1 − 1)d)? = k + 1 2 (2a + kd) k 2 (2a + (k − 1)d) Using P (k) +(a + kd) k(2a + (k − 1)d) + 2(a + kd) 2 2ka + 2a + k2d + kd 2? =? = = (k + 1)(2a + kd) 2 2ka + k2d + 2a + kd 2 2ka + 2a + k2d + kd 2 Since all of our steps on both sides of the string of equations are reversible, we conclude that the two sides of the equation are equivalent and hence, P (k + 1) is true. By the Principle of Mathematical Induction, we have that P (n) is true for all natural numbers n. 2. We let P (n) be the formula (z)n = zn.
The base case P (1) is (z)1 = z1, which reduces to z = z which is true. We now assume P (k) is true, that is, we assume (z)k = zk and attempt to show that P (k + 1) is true. Since (z)k+1 = (z)k z, we can use the induction hypothesis and 676 Sequences and the Binomial Theorem write (z)k = zk. Hence, (z)k+1 = (z)k z = zk z. We now use the product rule for conjugates4 to write zk z = zkz = zk+1. This establishes (z)k+1 = zk+1, so that P (k + 1) is true. Hence, by the Principle of Mathematical Induction, (z)n = zn for all n ≥ 1. 3. The ﬁrst wrinkle we encounter in this problem is that we are asked to prove this formula for n > 5 instead of n ≥ 1. Since n is a natural number, this means our base step occurs at n = 6. We can still use the PMI in this case, but our conclusion will be that the formula is valid for all n ≥ 6. We let P (n) be the inequality 3n > 100n, and check that P (6) is true. Comparing 36 = 729 and 100(6) = 600, we see 36 > 100(6) as required. Next, we assume that P (k) is true, that is we assume 3k > 100k. We need to show that P (k + 1) is true, that is, we need to show 3k+1 > 100(k + 1). Since 3k+1 = 3 · 3k, the induction hypothesis gives 3k+1 = 3 · 3k > 3(100k) = 300k. We are done if we can show 300k > 100(k + 1) for k ≥ 6. Solving 300k > 100(k + 1) we get k > 1 2. Since k ≥ 6, we know this is true. Putting all of this together, we have 3k+1 = 3 · 3k > 3(100k) = 300k > 100(k + 1), and hence P (k + 1) is true. By induction, 3
n > 100n for all n ≥ 6. 4. To prove this determinant property, we use induction on n, where we take P (n) to be that the property we wish to prove is true for all n × n matrices. For the base case, we note that if A is a 1 × 1 matrix, then A = [a] so A = [ca]. By deﬁnition, det(A) = a and det(A) = ca so we have det(A) = c det(A) as required. Now suppose that the property we wish to prove is true for all k × k matrices. Let A be a (k + 1) × (k + 1) matrix. We have two cases, depending on whether or not the row R being replaced is the ﬁrst row of A. Case 1: The row R being replaced is the ﬁrst row of A. By deﬁnition, det(A) = n p=1 1pC a 1p 1p = (−1)(1+p) det A where the 1p cofactor of A is C 1p is the k × k matrix obtained by deleting the 1st row and pth column of A.5 Since the ﬁrst row of A is c times the ﬁrst 1p = c a1p. In addition, since the remaining rows of A are identical to row of A, we have a 1p = A1p. (To obtain these matrices, the ﬁrst row of A is removed.) Hence those of A, A det A 1p = det (A1p), so that C 1p = C1p. As a result, we get and A 1p det(A) = n p=1 1pC a 1p = n p=1 n c a1pC1p = c p=1 a1pC1p = c det(A), as required. Hence, P (k + 1) is true in this case, which means the result is true in this case for all natural numbers n ≥ 1. (You’ll note that we did not use the induction hypothesis at all in this case. It is possible to restructure the proof so that induction is only used where 4See Exercise 54 in Section 3.4. 5See Section 8.5 for a review of
this notation. 9.3 Mathematical Induction 677 it is needed. While mathematically more elegant, it is less intuitive, and we stand by our approach because of its pedagogical value.) Case 2: The row R being replaced is the not the ﬁrst row of A. By deﬁnition, det(A) = n p=1 1pC a 1p, where in this case, a 1p = a1p, since the ﬁrst rows of A and A are the same. The matrices 1p and A1p, on the other hand, are diﬀerent but in a very predictable way − the row in A A 1p which corresponds to the row cR in A is exactly c times the row in A1p which corresponds to the row R in A. In other words, A 1p and A1p are k × k matrices which satisfy the induction hypothesis. Hence, we know det A 1p = c det (A1p) and C 1p = c C1p. We get det(A) = n p=1 1pC a 1p = n p=1 n a1pc C1p = c p=1 a1pC1p = c det(A), which establishes P (k + 1) to be true. Hence by induction, we have shown that the result holds in this case for n ≥ 1 and we are done. While we have used the Principle of Mathematical Induction to prove some of the formulas we have merely motivated in the text, our main use of this result comes in Section 9.4 to prove the celebrated Binomial Theorem. The ardent Mathematics student will no doubt see the PMI in many courses yet to come. Sometimes it is explicitly stated and sometimes it remains hidden in the background. If ever you see a property stated as being true ‘for all natural numbers n’, it’s a solid bet that the formal proof requires the Principle of Mathematical Induction. 678 Sequences and the Binomial Theorem 9.3.1 Exercises In Exercises 1 - 7, prove each assertion using the Principle of Mathematical Induction. 1. 2. n j=1 n j=1 j2 = n(n + 1)(2n + 1) 6 j3 = n2(n + 1)2 4 3. 2n > 500n for
n > 12 4. 3n ≥ n3 for n ≥ 4 5. Use the Product Rule for Absolute Value to show \|xn\| = \|x\|n for all real numbers x and all natural numbers n ≥ 1 6. Use the Product Rule for Logarithms to show log (xn) = n log(x) for all real numbers x > 0 and all natural numbers n ≥ 1. an a 0 0 0 b 0 bn n = for n ≥ 1. 7. 8. Prove Equations 9.1 and 9.2 for the case of geometric sequences. That is: (a) For the sequence a1 = a, an+1 = ran, n ≥ 1, prove an = arn−1, n ≥ 1. (b) n j=1 arn−1 = a 1 − rn 1 − r, if r = 1, n j=1 arn−1 = na, if r = 1. 9. Prove that the determinant of a lower triangular matrix is the product of the entries on the main diagonal. (See Exercise 8.3.1 in Section 8.3.) Use this result to then show det (In) = 1 where In is the n × n identity matrix. 10. Discuss the classic ‘paradox’ All Horses are the Same Color problem with your classmates. 9.3 Mathematical Induction 679 9.3.2 Selected Answers 1. Let P (n) be the sentence n j=1 j2 = n(n + 1)(2n + 1) 6. For the base case, n = 1, we get 1 j=1 j2? = (1)(1 + 1)(2(1) + 1) 6 12 = 1 We now assume P (k) is true and use it to show P (k + 1) is true. We have k+1 j=1 j2? = (k + 1)((k + 1) + 1)(2(k + 1) + 1) 6 k j=1 j2 + (k + 1)2 k(k + 1)(2k + 1) 6 Using P (k) +(k + 1)2 k(k + 1)(2k + 1) 6 + 6(k + 1)2 6 k(k + 1)(2k + 1) + 6(k + 1)2 6 (k + 1)(k(2
k + 1) + 6(k + 1)) 6 (k + 1) 2k2 + 7k + 6 6 (k + 1)(k + 2)(2k + 3k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 By induction, n j=1 j2 = n(n + 1)(2n + 1) 6 is true for all natural numbers n ≥ 1. 4. Let P (n) be the sentence 3n > n3. Our base case is n = 4 and we check 34 = 81 and 43 = 64 so that 34 > 43 as required. We now assume P (k) is true, that is 3k > k3, and try to show P (k + 1) is true. We note that 3k+1 = 3 · 3k > 3k3 and so we are done if we can show 3k3 > (k + 1)3 for k ≥ 4. We can solve the inequality 3x3 > (x + 1)3 using the techniques of Section 5.3, and doing so gives us x > 1 ≈ 2.26. Hence, for k ≥ 4, 3√ 3k+1 = 3 · 3k > 3k3 > (k + 1)3 so that 3k+1 > (k + 1)3. By induction, 3n > n3 is true for all natural numbers n ≥ 4. 3−1 680 Sequences and the Binomial Theorem 6. Let P (n) be the sentence log (xn) = n log(x). For the duration of this argument, we assume x > 0. The base case P (1) amounts checking that log x1 = 1 log(x) which is clearly true. Next we assume P (k) is true, that is log xk = k log(x) and try to show P (k + 1) is true. Using the Product Rule for Logarithms along with the induction hypothesis, we get xk+
1 log = log xk · x = log xk + log(x) = k log(x) + log(x) = (k + 1) log(x) Hence, log xk+1 = (k + 1) log(x). By induction log (xn) = n log(x) is true for all x > 0 and all natural numbers n ≥ 1. 9. Let A be an n × n lower triangular matrix. We proceed to prove the det(A) is the product of the entries along the main diagonal by inducting on n. For n = 1, A = [a] and det(A) = a, so the result is (trivially) true. Next suppose the result is true for k × k lower triangular matrices. Let A be a (k + 1) × (k + 1) lower triangular matrix. Expanding det(A) along the ﬁrst row, we have det(A) = n p=1 a1pC1p Since a1p = 0 for 2 ≤ p ≤ k + 1, this simpliﬁes det(A) = a11C11. By deﬁnition, we know that C11 = (−1)1+1 det (A11) = det (A11) where A11 is k × k matrix obtained by deleting the ﬁrst row and ﬁrst column of A. Since A is lower triangular, so is A11 and, as such, the induction hypothesis applies to A11. In other words, det (A11) is the product of the entries along A11’s main diagonal. Now, the entries on the main diagonal of A11 are the entries a22, a33,..., a(k+1)(k+1) from the main diagonal of A. Hence, det(A) = a11 det (A11) = a11 a22a33 · · · a(k+1)(k+1) = a11a22a33 · · · a(k+1)(k+1) We have det(A) is the product of the entries along its main diagonal. This shows P (k + 1) is true, and, hence, by induction, the result holds for all n × n upper triangular matrices. The n × n identity matrix In is a lower triangular matrix whose main
diagonal consists of all 1’s. Hence, det (In) = 1, as required. 9.4 The Binomial Theorem 681 9.4 The Binomial Theorem In this section, we aim to prove the celebrated Binomial Theorem. Simply stated, the Binomial Theorem is a formula for the expansion of quantities (a + b)n for natural numbers n. In Elementary and Intermediate Algebra, you should have seen speciﬁc instances of the formula, namely (a + b)1 = a + b (a + b)2 = a2 + 2ab + b2 (a + b)3 = a3 + 3a2b + 3ab2 + b3 If we wanted the expansion for (a + b)4 we would write (a + b)4 = (a + b)(a + b)3 and use the formula that we have for (a+b)3 to get (a+b)4 = (a+b) a3 + 3a2b + 3ab2 + b3 = a4+4a3b+6a2b2+4ab3+b4. Generalizing this a bit, we see that if we have a formula for (a + b)k, we can obtain a formula for (a + b)k+1 by rewriting the latter as (a + b)k+1 = (a + b)(a + b)k. Clearly this means Mathematical Induction plays a major role in the proof of the Binomial Theorem.1 Before we can state the theorem we need to revisit the sequence of factorials which were introduced in Example 9.1.1 number 6 in Section 9.1. Deﬁnition 9.4. Factorials: For a whole number n, n factorial, denoted n!, is the term fn of the sequence f0 = 1, fn = n · fn−1, n ≥ 1. Recall this means 0! = 1 and n! = n(n − 1)! for n ≥ 1. Using the recursive deﬁnition, we get: 1! = 1 · 0! = 1 · 1 = 1, 2! = 2 · 1! = 2 · 1 = 2, 3! = 3 · 2! = 3 · 2 · 1 = 6 and 4! = 4 · 3! = 4 · 3 · 2 · 1 = 24
. Informally, n! = n · (n − 1) · (n − 2) · · · 2 · 1 with 0! = 1 as our ‘base case.’ Our ﬁrst example familiarizes us with some of the basic computations involving factorials. Example 9.4.1. 1. Simplify the following expressions. (a) 3! 2! 0! (b) 7! 5! (c) 1000! 998! 2! (d) (k + 2)! (k − 1)!, k ≥ 1 2. Prove n! > 3n for all n ≥ 7. Solution. 1. We keep in mind the mantra, “When in doubt, write it out!” as we simplify the following. (a) We have been programmed to react with alarm to the presence of a 0 in the denominator, but in this case 0! = 1, so the fraction is deﬁned after all. As for the numerator, 3! = 3 · 2 · 1 = 6 and 2! = 2 · 1 = 2, so we have 3! 2! 0! = (6)(2) 1 = 12. 1It’s pretty much the reason Section 9.3 is in the book. 682 Sequences and the Binomial Theorem (b) We have 7 = 5040 while 5 = 120. Dividing, we get 120 = 42. While this is correct, we note that we could have saved ourselves some 7! 5! = 5040 of time had we proceeded as follows 7! 5 = 42 In fact, should we want to fully exploit the recursive nature of the factorial, we can write 7! 5! = 7 · 6 · 5! 5! = 7 · 6 · 5! 5! = 42 (c) Keeping in mind the lesson we learned from the previous problem, we have 1000! 998! 2! = 1000 · 999 · 998! 998! · 2! = 1000 · 999 · 998! 998! · 2! = 999000 2 = 499500 (d) This problem continues the theme which we have seen in the previous two problems. We ﬁrst note that since k + 2 is larger than k − 1, (k + 2)! contains all of the factors of (k − 1)! and as a result we can get the (k − 1)! to cancel from the denomin
ator. To see this, we begin by writing out (k + 2)! starting with (k + 2) and multiplying it by the numbers which precede it until we reach (k − 1): (k + 2)! = (k + 2)(k + 1)(k)(k − 1)!. As a result, we have (k + 2)! (k − 1)! = (k + 2)(k + 1)(k)(k − 1)! (k − 1)! = (k + 2)(k + 1)(k) (k − 1)! (k − 1)! = k(k + 1)(k + 2) The stipulation k ≥ 1 is there to ensure that all of the factorials involved are deﬁned. 2. We proceed by induction and let P (n) be the inequality n! > 3n. The base case here is n = 7 and we see that 7! = 5040 is larger than 37 = 2187, so P (7) is true. Next, we assume that P (k) is true, that is, we assume k! > 3k and attempt to show P (k + 1) follows. Using the properties of the factorial, we have (k + 1)! = (k + 1)k! and since k! > 3k, we have (k + 1)! > (k + 1)3k. Since k ≥ 7, k + 1 ≥ 8, so (k + 1)3k ≥ 8 · 3k > 3 · 3k = 3k+1. Putting all of this together, we have (k + 1)! = (k + 1)k! > (k + 1)3k > 3k+1 which shows P (k + 1) is true. By the Principle of Mathematical Induction, we have n! > 3n for all n ≥ 7. Of all of the mathematical animals we have discussed in the text, factorials grow most quickly. In problem 2 of Example 9.4.1, we proved that n! overtakes 3n at n = 7. ‘Overtakes’ may be too polite a word, since n! thoroughly trounces 3n for n ≥ 7, as any reasonable set of data will show. It can be shown that for any real number x > 0, not only does n! eventually overtake xn, but the ratio xn Applications of factorials in
the wild often involve counting arrangements. For example, if you have ﬁfty songs on your mp3 player and wish arrange these songs in a playlist in which the order of the n! → 0 as n → ∞.2 2This fact is far more important than you could ever possibly imagine. 9.4 The Binomial Theorem 683 songs matters, it turns out that there are 50! diﬀerent possible playlists. If you wish to select only ten of the songs to create a playlist, then there are 50! 40! such playlists. If, on the other hand, you just want to select ten song ﬁles out of the ﬁfty to put on a ﬂash memory card so that now the order 50! 40!10! ways to achieve this.3 While some of these ideas are explored no longer matters, there are in the Exercises, the authors encourage you to take courses such as Finite Mathematics, Discrete Mathematics and Statistics. We introduce these concepts here because this is how the factorials make their way into the Binomial Theorem, as our next deﬁnition indicates. Deﬁnition 9.5. Binomial Coeﬃcients: Given two whole numbers n and j with n ≥ j, the binomial coeﬃcient n j (read, n choose j) is the whole number given by n j = n! j!(n − j)! The name ‘binomial coeﬃcient’ will be justiﬁed shortly. For now, we can physically interpret as the number of ways to select j items from n items where the order of the items selected is n j unimportant. For example, suppose you won two free tickets to a special screening of the latest Hollywood blockbuster and have ﬁve good friends each of whom would love to accompany you to ways to choose who goes with you. Applying Deﬁnition 9.5, we get the movies. There are 5 2 5 2 = 5! 2!(5 − 2)! = 5! 2!3! = 5 · 4 2 = 10 So there are 10 diﬀerent ways to distribute those two tickets among ﬁve friends. (Some will see it as 10 ways to decide which three friends have to stay home.) The reader is encouraged to verify this by actually taking the time to list all of
the possibilities. We now state anf prove a theorem which is crucial to the proof of the Binomial Theorem. Theorem 9.3. For natural numbers n and j with n ≥ j = The proof of Theorem 9.3 is purely computational and uses the deﬁnition of binomial coeﬃcients, the recursive property of factorials and common denominators. 3For reference, 50! 50! 40! 50! 40!10! = 30414093201713378043612608166064768844377641568960512000000000000, = 37276043023296000, and = 10272278170 684 Sequences and the Binomial Theorem! (j − 1)!(n − (j − 1))! + n! j!(n − j)! n! (j − 1)!(n − j + 1)! + n! j!(n − j)! n! (j − 1)!(n − j + 1)(n − j)! + n! j(j − 1)!(n − j)! n! j j(j − 1)!(n − j + 1)(n − j)! + n!(n − j + 1) j(j − 1)!(n − j + 1)(n − j)! n! j j!(n − j + 1)! + n!(n − j + 1) j!(n − j + 1)! n! j + n!(n − j + 1) j!(n − j + 1)! n! (j + (n − j + 1)) j!(n − j + 1)! (n + 1)n! j!(n + 1 − j))! (n + 1)! j!((n + 1) − j))! n + 1 j We are now in position to state and prove the Binomial Theorem where we see that binomial coeﬃcients are just that - coeﬃcients in the binomial expansion. Theorem 9.4. Binomial Theorem: For nonzero real numbers a and b, (a + b)n = an−jbj n j=0 n j for all natural numbers n. To get a feel of what this theorem is saying and how it really isn’t as hard to remember as it may ﬁrst appear, let’
s consider the speciﬁc case of n = 4. According to the theorem, we have 9.4 The Binomial Theorem 685 (a + b)4 = 4 a4−jbj 4 j j=0 4 0 4 0 = = a4−0b0 + a4−1b1 + 4 1 a4−2b2 + 4 2 a4−3b3 + a4−4b4 4 4 a4 + a3b + 4 1 4 2 a2b2 + 4 3 ab3 + 4 3 4 4 b4 We forgo the simpliﬁcation of the coeﬃcients in order to note the pattern in the expansion. First note that in each term, the total of the exponents is 4 which matched the exponent of the binomial (a + b)4. The exponent on a begins at 4 and decreases by one as we move from one term to the next while the exponent on b starts at 0 and increases by one each time. Also note that the binomial coeﬃcients themselves have a pattern. The upper number, 4, matches the exponent on the binomial (a + b)4 whereas the lower number changes from term to term and matches the exponent of b in that term. This is no coincidence and corresponds to the kind of counting we discussed earlier. If we think of obtaining (a + b)4 by multiplying (a + b)(a + b)(a + b)(a + b), our answer is the sum of all possible products with exactly four factors - some a, some b. If we wish to count, for instance, the number of ways we obtain 1 factor of b out of a total of 4 possible factors, thereby forcing the remaining 3 factors to be a, the answer is 4 a3b is in the expansion. The. Hence, the term 4 1 1 other terms which appear cover the remaining cases. While this discussion gives an indication as to why the theorem is true, a formal proof requires Mathematical Induction.4 To prove the Binomial Theorem, we let P (n) be the expansion formula given in the statement of the theorem and we note that P (1) is true since (a + b)1 a + b? =? = 1 a1−jbj 1 j j=0 1 0 a1−0b0 + a1−1b1 1 1 a + b = a +
b Now we assume that P (k) is true. That is, we assume that we can expand (a + b)k using the formula given in Theorem 9.4 and attempt to show that P (k + 1) is true. 4and a fair amount of tenacity and attention to detail. 686 Sequences and the Binomial Theorem (a + b)k+1 = (a + b)(a + b)k k = (a + b) k j ak−jbj j=0 = a k j=0 k j ak−jbj + b = k j=0 k j ak+1−jbj + k j=0 k j=0 k j ak−jbj ak−jbj+1 k j Our goal is to combine as many of the terms as possible within the two summations. As the counter j in the ﬁrst summation runs from 0 through k, we get terms involving ak+1, akb, ak−1b2,..., abk. In the second summation, we get terms involving akb, ak−1b2,..., abk, bk+1. In other words, apart from the ﬁrst term in the ﬁrst summation and the last term in the second summation, we have terms common to both summations. Our next move is to ‘kick out’ the terms which we cannot combine and rewrite the summations so that we can combine them. To that end, we note k j=0 k j ak+1−jbj = ak+1 + k j=1 k j ak+1−jbj ak−jbj+1 = k j=0 k j k−1 j=0 k j ak−jbj+1 + bk+1 and so that (a + b)k+1 = ak+1 + ak+1−jbj + k j=1 k j k−1 j=0 k j ak−jbj+1 + bk+1 We now wish to write ak+1−jbj + k j=1 k j k−1 j=0 k j ak−jbj+1 as a single summation. The wrinkle is that the ﬁrst summation starts with j = 1, while the second starts with j = 0. Even though the sums produce terms with the same
powers of a and b, they do so for diﬀerent values of j. To resolve this, we need to shift the index on the second summation so that the index j starts at j = 1 instead of j = 0 and we make use of Theorem 9.1 in the process. 9.4 The Binomial Theorem 687 k−1 j=0 k j ak−jbj+1 = k−1+1 k j − 1 j=0+1 ak−(j−1)b(j−1)+1 = k k j − 1 j=1 ak+1−jbj We can now combine our two sums using Theorem 9.1 and simplify using Theorem 9.3 k j=1 k j ak+1−jbj + ak−jbj+1 = k−1 j=0 k j = = k j=1 k j=1 k j=1 ak+1−jbj + k k ak+1−jbj k j j=1 j − 1 ak+1−jbj ak+1−jbj Using this and the fact that k+1 0 = 1 and k+1 k+1 = 1, we get (a + b)k+1 = ak+1 + k k + 1 j ak+1−jbj + bk+1 j=1 k + 1 0 ak+1b0 + k+1 j=0 k + 1 j a(k+1)−jbj = = k j=1 k + 1 j ak+1−jbj + k + 1 k + 1 a0bk+1 which shows that P (k + 1) is true. Hence, by induction, we have established that the Binomial Theorem holds for all natural numbers n. Example 9.4.2. Use the Binomial Theorem to ﬁnd the following. 1. (x − 2)4 2. 2.13 3. The term containing x3 in the expansion (2x + y)5 Solution. 1. Since (x − 2)4 = (x + (−2))4, we identify a = x, b = −2 and n = 4 and obtain 688 Sequences and the Binomial Theorem (x − 2)4 = 4 x4−j(−2)j 4 j j=0 4 0 = x4−0(−2)0 + x4−
1(−2)1 + 4 1 4 2 x4−2(−2)2 + 4 3 x4−3(−2)3 + x4−4(−2)4 4 4 = x4 − 8x3 + 24x2 − 32x + 16 2. At ﬁrst this problem seem misplaced, but we can write 2.13 = (2 + 0.1)3. Identifying a = 2, b = 0.1 = 1 10 and n = 3, we get 3 23−j 3 j j 1 10 2 + 3 1 10 = = j=0 3 0 23−0 0 1 10 3 1 + 23−1 1 1 10 3 2 + 23−2 2 1 10 3 3 + 23−3 3 1 10 = 8 + 12 10 + 6 100 + 1 1000 = 8 + 1.2 + 0.06 + 0.001 = 9.261 3. Identifying a = 2x, b = y and n = 5, the Binomial Theorem gives (2x + y)5 = (2x)5−jyj 5 j=0 5 j Since we are concerned with only the term containing x3, there is no need to expand the entire sum. The exponents on each term must add to 5 and if the exponent on x is 3, the exponent on y must be 2. Plucking out the term j = 2, we get 5 2 (2x)5−2y2 = 10(2x)3y2 = 80x3y2 We close this section with Pascal’s Triangle, named in honor of the mathematician Blaise Pascal. Pascal’s Triangle is obtained by arranging the binomial coeﬃcients in the triangular fashion below. 9.4 The Binomial Theorem 689... for all whole numbers n, we get that each row of Pascal’s Triangle = 1 and n Since n n 0 begins and ends with 1. To generate the numbers in the middle of the rows (from the third row onwards), we take advantage of the additive relationship expressed in Theorem 9.3. For instance, and so forth. This relationship is indicated by the arrows in the + array above. With these two facts in hand, we can quickly generate Pascal’s Triangle. We start with the ﬁrst two rows, 1 and 1 1. From that point on, each successive row begins
and ends with 1 and the middle numbers are generated using Theorem 9.3. Below we attempt to demonstrate this building process to generate the ﬁrst ﬁve rows of Pascal’s Triangle −−−−−−→ −−−−−−→ −−−−−−→ 690 Sequences and the Binomial Theorem To see how we can use Pascal’s Triangle to expedite the Binomial Theorem, suppose we wish to for j = 0, 1, 2, 3, 4 and are the numbers which expand (3x − y)4. The coeﬃcients we need are 4 j form the ﬁfth row of Pascal’s Triangle. Since we know that the exponent of 3x in the ﬁrst term is 4 and then decreases by one as we go from left to right while the exponent of −y starts at 0 in the ﬁrst term and then increases by one as we move from left to right, we quickly obtain (3x − y)4 = (1)(3x)4 + (4)(3x)3(−y) + (6)(3x)2(−y)2 + 4(3x)(−y)3 + 1(−y)4 = 81x4 − 108x3y + 54x2y2 − 12xy3 + y4 We would like to stress that Pascal’s Triangle is a very quick method to expand an entire binomial. If only a term (or two or three) is required, then the Binomial Theorem is deﬁnitely the way to go. 9.4 The Binomial Theorem 691 9.4.1 Exercises In Exercises 1 - 9, simplify the given expression. 1. (3!)2 4. 7. 9! 4!3!2! 8 3 2. 5. 8. 10! 7! (n + 1)! n! 117 0, n ≥ 0. 3. 6. 9. 7! 233! (k − 1)! (k + 2)! n n − 2, k ≥ 1., n ≥ 2 In Exercises 10 - 13, use Pascal’s Triangle to expand the given binomial. 10. (x + 2)5 11. (2x − 1)4 12. 1 3 x + y23 13. x − x−14 In
Exercises 14 - 17, use Pascal’s Triangle to simplify the given power of a complex number. 14. (1 + 2i)4 √ 16. 3 i 3 2 + 1 2 15. −1 + i √ √ 33 √ 4 i 2 2 − 2 2 17. In Exercises 18 - 22, use the Binomial Theorem to ﬁnd the indicated term. 18. The term containing x3 in the expansion (2x − y)5 19. The term containing x117 in the expansion (x + 2)118 √ 7 2 in the expansion ( 20. The term containing x 21. The term containing x−7 in the expansion 2x − x−35 22. The constant term in the expansion x + x−18 23. Use the Prinicple of Mathematical Induction to prove n! > 2n for n ≥ 4. x − 3)8 24. Prove n j=0 n j = 2n for all natural numbers n. (HINT: Use the Binomial Theorem!) 25. With the help of your classmates, research Patterns and Properties of Pascal’s Triangle. 26. You’ve just won three tickets to see the new ﬁlm, ‘8.9.’ Five of your friends, Albert, Beth, Chuck, Dan, and Eugene, are interested in seeing it with you. With the help of your classmates, list all the possible ways to distribute your two extra tickets among your ﬁve friends. Now suppose you’ve come down with the ﬂu. List all the diﬀerent ways you can distribute the three tickets among these ﬁve friends. How does this compare with the ﬁrst list you made?? = 5 What does this have to do with the fact that 5 3 2 692 Sequences and the Binomial Theorem 9.4.2 Answers 1. 36 4. 1260 7. 56 2. 720 5. n + 1 8. 1 3. 105 6. 1 k(k+1)(k+2) 9. n(n−1) 2 10. (x + 2)5 = x5 + 10x4 + 40x3 + 80x2 + 80x + 32 11. (2x − 1)4 = 16x4 − 32x3 + 24x2 − 8x + 1 12. 1 3
x + y23 13. x − x−14 = 1 27 x3 + 1 3 x2y2 + xy4 + y6 = x4 − 4x2 + 6 − 4x−2 + x−4 14. −7 − 24i 15. 8 16. i 17. −1 18. 80x3y2 19. 236x117 20. −24x 7 2 21. −40x−7 22. 70 Chapter 10 Foundations of Trigonometry 10.1 Angles and their Measure This section begins our study of Trigonometry and to get started, we recall some basic deﬁnitions from Geometry. A ray is usually described as a ‘half-line’ and can be thought of as a line segment in which one of the two endpoints is pushed oﬀ inﬁnitely distant from the other, as pictured below. The point from which the ray originates is called the initial point of the ray. P A ray with initial point P. When two rays share a common initial point they form an angle and the common initial point is called the vertex of the angle. Two examples of what are commonly thought of as angles are P An angle with vertex P. Q An angle with vertex Q. However, the two ﬁgures below also depict angles - albeit these are, in some sense, extreme cases. In the ﬁrst case, the two rays are directly opposite each other forming what is known as a straight angle; in the second, the rays are identical so the ‘angle’ is indistinguishable from the ray itself. P A straight angle. Q The measure of an angle is a number which indicates the amount of rotation that separates the rays of the angle. There is one immediate problem with this, as pictured below. 694 Foundations of Trigonometry Which amount of rotation are we attempting to quantify? What we have just discovered is that we have at least two angles described by this diagram.1 Clearly these two angles have diﬀerent measures because one appears to represent a larger rotation than the other, so we must label them diﬀerently. In this book, we use lower case Greek letters such as α (alpha), β (beta), γ (gamma) and θ (theta) to label angles. So, for instance, we have β α One commonly used system to measure angles is degree measure. Quantities measured in
degrees are denoted by the familiar ‘◦’ symbol. One complete revolution as shown below is 360◦, and parts of a revolution are measured proportionately.2 Thus half of a revolution (a straight angle) measures 1 2 (360◦) = 180◦, a quarter of a revolution (a right angle) measures 1 4 (360◦) = 90◦ and so on. One revolution ↔ 360◦ 180◦ 90◦ Note that in the above ﬁgure, we have used the small square ‘ ’ to denote a right angle, as is commonplace in Geometry. Recall that if an angle measures strictly between 0◦ and 90◦ it is called an acute angle and if it measures strictly between 90◦ and 180◦ it is called an obtuse angle. It is important to note that, theoretically, we can know the measure of any angle as long as we 1The phrase ‘at least’ will be justiﬁed in short order. 2The choice of ‘360’ is most often attributed to the Babylonians. 10.1 Angles and their Measure 695 know the proportion it represents of entire revolution.3 For instance, the measure of an angle which represents a rotation of 2 3 (360◦) = 240◦, the measure of an angle which constitutes only 1 12 (360◦) = 30◦ and an angle which indicates no rotation at all is measured as 0◦. 3 of a revolution would measure 2 12 of a revolution measures 1 240◦ 30◦ 0◦ Using our deﬁnition of degree measure, we have that 1◦ represents the measure of an angle which 1 constitutes 360 of a revolution. Even though it may be hard to draw, it is nonetheless not diﬃcult to imagine an angle with measure smaller than 1◦. There are two ways to subdivide degrees. The ﬁrst, and most familiar, is decimal degrees. For example, an angle with a measure of 30.5◦ would represent a rotation halfway between 30◦ and 31◦, or equivalently, 30.5 720 of a full rotation. This make sense.4 The second way can be taken to the limit using Calculus so that measures like to divide degrees is the Degree - Minute - Second (DMS) system. In this system, one degree is divided equally
into sixty minutes, and in turn, each minute is divided equally into sixty seconds.5 In symbols, we write 1◦ = 60 and 1 = 60, from which it follows that 1◦ = 3600. To convert a measure of 42.125◦ to the DMS system, we start by noting that 42.125◦ = 42◦ + 0.125◦. Converting the partial amount of degrees to minutes, we ﬁnd 0.125◦ 60 = 7.5 = 7 + 0.5. Converting the partial amount of minutes to seconds gives 0.5 60 = 30. Putting it all together yields 360 = 61 2 √ 1◦ ◦ 1 42.125◦ = 42◦ + 0.125◦ = 42◦ + 7.5 = 42◦ + 7 + 0.5 = 42◦ + 7 + 30 = 42◦730 On the other hand, to convert 117◦1545 to decimal degrees, we ﬁrst compute 15 1◦ 60 45 1◦ 3600. Then we ﬁnd = 1 80 ◦ ◦ = 1 4 and 3This is how a protractor is graded. 4Awesome math pun aside, this is the same idea behind deﬁning irrational exponents in Section 6.1. 5Does this kind of system seem familiar? 696 Foundations of Trigonometry 117◦1545 = 117◦ + 15 + 45 ◦ ◦ + 1 80 = 117◦ + 1 4 = 9381 80 = 117.2625◦ ◦ Recall that two acute angles are called complementary angles if their measures add to 90◦. Two angles, either a pair of right angles or one acute angle and one obtuse angle, are called supplementary angles if their measures add to 180◦. In the diagram below, the angles α and β are supplementary angles while the pair γ and θ are complementary angles. β α θ γ Supplementary Angles Complementary Angles In practice, the distinction between the angle itself and its measure is blurred so that the sentence ‘α is an angle measuring 42◦’ is often abbreviated as ‘α = 42◦.’ It is now time for an example. Example 10.1.1. Let α = 111.371◦ and β = 37◦2817. 1.
Convert α to the DMS system. Round your answer to the nearest second. 2. Convert β to decimal degrees. Round your answer to the nearest thousandth of a degree. 3. Sketch α and β. 4. Find a supplementary angle for α. 5. Find a complementary angle for β. Solution. 1. To convert α to the DMS system, we start with 111.371◦ = 111◦ + 0.371◦. Next we convert 0.371◦ 60 1◦ = 22.26. Writing 22.26 = 22 + 0.26, we convert 0.26 60 = 15.6. Hence, 1 111.371◦ = 111◦ + 0.371◦ = 111◦ + 22.26 = 111◦ + 22 + 0.26 = 111◦ + 22 + 15.6 = 111◦2215.6 Rounding to seconds, we obtain α ≈ 111◦2216. 10.1 Angles and their Measure 697 2. To convert β to decimal degrees, we convert 28 1◦ 60 = 7 15 ◦ and 17 1◦ 3600 = 17 3600 ◦. Putting it all together, we have 37◦2817 = 37◦ + 28 + 17 + 17 3600 ◦ = 37◦ + 7 15 ◦ = 134897 3600 ≈ 37.471◦ ◦ 3. To sketch α, we ﬁrst note that 90◦ < α < 180◦. If we divide this range in half, we get 90◦ < α < 135◦, and once more, we have 90◦ < α < 112.5◦. This gives us a pretty good estimate for α, as shown below.6 Proceeding similarly for β, we ﬁnd 0◦ < β < 90◦, then 0◦ < β < 45◦, 22.5◦ < β < 45◦, and lastly, 33.75◦ < β < 45◦. Angle α Angle β 4. To ﬁnd a supplementary angle for α, we seek an angle θ so that α + θ = 180◦. We get θ = 180◦ − α = 180◦ − 111.371◦ = 68.629◦. 5. To ﬁnd a complementary angle for β, we seek an
angle γ so that β + γ = 90◦. We get γ = 90◦ − β = 90◦ − 37◦2817. While we could reach for the calculator to obtain an approximate answer, we choose instead to do a bit of sexagesimal7 arithmetic. We ﬁrst rewrite 90◦ = 90◦00 = 89◦600 = 89◦5960. In essence, we are ‘borrowing’ 1◦ = 60 from the degree place, and then borrowing 1 = 60 from the minutes place.8 This yields, γ = 90◦ − 37◦2817 = 89◦5960 − 37◦2817 = 52◦3143. Up to this point, we have discussed only angles which measure between 0◦ and 360◦, inclusive. Ultimately, we want to use the arsenal of Algebra which we have stockpiled in Chapters 1 through 9 to not only solve geometric problems involving angles, but also to extend their applicability to other real-world phenomena. A ﬁrst step in this direction is to extend our notion of ‘angle’ from merely measuring an extent of rotation to quantities which can be associated with real numbers. To that end, we introduce the concept of an oriented angle. As its name suggests, in an oriented 6If this process seems hauntingly familiar, it should. Compare this method to the Bisection Method introduced in Section 3.3. 7Like ‘latus rectum,’ this is also a real math term. 8This is the exact same kind of ‘borrowing’ you used to do in Elementary School when trying to ﬁnd 300 − 125. Back then, you were working in a base ten system; here, it is base sixty. 698 Foundations of Trigonometry angle, the direction of the rotation is important. We imagine the angle being swept out starting from an initial side and ending at a terminal side, as shown below. When the rotation is counter-clockwise9 from initial side to terminal side, we say that the angle is positive; when the rotation is clockwise, we say that the angle is negative. Sid e T er m in al Initial Side Initial Side A positive angle, 45◦ A negative angle, −45◦ At this point, we also extend our allowable rotations to include angles which encompass more than one revolution. For example, to sketch an
angle with measure 450◦ we start with an initial side, rotate counter-clockwise one complete revolution (to take care of the ‘ﬁrst’ 360◦) then continue with an additional 90◦ counter-clockwise rotation, as seen below. 450◦ To further connect angles with the Algebra which has come before, we shall often overlay an angle diagram on the coordinate plane. An angle is said to be in standard position if its vertex is the origin and its initial side coincides with the positive x-axis. Angles in standard position are classiﬁed according to where their terminal side lies. For instance, an angle in standard position whose terminal side lies in Quadrant I is called a ‘Quadrant I angle’. If the terminal side of an angle lies on one of the coordinate axes, it is called a quadrantal angle. Two angles in standard position are called coterminal if they share the same terminal side.10 In the ﬁgure below, α = 120◦ and β = −240◦ are two coterminal Quadrant II angles drawn in standard position. Note that α = β + 360◦, or equivalently, β = α − 360◦. We leave it as an exercise to the reader to verify that coterminal angles always diﬀer by a multiple of 360◦.11 More precisely, if α and β are coterminal angles, then β = α + 360◦ · k where k is an integer.12 9‘widdershins’ 10Note that by being in standard position they automatically share the same initial side which is the positive x-axis. 11It is worth noting that all of the pathologies of Analytic Trigonometry result from this innocuous fact. 12Recall that this means k = 0, ±1, ±2,.... m 10.1 Angles and their Measure 699 y 4 3 2 1 α = 120◦ −4 −3 −2 −1 −1 1 2 3 4 x β = −240◦ −2 −3 −4 Two coterminal angles, α = 120◦ and β = −240◦, in standard position. Example 10.1.2. Graph each of the (oriented) angles below in standard position and classify them according to where their terminal side lies. Find three coterminal angles
, at least one of which is positive and one of which is negative. 1. α = 60◦ 2. β = −225◦ 3. γ = 540◦ 4. φ = −750◦ Solution. 1. To graph α = 60◦, we draw an angle with its initial side on the positive x-axis and rotate counter-clockwise 60◦ 6 of a revolution. We see that α is a Quadrant I angle. To ﬁnd angles which are coterminal, we look for angles θ of the form θ = α + 360◦ · k, for some integer k. When k = 1, we get θ = 60◦+360◦ = 420◦. Substituting k = −1 gives θ = 60◦−360◦ = −300◦. Finally, if we let k = 2, we get θ = 60◦ + 720◦ = 780◦. 360◦ = 1 2. Since β = −225◦ is negative, we start at the positive x-axis and rotate clockwise 225◦ 8 of a revolution. We see that β is a Quadrant II angle. To ﬁnd coterminal angles, we proceed as before and compute θ = −225◦ + 360◦ · k for integer values of k. We ﬁnd 135◦, −585◦ and 495◦ are all coterminal with −225◦. 360◦ = 5 y 4 3 2 1 α = 60◦ y 4 3 2 1 −4 −3 −2 −1 −1 1 2 3 4 x −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 β = −225◦ −2 −3 −4 α = 60◦ in standard position. β = −225◦ in standard position. 700 Foundations of Trigonometry 3. Since γ = 540◦ is positive, we rotate counter-clockwise from the positive x-axis. One full revolution accounts for 360◦, with 180◦, or 1 2 of a revolution remaining. Since the terminal side of γ lies on the negative x-axis, γ is a quadrantal angle. All angles coterminal with γ are of the form θ = 540◦ + 360◦ ·
k, where k is an integer. Working through the arithmetic, we ﬁnd three such angles: 180◦, −180◦ and 900◦. 4. The Greek letter φ is pronounced ‘fee’ or ‘ﬁe’ and since φ is negative, we begin our rotation clockwise from the positive x-axis. Two full revolutions account for 720◦, with just 30◦ or 1 12 of a revolution to go. We ﬁnd that φ is a Quadrant IV angle. To ﬁnd coterminal angles, we compute θ = −750◦ + 360◦ · k for a few integers k and obtain −390◦, −30◦ and 330◦. γ = 5404 −3 −2 −1 −1 1 2 3 4 x −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 φ = −750◦ −2 −3 −4 γ = 540◦ in standard position. φ = −750◦ in standard position. Note that since there are inﬁnitely many integers, any given angle has inﬁnitely many coterminal angles, and the reader is encouraged to plot the few sets of coterminal angles found in Example 10.1.2 to see this. We are now just one step away from completely marrying angles with the real numbers and the rest of Algebra. To that end, we recall this deﬁnition from Geometry. Deﬁnition 10.1. The real number π is deﬁned to be the ratio of a circle’s circumference to its diameter. In symbols, given a circle of circumference C and diameter d, π = C d While Deﬁnition 10.1 is quite possibly the ‘standard’ deﬁnition of π, the authors would be remiss if we didn’t mention that buried in this deﬁnition is actually a theorem. As the reader is probably aware, the number π is a mathematical constant - that is, it doesn’t matter which circle is selected, the ratio of its circumference to its diameter will have the same value as any other circle. While this is indeed true, it is far from obvious and leads to a counterintuitive scenario
which is explored in the Exercises. Since the diameter of a circle is twice its radius, we can quickly rearrange the equation in Deﬁnition 10.1 to get a formula more useful for our purposes, namely: 2π = C r 10.1 Angles and their Measure 701 This tells us that for any circle, the ratio of its circumference to its radius is also always constant; in this case the constant is 2π. Suppose now we take a portion of the circle, so instead of comparing the entire circumference C to the radius, we compare some arc measuring s units in length to the radius, as depicted below. Let θ be the central angle subtended by this arc, that is, an angle whose vertex is the center of the circle and whose determining rays pass through the endpoints of the arc. Using proportionality arguments, it stands to reason that the ratio should also be a constant among all circles, and it is this ratio which deﬁnes the radian measure of an angle. s r θ r s r The radian measure of θ is s r. To get a better feel for radian measure, we note that an angle with radian measure 1 means the corresponding arc length s equals the radius of the circle r, hence s = r. When the radian measure is 2, we have s = 2r; when the radian measure is 3, s = 3r, and so forth. Thus the radian measure of an angle θ tells us how many ‘radius lengths’ we need to sweep out along the circle to subtend the angle θ has radian measure 1 β has radian measure 4 Since one revolution sweeps out the entire circumference 2πr, one revolution has radian measure 2πr r = 2π. From this we can ﬁnd the radian measure of other central angles using proportions, 702 Foundations of Trigonometry just like we did with degrees. For instance, half of a revolution has radian measure 1 2 (2π) = π, a quarter revolution has radian measure 1 4 (2π) = π 2, and so forth. Note that, by deﬁnition, the radian measure of an angle is a length divided by another length so that these measurements are actually dimensionless and are considered ‘pure’ numbers. For this reason, we do not use any symbols to denote radian measure, but we use the
word ‘radians’ to denote these dimensionless units as needed. For instance, we say one revolution measures ‘2π radians,’ half of a revolution measures ‘π radians,’ and so forth. As with degree measure, the distinction between the angle itself and its measure is often blurred in practice, so when we write ‘θ = π 2 radians.13 We extend radian measure to oriented angles, just as we did with degrees beforehand, so that a positive measure indicates counter-clockwise rotation and a negative measure indicates clockwise rotation.14 Much like before, two positive angles α and β are supplementary if α + β = π and complementary if α + β = π 2. Finally, we leave it to the reader to show that when using radian measure, two angles α and β are coterminal if and only if β = α + 2πk for some integer k. 2 ’, we mean θ is an angle which measures π Example 10.1.3. Graph each of the (oriented) angles below in standard position and classify them according to where their terminal side lies. Find three coterminal angles, at least one of which is positive and one of which is negative. 1. α = π 6 Solution. 2. β = − 4π 3 3. γ = 9π 4 4. φ = − 5π 2 1. The angle α = π 6 is positive, so we draw an angle with its initial side on the positive x-axis and rotate counter-clockwise (π/6) 12 of a revolution. Thus α is a Quadrant I angle. Coterminal angles θ are of the form θ = α + 2π · k, for some integer k. To make the arithmetic a bit easier, we note that 2π = 12π 6, thus when k = 1, we get θ = π 6. Substituting 6 = 25π k = −1 gives θ = π 6. 6 + 12π 6 and when we let k = 2, we get θ = π 6 = 13π 6 + 24π 6 = − 11π 2π = 1 6 − 12π 2. Since β = − 4π 3 is negative, we start at the positive x-axis and rotate clockwise (4π/3) 3 of a revolution. We ﬁnd β to
be a Quadrant II angle. To ﬁnd coterminal angles, we proceed as before using 2π = 6π 3 · k for integer values of k. We obtain 2π 3 + 6π 3, − 10π 3, and compute θ = − 4π 2π = 2 3 as coterminal angles. 3 and 8π 13The authors are well aware that we are now identifying radians with real numbers. We will justify this shortly. 14This, in turn, endows the subtended arcs with an orientation as well. We address this in short order. 10.1 Angles and their Measure 703 y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 −3 −2 −1 −1 1 2 3 4 x β = − 4π 3 −2 −3 −4 α = π 6 in standard position. β = − 4π 3 in standard position. 3. Since γ = 9π 4 revolution accounts for 2π = 8π We have γ as a Quadrant I angle. All angles coterminal with γ are of the form θ = 9π where k is an integer. Working through the arithmetic, we ﬁnd: π is positive, we rotate counter-clockwise from the positive x-axis. One full 8 of a revolution remaining. 4 · k, 4 of the radian measure with π 4 + 8π 4 or 1 4, − 7π 4 and 17π 4. 4. To graph φ = − 5π 2, we begin our rotation clockwise from the positive x-axis. As 2π = 4π 2, after one full revolution clockwise, we have π 4 of a revolution remaining. Since the terminal side of φ lies on the negative y-axis, φ is a quadrantal angle. To ﬁnd coterminal angles, we compute θ = − 5π 2 · k for a few integers k and obtain − π 2 and 7π 2. 2 + 4π 2 or 1 2, 3π = − 5π 2 −4 −3 −2 −1 −1 1 2 3 4 x −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 γ = 9π 4 −2 −3 −4 γ = 9π 4 in standard position. φ = − 5π
2 in standard position. It is worth mentioning that we could have plotted the angles in Example 10.1.3 by ﬁrst converting them to degree measure and following the procedure set forth in Example 10.1.2. While converting back and forth from degrees and radians is certainly a good skill to have, it is best that you learn to ‘think in radians’ as well as you can ‘think in degrees’. The authors would, however, be 704 Foundations of Trigonometry derelict in our duties if we ignored the basic conversion between these systems altogether. Since one revolution counter-clockwise measures 360◦ and the same angle measures 2π radians, we can use the proportion 2π radians, as the conversion factor between the two systems. For example, to convert 60◦ to radians we ﬁnd 60◦ π radians 3 radians, or 180◦ simply π π radian. For = −150◦.15 Of particular interest is the example, − 5π fact that an angle which measures 1 in radian measure is equal to 180◦ = π 3. To convert from radian measure back to degrees, we multiply by the ratio 6 radians 180◦ 6 radians is equal to − 5π, or its reduced equivalent, π radians π radians 180◦ 360◦ 180◦ π ≈ 57.2958◦. We summarize these conversions below. Equation 10.1. Degree - Radian Conversion: To convert degree measure to radian measure, multiply by To convert radian measure to degree measure, multiply by π radians 180◦ 180◦ π radians In light of Example 10.1.3 and Equation 10.1, the reader may well wonder what the allure of radian measure is. The numbers involved are, admittedly, much more complicated than degree measure. The answer lies in how easily angles in radian measure can be identiﬁed with real numbers. Consider the Unit Circle, x2 +y2 = 1, as drawn below, the angle θ in standard position and the corresponding arc measuring s units in length. By deﬁnition, and the fact that the Unit Circle has radius 1, the radian measure of θ is = s so that, once again blurring the distinction between an angle and its measure, we have
θ = s. In order to identify real numbers with oriented angles, we make good use of this fact by essentially ‘wrapping’ the real number line around the Unit Circle and associating to each real number t an oriented arc on the Unit Circle with initial point (1, 0). s 1 s r = Viewing the vertical line x = 1 as another real number line demarcated like the y-axis, given a real number t > 0, we ‘wrap’ the (vertical) interval [0, t] around the Unit Circle in a counter-clockwise fashion. The resulting arc has a length of t units and therefore the corresponding angle has radian measure equal to t. If t < 0, we wrap the interval [t, 0] clockwise around the Unit Circle. Since we have deﬁned clockwise rotation as having negative radian measure, the angle determined by this arc has radian measure equal to t. If t = 0, we are at the point (1, 0) on the x-axis which corresponds to an angle with radian measure 0. In this way, we identify each real number t with the corresponding angle with radian measure t. 15Note that the negative sign indicates clockwise rotation in both systems, and so it is carried along accordingly. 10.1 Angles and their Measure 705 On the Unit Circle, θ = s. Identifying t > 0 with an angle. Identifying t < 0 with an angle. Example 10.1.4. Sketch the oriented arc on the Unit Circle corresponding to each of the following real numbers. 1. t = 3π 4 Solution. 2. t = −2π 3. t = −2 4. t = 117 1. The arc associated with t = 3π 4 is 3 4 is the arc on the Unit Circle which subtends the angle 3π 4 in 8 of a revolution, we have an arc which begins at the point (1, 0) radian measure. Since 3π proceeds counter-clockwise up to midway through Quadrant II. 2. Since one revolution is 2π radians, and t = −2π is negative, we graph the arc which begins at (1, 0) and proceeds clockwise for one full revolution = 3π 4 t = −2π 3. Like t = −2π, t = −2 is negative, so we begin our arc at (1, 0) and proceed clockwise
around the unit circle. Since π ≈ 3.14 and π 2 ≈ 1.57, we ﬁnd that rotating 2 radians clockwise from the point (1, 0) lands us in Quadrant III. To more accurately place the endpoint, we proceed as we did in Example 10.1.1, successively halving the angle measure until we ﬁnd 5π 8 ≈ 1.96 which tells us our arc extends just a bit beyond the quarter mark into Quadrant III. 706 Foundations of Trigonometry 4. Since 117 is positive, the arc corresponding to t = 117 begins at (1, 0) and proceeds counterclockwise. As 117 is much greater than 2π, we wrap around the Unit Circle several times before ﬁnally reaching our endpoint. We approximate 117 2π as 18.62 which tells us we complete 18 revolutions counter-clockwise with 0.62, or just shy of 5 8 of a revolution to spare. In other words, the terminal side of the angle which measures 117 radians in standard position is just short of being midway through Quadrant III2 t = 117 10.1.1 Applications of Radian Measure: Circular Motion Now that we have paired angles with real numbers via radian measure, a whole world of applications awaits us. Our ﬁrst excursion into this realm comes by way of circular motion. Suppose an object is moving as pictured below along a circular path of radius r from the point P to the point Q in an amount of time t. Q θ r s P Here s represents a displacement so that s > 0 means the object is traveling in a counter-clockwise direction and s < 0 indicates movement in a clockwise direction. Note that with this convention the formula we used to deﬁne radian measure, namely θ =, still holds since a negative value of s incurred from a clockwise displacement matches the negative we assign to θ for a clockwise rotation. In Physics, the average velocity of the object, denoted v and read as ‘v-bar’, is deﬁned as the average rate of change of the position of the object with respect to time.16 As a result, we s r 16See Deﬁnition 2.3 in Section 2.1 for a review of this concept. 10.1 Angles and their Measure 707 s t = time. The quantity
v has units of length have v = displacement time and conveys two ideas: the direction in which the object is moving and how fast the position of the object is changing. The contribution of direction in the quantity v is either to make it positive (in the case of counter-clockwise motion) or negative (in the case of clockwise motion), so that the quantity \|v\| quantiﬁes how fast the object is moving - it is the speed of the object. Measuring θ in radians we have θ = thus s = rθ and s r v = s t = rθ t = r · θ t θ t is called the average angular velocity of the object. It is denoted by ω and is The quantity read ‘omega-bar’. The quantity ω is the average rate of change of the angle θ with respect to time and thus has units radians time. If ω is constant throughout the duration of the motion, then it can be shown17 that the average velocities involved, namely v and ω, are the same as their instantaneous counterparts, v and ω, respectively. In this case, v is simply called the ‘velocity’ of the object and is the instantaneous rate of change of the position of the object with respect to time.18 Similarly, ω is called the ‘angular velocity’ and is the instantaneous rate of change of the angle with respect to time. If the path of the object were ‘uncurled’ from a circle to form a line segment, then the velocity of the object on that line segment would be the same as the velocity on the circle. For this reason, the quantity v is often called the linear velocity of the object in order to distinguish it from the angular velocity, ω. Putting together the ideas of the previous paragraph, we get the following. Equation 10.2. Velocity for Circular Motion: For an object moving on a circular path of radius r with constant angular velocity ω, the (linear) velocity of the object is given by v = rω. time. Thus the left hand side of the equation v = rω has units length We need to talk about units here. The units of v are length time, the units of r are length only, and the units of ω are radians time, whereas time = length·radians the right hand side has units length · rad
ians. The supposed contradiction in units is resolved by remembering that radians are a dimensionless quantity and angles in radian measure are identiﬁed with real numbers so that the units length·radians time. We are long overdue for an example. reduce to the units length time time Example 10.1.5. Assuming that the surface of the Earth is a sphere, any point on the Earth can be thought of as an object traveling on a circle which completes one revolution in (approximately) 24 hours. The path traced out by the point during this 24 hour period is the Latitude of that point. Lakeland Community College is at 41.628◦ north latitude, and it can be shown19 that the radius of the earth at this Latitude is approximately 2960 miles. Find the linear velocity, in miles per hour, of Lakeland Community College as the world turns. Solution. To use the formula v = rω, we ﬁrst need to compute the angular velocity ω. The earth π makes one revolution in 24 hours, and one revolution is 2π radians, so ω = 2π radians 12 hours, 24 hours = 17You guessed it, using Calculus... 18See the discussion on Page 161 for more details on the idea of an ‘instantaneous’ rate of change. 19We will discuss how we arrived at this approximation in Example 10.2.6. 708 Foundations of Trigonometry where, once again, we are using the fact that radians are real numbers and are dimensionless. (For simplicity’s sake, we are also assuming that we are viewing the rotation of the earth as counterclockwise so ω > 0.) Hence, the linear velocity is v = 2960 miles · π 12 hours ≈ 775 miles hour It is worth noting that the quantity 1 revolution in Example 10.1.5 is called the ordinary frequency 24 hours of the motion and is usually denoted by the variable f. The ordinary frequency is a measure of how often an object makes a complete cycle of the motion. The fact that ω = 2πf suggests that ω is also a frequency. Indeed, it is called the angular frequency of the motion. On a related note, the quantity T = is called the period of the motion and is the amount of time it takes for the 1 f object to complete one cycle of the motion. In the scenario of Example 10.1.5,
the period of the motion is 24 hours, or one day. The concepts of frequency and period help frame the equation v = rω in a new light. That is, if ω is ﬁxed, points which are farther from the center of rotation need to travel faster to maintain the same angular frequency since they have farther to travel to make one revolution in one period’s time. The distance of the object to the center of rotation is the radius of the circle, r, and is the ‘magniﬁcation factor’ which relates ω and v. We will have more to say about frequencies and periods in Section 11.1. While we have exhaustively discussed velocities associated with circular motion, we have yet to discuss a more natural question: if an object is moving on a circular path of radius r with a ﬁxed angular velocity (frequency) ω, what is the position of the object at time t? The answer to this question is the very heart of Trigonometry and is answered in the next section. 10.1 Angles and their Measure 709 10.1.2 Exercises In Exercises 1 - 4, convert the angles into the DMS system. Round each of your answers to the nearest second. 1. 63.75◦ 2. 200.325◦ 3. −317.06◦ 4. 179.999◦ In Exercises 5 - 8, convert the angles into decimal degrees. Round each of your answers to three decimal places. 5. 125◦50 6. −32◦1012 7. 502◦35 8. 237◦5843 In Exercises 9 - 28, graph the oriented angle in standard position. Classify each angle according to where its terminal side lies and then give two coterminal angles, one of which is positive and the other negative. 9. 330◦ 10. −135◦ 13. −270◦ 17. 3π 4 21. − π 2 25. −2π 14. 5π 6 18. − π 3 22. 7π 6 26. − π 4 11. 120◦ 15. − 11π 3 19. 7π 2 23. − 5π 3 27. 15π 4 12. 405◦ 16. 20. 5π 4 π 4 24. 3π 28. − 13π 6 In Exercises 29 - 36, convert
the angle from degree measure into radian measure, giving the exact value in terms of π. 29. 0◦ 33. −315◦ 30. 240◦ 34. 150◦ 31. 135◦ 35. 45◦ 32. −270◦ 36. −225◦ In Exercises 37 - 44, convert the angle from radian measure into degree measure. 37. π 41. π 3 38. − 2π 3 42. 5π 3 39. 7π 6 43. − π 6 40. 44. 11π 6 π 2 710 Foundations of Trigonometry In Exercises 45 - 49, sketch the oriented arc on the Unit Circle which corresponds to the given real number. 45. t = 5π 6 46. t = −π 47. t = 6 48. t = −2 49. t = 12 50. A yo-yo which is 2.25 inches in diameter spins at a rate of 4500 revolutions per minute. How fast is the edge of the yo-yo spinning in miles per hour? Round your answer to two decimal places. 51. How many revolutions per minute would the yo-yo in exercise 50 have to complete if the edge of the yo-yo is to be spinning at a rate of 42 miles per hour? Round your answer to two decimal places. 52. In the yo-yo trick ‘Around the World,’ the performer throws the yo-yo so it sweeps out a vertical circle whose radius is the yo-yo string. If the yo-yo string is 28 inches long and the yo-yo takes 3 seconds to complete one revolution of the circle, compute the speed of the yo-yo in miles per hour. Round your answer to two decimal places. 53. A computer hard drive contains a circular disk with diameter 2.5 inches and spins at a rate of 7200 RPM (revolutions per minute). Find the linear speed of a point on the edge of the disk in miles per hour. 54. A rock got stuck in the tread of my tire and when I was driving 70 miles per hour, the rock came loose and hit the inside of the wheel well of the car. How fast, in miles per hour, was the rock traveling when it came out of the tread? (The tire has a diameter of 23 inches.) 55. The Giant Wheel at Cedar Point is a circle with diameter 128 feet which sits on an 8 foot tall platform making its overall height
is 136 feet. (Remember this from Exercise 17 in Section 7.2?) It completes two revolutions in 2 minutes and 7 seconds.20 Assuming the riders are at the edge of the circle, how fast are they traveling in miles per hour? 56. Consider the circle of radius r pictured below with central angle θ, measured in radians, and subtended arc of length s. Prove that the area of the shaded sector is A = 1 (Hint: Use the proportion s circumference of the circle.) A area of the circle = 2 r2θ. r s θ r 20Source: Cedar Point’s webpage. 10.1 Angles and their Measure 711 In Exercises 57 - 62, use the result of Exercise 56 to compute the areas of the circular sectors with the given central angles and radii. 57. θ = π 6, r = 12 60. θ = π, r = 1 58. θ = 5π 4, r = 100 61. θ = 240◦, r = 5 59. θ = 330◦, r = 9.3 62. θ = 1◦, r = 117 63. Imagine a rope tied around the Earth at the equator. Show that you need to add only 2π feet of length to the rope in order to lift it one foot above the ground around the entire equator. (You do NOT need to know the radius of the Earth to show this.) 64. With the help of your classmates, look for a proof that π is indeed a constant. 712 Foundations of Trigonometry 10.1.3 Answers 1. 63◦45 5. 125.833◦ 2. 200◦1930 3. −317◦336 4. 179◦5956 6. −32.17◦ 7. 502.583◦ 8. 237.979◦ 9. 330◦ is a Quadrant IV angle coterminal with 690◦ and −30◦ 10. −135◦ is a Quadrant III angle coterminal with 225◦ and −4954 −3 −2 −1 −1 1 2 3 4 x −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 −2 −3 −4 11. 120◦ is a Quadrant II angle coterminal with 480◦ and −240�
� 12. 405◦ is a Quadrant I angle coterminal with 45◦ and −3154 −3 −2 −1 −1 1 2 3 4 x −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 −2 −3 −4 13. −270◦ lies on the positive y-axis 14. coterminal with 90◦ and −630◦ is a Quadrant II angle 5π 6 coterminal with 17π 6 and − 7π 4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x −4 −3 −2 −1 −1 −2 −3 −4 10.1 Angles and their Measure 713 15. − 11π 3 is a Quadrant I angle 5π π 3 3 and − coterminal with y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 17. 19. is a Quadrant II angle 3π 4 coterminal with 11π 4 and − 5π 4 −3 −2 −1 −1 −2 −3 −4 7π 2 coterminal with 3π 2 and − π 2 y 4 3 2 1 16. is a Quadrant III angle 5π 4 coterminal with 13π 4 and − 3π 4 −3 −2 −1 −1 −2 −3 −4 18. − π 3 is a Quadrant IV angle 7π 3 and − 5π 3 coterminal with y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x π 4 coterminal with 9π 4 and − 7π 4 y 4 3 2 1 lies on the negative y-axis 20. is a Quadrant I angle −4 −3 −2 −1 −1 1 2 3 4 x −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 −2 −3 −4 714 21. − π 2 lies on the negative y-axis 5π 3π 2 2 and − coterminal with y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x Foundations of Trigonometry 22. is a Quadrant III angle 7π 6 coter
minal with 19π 6 and − 5π 4 −3 −2 −1 −1 −2 −3 −4 23. − 5π 3 is a Quadrant I angle 24. 3π lies on the negative x-axis coterminal with y 4 3 2 1 π 3 and − 11π 3 coterminal with π and −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x −4 −3 −2 −1 −1 −2 −3 −4 25. −2π lies on the positive x-axis 26. − π 4 is a Quadrant IV angle coterminal with 2π and −4π coterminal with 7π 4 and − 9π 4 −4 −3 −2 −1 −1 1 2 3 4 x −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 −2 −3 −4 10.1 Angles and their Measure 715 27. 15π 4 is a Quadrant IV angle π 4 and − 7π 4 coterminal with y 28. − 13π 6 is a Quadrant IV angle π 11π 6 6 and − coterminal with y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 29. 0 33. − 7π 4 37. 180◦ 41. 60◦ 45. t = 5π 6 47. t = 6 30. 34. 4π 3 5π 6 38. −120◦ 42. 300◦ 1 x y 1 y 1 31. 35. 3π 4 π 4 39. 210◦ 43. −30◦ 46. t = −π 48. t = −2 32. − 36. − 3π 2 5π 4 40. 330◦ 44. 90 716 Foundations of Trigonometry 49. t = 12 (between 1 and 2 revolutions) y 1 1 x 50. About 30.12 miles per hour 51. About 6274.52 revolutions per minute 52. About 3.33 miles per hour 53. About 53.55 miles per hour 54. 70 miles per hour 55. About 4.32 miles per hour 57. 12π square units 59. 79.2825π ≈ 249.07 square units 61. 50
π 3 square units 58. 6250π square units 60. π 2 square units 62. 38.025π ≈ 119.46 square units 10.2 The Unit Circle: Cosine and Sine 717 10.2 The Unit Circle: Cosine and Sine In Section 10.1.1, we introduced circular motion and derived a formula which describes the linear velocity of an object moving on a circular path at a constant angular velocity. One of the goals of this section is describe the position of such an object. To that end, consider an angle θ in standard position and let P denote the point where the terminal side of θ intersects the Unit Circle. By associating the point P with the angle θ, we are assigning a position on the Unit Circle to the angle θ. The x-coordinate of P is called the cosine of θ, written cos(θ), while the y-coordinate of P is called the sine of θ, written sin(θ).1 The reader is encouraged to verify that these rules used to match an angle with its cosine and sine do, in fact, satisfy the deﬁnition of a function. That is, for each angle θ, there is only one associated value of cos(θ) and only one associated value of sin(θ). y 1 θ 1 x y 1 P (cos(θ), sin(θ)) θ 1 x Example 10.2.1. Find the cosine and sine of the following angles. 1. θ = 270◦ 2. θ = −π 3. θ = 45◦ 4. θ = π 6 5. θ = 60◦ Solution. 1. To ﬁnd cos (270◦) and sin (270◦), we plot the angle θ = 270◦ in standard position and ﬁnd the point on the terminal side of θ which lies on the Unit Circle. Since 270◦ represents 3 4 of a counter-clockwise revolution, the terminal side of θ lies along the negative y-axis. Hence, the point we seek is (0, −1) so that cos (270◦) = 0 and sin (270◦) = −1. 2. The angle θ = −π represents one half of a clockwise revolution so its terminal side lies on the negative x-
axis. The point on the Unit Circle that lies on the negative x-axis is (−1, 0) which means cos(−π) = −1 and sin(−π) = 0. 1The etymology of the name ‘sine’ is quite colorful, and the interested reader is invited to research it; the ‘co’ in ‘cosine’ is explained in Section 10.4. 718 Foundations of Trigonometry y 1 θ = 270◦ P (0, −1) y 1 P (−1, 0) 1 x 1 x θ = −π Finding cos (270◦) and sin (270◦) Finding cos (−π) and sin (−π) 3. When we sketch θ = 45◦ in standard position, we see that its terminal does not lie along any of the coordinate axes which makes our job of ﬁnding the cosine and sine values a bit more diﬃcult. Let P (x, y) denote the point on the terminal side of θ which lies on the Unit Circle. By deﬁnition, x = cos (45◦) and y = sin (45◦). If we drop a perpendicular line segment from P to the x-axis, we obtain a 45◦ − 45◦ − 90◦ right triangle whose legs have lengths x and y units. From Geometry,2 we get y = x. Since P (x, y) lies on the Unit Circle, we have √ 2 x2 + y2 = 1. Substituting y = x into this equation yields 2x2 = 1, or x = ± 2. √ 2 Since P (x, y) lies in the ﬁrst quadrant, x > 0, so x = cos (45◦) = 2 and with y = x we have y = sin (45◦) = = 45◦ P (x, y) x 1 P (x, y) 45◦ y θ = 45◦ x 2Can you show this? 10.2 The Unit Circle: Cosine and Sine 719 4. As before, the terminal side of θ = π 6 does not lie on any of the coordinate axes, so we proceed using a triangle approach. Letting P (x, y) denote the point on the terminal side of θ which lies on
the Unit Circle, we drop a perpendicular line segment from P to the x-axis to form = 1 a 30◦ − 60◦ − 90◦ right triangle. After a bit of Geometry3 we ﬁnd y = 1 2. Since P (x, y) lies on the Unit Circle, we substitute y = 1 4, or x = ± 2 into x2 + y2 = 1 to get x2 = 3 2. Here, x > 0 so x = cos π 2 so sin x, y) x 1 P (x, y) 60◦ y θ = π 6 = 30◦ x 5. Plotting θ = 60◦ in standard position, we ﬁnd it is not a quadrantal angle and set about using a triangle approach. Once again, we get a 30◦ − 60◦ − 90◦ right triangle and, after the usual computations, ﬁnd x = cos (60◦) = 1 2 and y = sin (60◦) = √ 3 2. y 1 P (x, y) θ = 60◦ x 1 P (x, y) 30◦ y θ = 60◦ x 3Again, can you show this? 720 Foundations of Trigonometry In Example 10.2.1, it was quite easy to ﬁnd the cosine and sine of the quadrantal angles, but for In these latter cases, we made good non-quadrantal angles, the task was much more involved. use of the fact that the point P (x, y) = (cos(θ), sin(θ)) lies on the Unit Circle, x2 + y2 = 1. If we substitute x = cos(θ) and y = sin(θ) into x2 + y2 = 1, we get (cos(θ))2 + (sin(θ))2 = 1. An unfortunate4 convention, which the authors are compelled to perpetuate, is to write (cos(θ))2 as cos2(θ) and (sin(θ))2 as sin2(θ). Rewriting the identity using this convention results in the following theorem, which is without a doubt one of the most important results in Trigonometry. Theorem 10.1. The Pythagorean Identity: For any angle θ, cos2(
θ) + sin2(θ) = 1. The moniker ‘Pythagorean’ brings to mind the Pythagorean Theorem, from which both the Distance Formula and the equation for a circle are ultimately derived.5 The word ‘Identity’ reminds us that, regardless of the angle θ, the equation in Theorem 10.1 is always true. If one of cos(θ) or sin(θ) is known, Theorem 10.1 can be used to determine the other, up to a (±) sign. If, in addition, we know where the terminal side of θ lies when in standard position, then we can remove the ambiguity of the (±) and completely determine the missing value as the next example illustrates. Example 10.2.2. Using the given information about θ, ﬁnd the indicated value. 1. If θ is a Quadrant II angle with sin(θ) = 3 5, ﬁnd cos(θ). 2. If π < θ < 3π 2 with cos(θ) = − √ 5 5, ﬁnd sin(θ). 3. If sin(θ) = 1, ﬁnd cos(θ). Solution. 1. When we substitute sin(θ) = 3 5 into The Pythagorean Identity, cos2(θ) + sin2(θ) = 1, we obtain cos2(θ) + 9 5. Since θ is a Quadrant II angle, its terminal side, when plotted in standard position, lies in Quadrant II. Since the x-coordinates are negative in Quadrant II, cos(θ) is too. Hence, cos(θ) = − 4 5. 25 = 1. Solving, we ﬁnd cos(θ) = ± 4 √ √ 5 5 2. Substituting cos(θ) = − 5. Since we 5 are given that π < θ < 3π 2, we know θ is a Quadrant III angle. Hence both its sine and cosine are negative and we conclude sin(θ) = − 2 into cos2(θ) + sin2(θ) = 1 gives sin(θ) = ± 2. When we substitute sin(θ) = 1 into cos2(θ) + sin
2(θ) = 1, we ﬁnd cos(θ) = 0. Another tool which helps immensely in determining cosines and sines of angles is the symmetry inherent in the Unit Circle. Suppose, for instance, we wish to know the cosine and sine of θ = 5π 6. We plot θ in standard position below and, as usual, let P (x, y) denote the point on the terminal side of θ which lies on the Unit Circle. Note that the terminal side of θ lies π 6 radians short of one half revolution. In Example 10.2.1, we determined that cos π 2. This means 6 2 and sin π = 1 = √ 6 3 4This is unfortunate from a ‘function notation’ perspective. See Section 10.6. 5See Sections 1.1 and 7.2 for details. 10.2 The Unit Circle: Cosine and Sine 721 that the point on the terminal side of the angle π. From the ﬁgure below, it is clear that the point P (x, y) we seek can be obtained by reﬂecting that point about the y-axis. Hence, cos 5π 6 6, when plotted in standard position, is 2 and sin 5π = x, y) θ = 5π = 5π is called the reference angle for the angle 5π In the above scenario, the angle π 6. In general, for a non-quadrantal angle θ, the reference angle for θ (usually denoted α) is the acute angle made between the terminal side of θ and the x-axis. If θ is a Quadrant I or IV angle, α is the angle between the terminal side of θ and the positive x-axis; if θ is a Quadrant II or III angle, α is the angle between the terminal side of θ and the negative x-axis. If we let P denote the point (cos(θ), sin(θ)), then P lies on the Unit Circle. Since the Unit Circle possesses symmetry with respect to the x-axis, y-axis and origin, regardless of where the terminal side of θ lies, there is a point Q symmetric with P which determines θ’s reference angle, α as seen below Reference angle α for a Quadrant I angle Reference angle α for a Quadrant II
angle 722 Foundations of Trigonometry Reference angle α for a Quadrant III angle Reference angle α for a Quadrant IV angle We have just outlined the proof of the following theorem. Theorem 10.2. Reference Angle Theorem. Suppose α is the reference angle for θ. Then cos(θ) = ± cos(α) and sin(θ) = ± sin(α), where the choice of the (±) depends on the quadrant in which the terminal side of θ lies. In light of Theorem 10.2, it pays to know the cosine and sine values for certain common angles. In the table below, we summarize the values which we consider essential and must be memorized. Cosine and Sine Values of Common Angles θ(degrees) 0◦ 30◦ 45◦ 60◦ 90◦ θ(radians) cos(θ) sin(θ Example 10.2.3. Find the cosine and sine of the following angles. 1. θ = 225◦ 2. θ = 11π 6 3. θ = − 5π 4 4. θ = 7π 3 Solution. 1. We begin by plotting θ = 225◦ in standard position and ﬁnd its terminal side overshoots the negative x-axis to land in Quadrant III. Hence, we obtain θ’s reference angle α by subtracting: α = θ − 180◦ = 225◦ − 180◦ = 45◦. Since θ is a Quadrant III angle, both cos(θ) < 0 and 10.2 The Unit Circle: Cosine and Sine sin(θ) < 0. The Reference Angle Theorem yields: cos (225◦) = − cos (45◦) = − sin (225◦) = − sin (45◦) = − √ 2 2. 723 √ 2 2 and 2. The terminal side of θ = 11π 6, when plotted in standard position, lies in Quadrant IV, just shy of the positive x-axis. To ﬁnd θ’s reference angle α, we subtract: α = 2π − θ = 2π − 11π 6 = π 6. Since θ is a Quadrant IV angle, cos(θ) > 0 and sin(θ) < 0,
so the Reference Angle Theorem gives: cos 11π 6 2 and sin 11π = − sin π 6 = cos = 225◦ 45 = 11π 6 Finding cos (225◦) and sin (225◦) Finding cos 11π 6 and sin 11π 6 3. To plot θ = − 5π 4, we rotate clockwise an angle of 5π 4 from the positive x-axis. The terminal side of θ, therefore, lies in Quadrant II making an angle of α = 5π 4 − π = π 4 radians with respect to the negative x-axis. Since θ is a Quadrant II angle, the Reference Angle Theorem gives: cos − 5π 4 = − cos π 4 = sin. Since the angle θ = 7π one full revolution followed by an additional α = 7π coterminal, cos 7π 3 2 and sin 7π = cos π 3 = 1 = sin π 3 3 3, we ﬁnd the terminal side of θ by rotating 3 − 2π = π 3 radians. Since θ and α are √ = 3 2. 2 = − 2 and sin − 5π 3 measures more than 2π = 6π y 1 π 4 1 x θ = − 5π 4 y 1 θ = 7π 3 π 3 1 x Finding cos − 5π 4 and sin − 5π 4 Finding cos 7π 3 and sin 7π 3 724 Foundations of Trigonometry 6 as a reference angle, those with a denominator of 4 have π The reader may have noticed that when expressed in radian measure, the reference angle for a non-quadrantal angle is easy to spot. Reduced fraction multiples of π with a denominator of 6 have π 4 as their reference angle, and those with a denominator of 3 have π 3 as their reference angle.6 The Reference Angle Theorem in conjunction with the table of cosine and sine values on Page 722 can be used to generate the following ﬁgure, which the authors feel should be committed to memory. y (0, 1 5π 6 2π 3 3π (−1, 0) 0, 2π (1, 0) x 7π 6 11π 5π 4 4π 3 − 1 2, − √ 3 2 7π 4 5π 3π 2 (0, −1) Important
Points on the Unit Circle 6For once, we have something convenient about using radian measure in contrast to the abstract theoretical nonsense about using them as a ‘natural’ way to match oriented angles with real numbers! 10.2 The Unit Circle: Cosine and Sine 725 The next example summarizes all of the important ideas discussed thus far in the section. Example 10.2.4. Suppose α is an acute angle with cos(α) = 5 13. 1. Find sin(α) and use this to plot α in standard position. 2. Find the sine and cosine of the following angles: (a) θ = π + α (b) θ = 2π − α (c) θ = 3π − α (d) θ = π 2 + α Solution. 1. Proceeding as in Example 10.2.2, we substitute cos(α) = 5 ﬁnd sin(α) = ± 12 Hence, sin(α) = 12 x-axis to the ray which contains the point (cos(α), sin(α)) = 5 13 into cos2(α) + sin2(α) = 1 and 13. Since α is an acute (and therefore Quadrant I) angle, sin(α) is positive. 13. To plot α in standard position, we begin our rotation on the positive. 13, 12 13 y 1 5 13, 12 13 α 1 x Sketching α 2. (a) To ﬁnd the cosine and sine of θ = π + α, we ﬁrst plot θ in standard position. We can imagine the sum of the angles π+α as a sequence of two rotations: a rotation of π radians followed by a rotation of α radians.7 We see that α is the reference angle for θ, so by 13 and sin(θ) = ± sin(α) = ± 12 The Reference Angle Theorem, cos(θ) = ± cos(α) = ± 5 13. Since the terminal side of θ falls in Quadrant III, both cos(θ) and sin(θ) are negative, hence, cos(θ) = − 5 13 and sin(θ) = − 12 13. 7Since π + α = α + π, θ may be plotted by reversing the order of rotations given here. You should do
this. 726 Foundations of Trigonometry Visualizing θ = π + α θ has reference angle α (b) Rewriting θ = 2π − α as θ = 2π + (−α), we can plot θ by visualizing one complete revolution counter-clockwise followed by a clockwise revolution, or ‘backing up,’ of α radians. We see that α is θ’s reference angle, and since θ is a Quadrant IV angle, the Reference Angle Theorem gives: cos(θ) = 5 13 and sin(θ) = − 12 13. y 1 y 1 θ θ 2π −α 1 x 1 x α Visualizing θ = 2π − α θ has reference angle α (c) Taking a cue from the previous problem, we rewrite θ = 3π − α as θ = 3π + (−α). The angle 3π represents one and a half revolutions counter-clockwise, so that when we ‘back up’ α radians, we end up in Quadrant II. Using the Reference Angle Theorem, we get cos(θ) = − 5 13 and sin(θ) = 12 13. 10.2 The Unit Circle: Cosine and Sine 727 y 1 −α 3π y 1 θ α 1 x 1 x Visualizing 3π − α θ has reference angle α (d) To plot θ = π 2 + α, we ﬁrst rotate π 2 radians and follow up with α radians. The reference angle here is not α, so The Reference Angle Theorem is not immediately applicable. (It’s important that you see why this is the case. Take a moment to think about this before reading on.) Let Q(x, y) be the point on the terminal side of θ which lies on the Unit Circle so that x = cos(θ) and y = sin(θ). Once we graph α in standard position, we use the fact that equal angles subtend equal chords to show that the dotted lines in the ﬁgure below are equal. Hence, x = cos(θ) = − 12 13. Similarly, we ﬁnd y = sin(θ) = 5 13. 13, 12 13 Q (x, y) α α 1 x 1 x Visualizing θ = �
� 2 + α Using symmetry to determine Q(x, y) 728 Foundations of Trigonometry Our next example asks us to solve some very basic trigonometric equations.8 Example 10.2.5. Find all of the angles which satisfy the given equation. 1. cos(θ) = 1 2 2. sin(θ) = − 1 2 3. cos(θ) = 0. Solution. Since there is no context in the problem to indicate whether to use degrees or radians, we will default to using radian measure in our answers to each of these problems. This choice will be justiﬁed later in the text when we study what is known as Analytic Trigonometry. In those sections to come, radian measure will be the only appropriate angle measure so it is worth the time to become “ﬂuent in radians” now. 1. If cos(θ) = 1 Unit Circle at x = 1 2, then the terminal side of θ, when plotted in standard position, intersects the 2. This means θ is a Quadrant I or IV angle with reference angle One solution in Quadrant I is θ = π coterminal with π IV case, we ﬁnd the solution to cos(θ) = 1 θ = 5π 3, we ﬁnd θ = π 3 + 2πk for integers k. 3, and since all other Quadrant I solutions must be 3 + 2πk for integers k.9 Proceeding similarly for the Quadrant 3, so our answer in this Quadrant is 2 here is 5π 2. If sin(θ) = − 1 2, then when θ is plotted in standard position, its terminal side intersects the Unit Circle at y = − 1 2. From this, we determine θ is a Quadrant III or Quadrant IV angle with reference angle π 6. 8We will study trigonometric equations more formally in Section 10.7. Enjoy these relatively straightforward exercises while they last! 9Recall in Section 10.1, two angles in radian measure are coterminal if and only if they diﬀer by an integer multiple of 2π. Hence to describe all angles coterminal with a given angle, we add 2πk for integers k = 0, ±1, ±2,.... 10.2
The Unit Circle: Cosine and Sine 729 In Quadrant III, one solution is 7π multiples of 2π: θ = 7π are of the form θ = 11π 6, so we capture all Quadrant III solutions by adding integer so all the solutions here 6 + 2πk. In Quadrant IV, one solution is 11π 6 + 2πk for integers k. 6 3. The angles with cos(θ) = 0 are quadrantal angles whose terminal sides, when plotted in standard position, lie along the y-axis While, technically speaking, π answers. If we follow the procedure set forth in the previous examples, we ﬁnd θ = π and θ = 3π θ = π 2 isn’t a reference angle we can nonetheless use it to ﬁnd our 2 + 2πk 2 + 2πk for integers, k. While this solution is correct, it can be shortened to 2 + πk for integers k. (Can you see why this works from the diagram?) One of the key items to take from Example 10.2.5 is that, in general, solutions to trigonometric equations consist of inﬁnitely many answers. To get a feel for these answers, the reader is encouraged to follow our mantra from Chapter 9 - that is, ‘When in doubt, write it out!’ This is especially important when checking answers to the exercises. For example, another Quadrant IV solution to 2 is θ = − π sin(θ) = − 1 6. Hence, the family of Quadrant IV answers to number 2 above could just have easily been written θ = − π 6 + 2πk for integers k. While on the surface, this family may look 730 Foundations of Trigonometry diﬀerent than the stated solution of θ = 11π they represent the same list of angles. 6 + 2πk for integers k, we leave it to the reader to show 10.2.1 Beyond the Unit Circle We began the section with a quest to describe the position of a particle experiencing circular motion. In deﬁning the cosine and sine functions, we assigned to each angle a position on the Unit Circle. In this subsection, we broaden our scope to include circles of radius r centered at the origin. Consider for the moment the acute angle θ drawn below in standard
position. Let Q(x, y) be the point on the terminal side of θ which lies on the circle x2 + y2 = r2, and let P (x, y) be the point on the terminal side of θ which lies on the Unit Circle. Now consider dropping perpendiculars from P and Q to create two right triangles, ∆OP A and ∆OQB. These triangles are similar,10 thus it follows that x 1 = r, so x = rx and, similarly, we ﬁnd y = ry. Since, by deﬁnition, x = cos(θ) and y = sin(θ), we get the coordinates of Q to be x = r cos(θ) and y = r sin(θ). By reﬂecting these points through the x-axis, y-axis and origin, we obtain the result for all non-quadrantal angles θ, and we leave it to the reader to verify these formulas hold for the quadrantal angles. x = r y r 1 Q (x, y) P (x, y) θ y 1 1 x r P (x, y) Q(x, y) = (r cos(θ), r sin(θ)) θ O A(x, 0) B(x, 0) x Not only can we describe the coordinates of Q in terms of cos(θ) and sin(θ) but since the radius of x2 + y2, we can also express cos(θ) and sin(θ) in terms of the coordinates of Q. the circle is r = These results are summarized in the following theorem. Theorem 10.3. If Q(x, y) is the point on the terminal side of an angle θ, plotted in standard position, which lies on the circle x2 + y2 = r2 then x = r cos(θ) and y = r sin(θ). Moreover, cos(θ) = x r = x x2 + y2 and sin(θ) = y r = y x2 + y2 10Do you remember why? 10.2 The Unit Circle: Cosine and Sine 731 Note that in the case of the Unit Circle we have r = our deﬁnitions of cos(θ) and sin(θ). x2 + y
2 = 1, so Theorem 10.3 reduces to Example 10.2.6. 1. Suppose that the terminal side of an angle θ, when plotted in standard position, contains the point Q(4, −2). Find sin(θ) and cos(θ). 2. In Example 10.1.5 in Section 10.1, we approximated the radius of the earth at 41.628◦ north latitude to be 2960 miles. Justify this approximation if the radius of the Earth at the Equator is approximately 3960 miles. Solution. 1. Using Theorem 10.3 with x = 4 and y = −2, we ﬁnd r = √ √ 5 5 5. 5 and sin(θ) = y that cos(θ) = x r = −4)2 + (−2)2 = √ 20 = 2 √ 5 so 2. Assuming the Earth is a sphere, a cross-section through the poles produces a circle of radius 3960 miles. Viewing the Equator as the x-axis, the value we seek is the x-coordinate of the point Q(x, y) indicated in the ﬁgure below. y 4 2 −4 −2 2 4 x Q(4, −2) −2 −4 y 3960 Q (x, y) 41.628◦ 3960 x The terminal side of θ contains Q(4, −2) A point on the Earth at 41.628◦N Using Theorem 10.3, we get x = 3960 cos (41.628◦). Using a calculator in ‘degree’ mode, we ﬁnd 3960 cos (41.628◦) ≈ 2960. Hence, the radius of the Earth at North Latitude 41.628◦ is approximately 2960 miles. 732 Foundations of Trigonometry Theorem 10.3 gives us what we need to describe the position of an object traveling in a circular path of radius r with constant angular velocity ω. Suppose that at time t, the object has swept out an angle measuring θ radians. If we assume that the object is at the point (r, 0) when t = 0, the angle θ is in standard position. By deﬁnition, ω = θ t which we rewrite as θ = ωt. According to Theorem
10.3, the location of the object Q(x, y) on the circle is found using the equations x = r cos(θ) = r cos(ωt) and y = r sin(θ) = r sin(ωt). Hence, at time t, the object is at the point (r cos(ωt), r sin(ωt)). We have just argued the following. Equation 10.3. Suppose an object is traveling in a circular path of radius r centered at the origin with constant angular velocity ω. If t = 0 corresponds to the point (r, 0), then the x and y coordinates of the object are functions of t and are given by x = r cos(ωt) and y = r sin(ωt). Here, ω > 0 indicates a counter-clockwise direction and ω < 0 indicates a clockwise direction. y r 1 Q (x, y) = (r cos(ωt), r sin(ωt)) θ = ωt 1 x r Equations for Circular Motion Example 10.2.7. Suppose we are in the situation of Example 10.1.5. Find the equations of motion of Lakeland Community College as the earth rotates. Solution. From Example 10.1.5, we take r = 2960 miles and and ω = π of motion are x = r cos(ωt) = 2960 cos π measured in miles and t is measured in hours. 12 hours. Hence, the equations 12 t, where x and y are 12 t and y = r sin(ωt) = 2960 sin π In addition to circular motion, Theorem 10.3 is also the key to developing what is usually called ‘right triangle’ trigonometry.11 As we shall see in the sections to come, many applications in trigonometry involve ﬁnding the measures of the angles in, and lengths of the sides of, right triangles. Indeed, we made good use of some properties of right triangles to ﬁnd the exact values of the cosine and sine of many of the angles in Example 10.2.1, so the following development shouldn’t be that much of a surprise. Consider the generic right triangle below with corresponding acute angle θ. The side with length a is called the side of the triangle adjacent to θ; the side with length b is called the side of the
triangle opposite θ; and the remaining side of length c (the side opposite the 11You may have been exposed to this in High School. 10.2 The Unit Circle: Cosine and Sine 733 right angle) is called the hypotenuse. We now imagine drawing this triangle in Quadrant I so that the angle θ is in standard position with the adjacent side to θ lying along the positive x-axis. y c θ P (a, b) x c b c a θ According to the Pythagorean Theorem, a2 + b2 = c2, so that the point P (a, b) lies on a circle of radius c. Theorem 10.3 tells us that cos(θ) = a c, so we have determined the cosine and sine of θ in terms of the lengths of the sides of the right triangle. Thus we have the following theorem. c and sin(θ) = b Theorem 10.4. Suppose θ is an acute angle residing in a right triangle. If the length of the side adjacent to θ is a, the length of the side opposite θ is b, and the length of the hypotenuse is c, then cos(θ) = and sin(θ) = a c b c. Example 10.2.8. Find the measure of the missing angle and the lengths of the missing sides of: 30◦ 7 Solution. The ﬁrst and easiest task is to ﬁnd the measure of the missing angle. Since the sum of angles of a triangle is 180◦, we know that the missing angle has measure 180◦ − 30◦ − 90◦ = 60◦. We now proceed to ﬁnd the lengths of the remaining two sides of the triangle. Let c denote the 7 length of the hypotenuse of the triangle. By Theorem 10.4, we have cos (30◦) = 7 cos(30◦). 2, we have, after the usual fraction gymnastics, c = 14 Since cos (30◦) =. At this point, we have two ways to proceed to ﬁnd the length of the side opposite the 30◦ angle, which we’ll denote b. We know the length of the adjacent side is 7 and the length of the hypotenuse is 14, so we 3 c, or c = √ √
√ 3 3 3 3 734 Foundations of Trigonometry could use the Pythagorean Theorem to ﬁnd the missing side and solve (7)2 + b2 = Alternatively, we could use Theorem 10.4, namely that sin (30◦) = b b = c sin (30◦) = 14 3 √ 3 3. The triangle with all of its data is recorded below. 2 = 7 · 1 √ 3 c. Choosing the latter, we ﬁnd 2 √ 14 3 3 for b. √ 3 c = 14 3 60◦ b = 7 √ 3 3 30◦ 7 We close this section by noting that we can easily extend the functions cosine and sine to real numbers by identifying a real number t with the angle θ = t radians. Using this identiﬁcation, we deﬁne cos(t) = cos(θ) and sin(t) = sin(θ). In practice this means expressions like cos(π) and sin(2) can be found by regarding the inputs as angles in radian measure or real numbers; the choice is the reader’s. If we trace the identiﬁcation of real numbers t with angles θ in radian measure to its roots on page 704, we can spell out this correspondence more precisely. For each real number t, we associate an oriented arc t units in length with initial point (1, 0) and endpoint P (cos(t), sin(t)). cos(t), sin(t)) θ = t 1 x In the same way we studied polynomial, rational, exponential, and logarithmic functions, we will study the trigonometric functions f (t) = cos(t) and g(t) = sin(t). The ﬁrst order of business is to ﬁnd the domains and ranges of these functions. Whether we think of identifying the real number t with the angle θ = t radians, or think of wrapping an oriented arc around the Unit Circle to ﬁnd coordinates on the Unit Circle, it should be clear that both the cosine and sine functions are deﬁned for all real numbers t. In other words, the domain of f (t) = cos(t) and of g(t) = sin(t) is (−∞, �
�). Since cos(t) and sin(t) represent x- and y-coordinates, respectively, of points on the Unit Circle, they both take on all of the values between −1 an 1, inclusive. In other words, the range of f (t) = cos(t) and of g(t) = sin(t) is the interval [−1, 1]. To summarize: 10.2 The Unit Circle: Cosine and Sine 735 Theorem 10.5. Domain and Range of the Cosine and Sine Functions: • The function f (t) = cos(t) • The function g(t) = sin(t) – has domain (−∞, ∞) – has range [−1, 1] – has domain (−∞, ∞) – has range [−1, 1] Suppose, as in the Exercises, we are asked to solve an equation such as sin(t) = − 1 2. As we have already mentioned, the distinction between t as a real number and as an angle θ = t radians is often blurred. Indeed, we solve sin(t) = − 1 2 in the exact same manner12 as we did in Example 10.2.5 number 2. Our solution is only cosmetically diﬀerent in that the variable used is t rather than θ: t = 7π 6 + 2πk for integers, k. We will study the cosine and sine functions in greater detail in Section 10.5. Until then, keep in mind that any properties of cosine and sine developed in the following sections which regard them as functions of angles in radian measure apply equally well if the inputs are regarded as real numbers. 6 + 2πk or t = 11π 12Well, to be pedantic, we would be technically using ‘reference numbers’ or ‘reference arcs’ instead of ‘reference angles’ – but the idea is the same. 736 Foundations of Trigonometry 10.2.2 Exercises In Exercises 1 - 20, ﬁnd the exact value of the cosine and sine of the given angle. 1. θ = 0 5. θ = 9. θ = 13. θ = 2π 3 5π 4 7π 4 17. θ = − 3π 4 2. θ = 6. θ = 10.
θ = 14. θ = π 4 3π 4 4π 3 23π 6 18. θ = − π 6 3. θ = π 3 7. θ = π 11. θ = 3π 2 15. θ = − 13π 2 19. θ = 10π 3 4. θ = 8. θ = 12. θ = π 2 7π 6 5π 3 16. θ = − 43π 6 20. θ = 117π In Exercises 21 - 30, use the results developed throughout the section to ﬁnd the requested value. 21. If sin(θ) = − 7 25 with θ in Quadrant IV, what is cos(θ)? 22. If cos(θ) = 23. If sin(θ) = 4 9 5 13 with θ in Quadrant I, what is sin(θ)? with θ in Quadrant II, what is cos(θ)? 24. If cos(θ) = − 2 11 with θ in Quadrant III, what is sin(θ)? 25. If sin(θ) = − 2 3 with θ in Quadrant III, what is cos(θ)? 26. If cos(θ) = 27. If sin(θ) = 28. If cos(θ) = 28 53 √ 2 5 √ 10 10 with θ in Quadrant IV, what is sin(θ)? 5 and π 2 < θ < π, what is cos(θ)? and 2π < θ < 5π 2 3π 2, what is sin(θ)?, what is cos(θ)? 29. If sin(θ) = −0.42 and π < θ < 30. If cos(θ) = −0.98 and π 2 < θ < π, what is sin(θ)? 10.2 The Unit Circle: Cosine and Sine 737 In Exercises 31 - 39, ﬁnd all of the angles which satisfy the given equation. 31. sin(θ) = 34. cos(θ) = 1 2 √ 2 2 32. cos(θ) = − √ 3 2 35. sin(θ) = √ 3 2 √ 3 2 33. sin(θ) =
0 36. cos(θ) = −1 39. cos(θ) = −1.001 37. sin(θ) = −1 38. cos(θ) = In Exercises 40 - 48, solve the equation for t. (See the comments following Theorem 10.5.) 40. cos(t) = 0 41. sin(t) = − 43. sin(t) = − 1 2 44. cos(t) = 1 2 √ 2 2 42. cos(t) = 3 45. sin(t) = −2 46. cos(t) = 1 47. sin(t) = 1 48. cos(t) = − √ 2 2 In Exercises 49 - 54, use your calculator to approximate the given value to three decimal places. Make sure your calculator is in the proper angle measurement mode! 49. sin(78.95◦) 50. cos(−2.01) 51. sin(392.994) 52. cos(207◦) 53. sin (π◦) 54. cos(e) In Exercises 55 - 58, ﬁnd the measurement of the missing angle and the lengths of the missing sides. (See Example 10.2.8) 55. Find θ, b, and c. 56. Find θ, a, and c. c θ b 30◦ 1 45◦ c a θ 3 738 Foundations of Trigonometry 57. Find α, a, and b. 58. Find β, a, and c. b α 8 a 33◦ a 48◦ 6 c β In Exercises 59 - 64, assume that θ is an acute angle in a right triangle and use Theorem 10.4 to ﬁnd the requested side. 59. If θ = 12◦ and the side adjacent to θ has length 4, how long is the hypotenuse? 60. If θ = 78.123◦ and the hypotenuse has length 5280, how long is the side adjacent to θ? 61. If θ = 59◦ and the side opposite θ has length 117.42, how long is the hypotenuse? 62. If θ = 5◦ and the hypotenuse has length 10, how long is the side opposite θ? 63. If θ = 5◦ and the hypot
enuse has length 10, how long is the side adjacent to θ? 64. If θ = 37.5◦ and the side opposite θ has length 306, how long is the side adjacent to θ? In Exercises 65 - 68, let θ be the angle in standard position whose terminal side contains the given point then compute cos(θ) and sin(θ). 65. P (−7, 24) 66. Q(3, 4) 67. R(5, −9) 68. T (−2, −11) In Exercises 69 - 72, ﬁnd the equations of motion for the given scenario. Assume that the center of the motion is the origin, the motion is counter-clockwise and that t = 0 corresponds to a position along the positive x-axis. (See Equation 10.3 and Example 10.1.5.) 69. A point on the edge of the spinning yo-yo in Exercise 50 from Section 10.1. Recall: The diameter of the yo-yo is 2.25 inches and it spins at 4500 revolutions per minute. 70. The yo-yo in exercise 52 from Section 10.1. Recall: The radius of the circle is 28 inches and it completes one revolution in 3 seconds. 71. A point on the edge of the hard drive in Exercise 53 from Section 10.1. Recall: The diameter of the hard disk is 2.5 inches and it spins at 7200 revolutions per minute. 10.2 The Unit Circle: Cosine and Sine 739 72. A passenger on the Big Wheel in Exercise 55 from Section 10.1. Recall: The diameter is 128 feet and completes 2 revolutions in 2 minutes, 7 seconds. 73. Consider the numbers: 0, 1, 2, 3, 4. Take the square root of each of these numbers, then divide each by 2. The resulting numbers should look hauntingly familiar. (See the values in the table on 722.) 74. Let α and β be the two acute angles of a right triangle. (Thus α and β are complementary angles.) Show that sin(α) = cos(β) and sin(β) = cos(α). The fact that co-functions of complementary angles are equal in this case is not an accident and a more general result will be given in Section 10.4. 75. In the scenario of Equation 10.3, we assumed that at t = 0
, the object was at the point (r, 0). If this is not the case, we can adjust the equations of motion by introducing a ‘time delay.’ If t0 > 0 is the ﬁrst time the object passes through the point (r, 0), show, with the help of your classmates, the equations of motion are x = r cos(ω(t − t0)) and y = r sin(ω(t − t0)). 740 Foundations of Trigonometry 10.2.3 Answers 1. cos(0) = 1, sin(0) = 0 1 2 = 3. cos 5. cos π 3 2π 3, sin π 3 = = − 1 2, sin 2π 3 √ 3 2 √ 3 2 = 7. cos(π) = −1, sin(π) = 0 9. cos 11. cos 13. cos 5π 4 3π 2 7π 4 √ 2 2 = −, sin 5π 4 = − √ 2 2 = 0, sin 3π 2 = −1 √ 2 2 =, sin 7π 4 = − √ 2 2 2. cos 4. cos 6. cos 8. cos 10. cos 12. cos π 4 π 2 3π 4 7π 6 4π 3 5π 3 √ 2 2 =, sin π 4 = √ 2 2 = 0, sin, sin, sin 3π 4 7π, sin 4π 3 = − √ 3 2 = 1 2, sin 5π 3 = − √ 3 2 14. cos 23π 6 √ 3 2, sin 23π 6 = − 1 2 = 15. cos − 13π 2 = 0, sin − 13π 2 = −1 16. cos − 43π 6 √ 3 2, sin − 43π 6 = 1 2 = − 17. cos 19. cos − 3π 4 10π 3 = − √ 2 2, sin − 3π 4 √ 2 2 = − 18. cos − π 6 = √ 3 2, sin −, sin 10π 3 = − √ 3 2 20. cos(117π) = −1, sin(117π) = 0 21. If sin(θ) = − 7 25 22. If cos(θ) = 23. If sin(θ) = 4 9 5 13 24. If cos(θ) =
− 2 11 25. If sin(θ) = − 2 3 26. If cos(θ) = 28 53 with θ in Quadrant IV, then cos(θ) = with θ in Quadrant I, then sin(θ) = with θ in Quadrant II, then cos(θ) = − 12 13. with θ in Quadrant III, then sin(θ) = − 24 25.. √ 65 9 √ 117 11. √ 5 3. with θ in Quadrant III, then cos(θ) = − with θ in Quadrant IV, then sin(θ) = − 45 53. 10.2 The Unit Circle: Cosine and Sine 741 27. If sin(θ) = 28. If cos(θ) = √ 2 5 5 √ 10 10 and π 2 < θ < π, then cos(θ) = − and 2π < θ <, then sin(θ) = √. 5 5 √ 3 10. 10 √ 5π 2 3π 2 29. If sin(θ) = −0.42 and π < θ <, then cos(θ) = − 0.8236 ≈ −0.9075. 30. If cos(θ) = −0.98 and < θ < π, then sin(θ) = √ 0.0396 ≈ 0.1990. π 2 31. sin(θ) = 1 2 32. cos(θ) = − √ 3 2 when θ = π 6 + 2πk or θ = 5π 6 + 2πk for any integer k. when θ = 5π 6 + 2πk or θ = 7π 6 + 2πk for any integer k. 33. sin(θ) = 0 when θ = πk for any integer k. 34. cos(θ) = 35. sin(θ) = √ 2 2 √ 3 2 when θ = when θ = π 4 π 3 + 2πk or θ = + 2πk or θ = 7π 4 2π 3 + 2πk for any integer k. + 2πk for any integer k. 36. cos(θ) = −1 when θ
= (2k + 1)π for any integer k. 37. sin(θ) = −1 when θ = 38. cos(θ) = √ 3 2 when θ = 3π 2 π 6 + 2πk for any integer k. + 2πk or θ = 11π 6 + 2πk for any integer k. 39. cos(θ) = −1.001 never happens 40. cos(t) = 0 when t = π 2 + πk for any integer k. 41. sin(t) = − √ 2 2 when t = 5π 4 + 2πk or t = 7π 4 + 2πk for any integer k. 42. cos(t) = 3 never happens. 43. sin(t) = − 1 2 when t = 7π 6 + 2πk or t = 11π 6 + 2πk for any integer k. 44. cos(t) = 1 2 when t = π 3 + 2πk or t = 5π 3 + 2πk for any integer k. 45. sin(t) = −2 never happens 46. cos(t) = 1 when t = 2πk for any integer k. 742 Foundations of Trigonometry π 2 + 2πk for any integer k. 47. sin(t) = 1 when t = 48. cos(t) = − √ 2 2 when t = 3π 4 + 2πk or t = 5π 4 + 2πk for any integer k. 49. sin(78.95◦) ≈ 0.981 50. cos(−2.01) ≈ −0.425 51. sin(392.994) ≈ −0.291 52. cos(207◦) ≈ −0.891 53. sin (π◦) ≈ 0.055 54. cos(e) ≈ −0.912 55. θ = 60◦, b = √ 3 3, c = 56. θ = 45◦, a = 3 57. α = 57◦, a = 8 cos(33◦) ≈ 6.709, b = 8 sin(33◦) ≈ 4.357 58. β = 42◦, c = 6 sin(48◦) ≈ 8.074, a = √
c2 − 62 ≈ 5.402 59. The hypotenuse has length 4 cos(12◦) ≈ 4.089. 60. The side adjacent to θ has length 5280 cos(78.123◦) ≈ 1086.68. 61. The hypotenuse has length 117.42 sin(59◦) ≈ 136.99. 62. The side opposite θ has length 10 sin(5◦) ≈ 0.872. 63. The side adjacent to θ has length 10 cos(5◦) ≈ 9.962. 64. The hypotenuse has length c = √ c2 − 3062 ≈ 398.797. 306 sin(37.5◦) 65. cos(θ) = − 7 25, sin(θ) = 24 25 ≈ 502.660, so the side adjacent to θ has length, sin(θ) = 4 5 66. cos(θ) = 67. cos(θ) = 3 5 √ 5, sin(θ) = − 68. cos(θ) = −, sin(θ) = − 106 106 √ 5 2 25 √ 106 9 106 √ 5 11 25 69. r = 1.125 inches, ω = 9000π radians minute, x = 1.125 cos(9000π t), y = 1.125 sin(9000π t). Here x and y are measured in inches and t is measured in minutes. 10.2 The Unit Circle: Cosine and Sine 743 70. r = 28 inches, ω = 2π 3 radians second, x = 28 cos 2π 3 t, y = 28 sin 2π 3 t. Here x and y are measured in inches and t is measured in seconds. 71. r = 1.25 inches, ω = 14400π radians minute, x = 1.25 cos(14400π t), y = 1.25 sin(14400π t). Here x and y are measured in inches and t is measured in minutes. 127 t, y = 64 sin 4π second, x = 64 cos 4π 72. r = 64 feet, ω = 4π 127 in feet and t is measured in seconds radians 127 t. Here x and y are measured 744 Foundations of Trigonometry 10.3 The Six Circular Functions and
Fundamental Identities In section 10.2, we deﬁned cos(θ) and sin(θ) for angles θ using the coordinate values of points on the Unit Circle. As such, these functions earn the moniker circular functions.1 It turns out that cosine and sine are just two of the six commonly used circular functions which we deﬁne below. Deﬁnition 10.2. The Circular Functions: Suppose θ is an angle plotted in standard position and P (x, y) is the point on the terminal side of θ which lies on the Unit Circle. The cosine of θ, denoted cos(θ), is deﬁned by cos(θ) = x. The sine of θ, denoted sin(θ), is deﬁned by sin(θ) = y. The secant of θ, denoted sec(θ), is deﬁned by sec(θ) = 1 x, provided x = 0. The cosecant of θ, denoted csc(θ), is deﬁned by csc(θ) = The tangent of θ, denoted tan(θ), is deﬁned by tan(θ) = 1 y y x The cotangent of θ, denoted cot(θ), is deﬁned by cot(θ) =, provided y = 0., provided x = 0. x y, provided y = 0. While we left the history of the name ‘sine’ as an interesting research project in Section 10.2, the names ‘tangent’ and ‘secant’ can be explained using the diagram below. Consider the acute angle θ below in standard position. Let P (x, y) denote, as usual, the point on the terminal side of θ which lies on the Unit Circle and let Q(1, y) denote the point on the terminal side of θ which lies on the vertical line x = 1. Q(1, y) = (1, tan(θ)) P (x, y) y 1 θ O A(x, 0) B(1, 0) x 1In Theorem 10.4 we also showed cosine and sine to be functions of an angle
residing in a right triangle so we could just as easily call them trigonometric functions. In later sections, you will ﬁnd that we do indeed use the phrase ‘trigonometric function’ interchangeably with the term ‘circular function’. 10.3 The Six Circular Functions and Fundamental Identities 745 y = 1 x which gives y = y The word ‘tangent’ comes from the Latin meaning ‘to touch,’ and for this reason, the line x = 1 is called a tangent line to the Unit Circle since it intersects, or ‘touches’, the circle at only one point, namely (1, 0). Dropping perpendiculars from P and Q creates a pair of similar triangles ∆OP A and ∆OQB. Thus y x = tan(θ), where this last equality comes from applying Deﬁnition 10.2. We have just shown that for acute angles θ, tan(θ) is the y-coordinate of the point on the terminal side of θ which lies on the line x = 1 which is tangent to the Unit Circle. Now the word ‘secant’ means ‘to cut’, so a secant line is any line that ‘cuts through’ a circle at two points.2 The line containing the terminal side of θ is a secant line since it intersects the Unit Circle in Quadrants I and III. With the point P lying on the Unit Circle, the length of the hypotenuse of ∆OP A is 1. If we let h denote the length of the hypotenuse of ∆OQB, we have from similar triangles that h x = sec(θ). Hence for an acute angle θ, sec(θ) is the length of the line segment which lies on the secant line determined by the terminal side of θ and ‘cuts oﬀ’ the tangent line x = 1. Not only do these observations help explain the names of these functions, they serve as the basis for a fundamental inequality needed for Calculus which we’ll explore in the Exercises. x, or h = 1 1 = 1 Of the six circular functions, only cosine and sine are deﬁned for all angles. Since cos(θ) = x and sin(θ) = y in De�
��nition 10.2, it is customary to rephrase the remaining four circular functions in terms of cosine and sine. The following theorem is a result of simply replacing x with cos(θ) and y with sin(θ) in Deﬁnition 10.2. Theorem 10.6. Reciprocal and Quotient Identities: sec(θ) = 1 cos(θ), provided cos(θ) = 0; if cos(θ) = 0, sec(θ) is undeﬁned. csc(θ) = 1 sin(θ), provided sin(θ) = 0; if sin(θ) = 0, csc(θ) is undeﬁned. tan(θ) = sin(θ) cos(θ), provided cos(θ) = 0; if cos(θ) = 0, tan(θ) is undeﬁned. cot(θ) = cos(θ) sin(θ), provided sin(θ) = 0; if sin(θ) = 0, cot(θ) is undeﬁned. It is high time for an example. Example 10.3.1. Find the indicated value, if it exists. 1. sec (60◦) 2. csc 7π 4 3. cot(3) 4. tan (θ), where θ is any angle coterminal with 3π 2. 5. cos (θ), where csc(θ) = − √ 5 and θ is a Quadrant IV angle. 6. sin (θ), where tan(θ) = 3 and π < θ < 3π 2. 2Compare this with the deﬁnition given in Section 2.1. 746 Solution. 1. According to Theorem 10.6, sec (60◦) = Foundations of Trigonometry 1 cos(60◦). Hence, sec (60◦) = 1 (1/2) = 2. 2. Since sin 7π 4 = − √ 2 2, csc 7π 4 = 1 sin( 7π 4 ) = 1 √ − 2/2 = − 2√ 2 √ = − 2. 3. Since θ =
3 radians is not one of the ‘common angles’ from Section 10.2, we resort to the calculator for a decimal approximation. Ensuring that the calculator is in radian mode, we ﬁnd cot(3) = cos(3) sin(3) ≈ −7.015. = 0 and sin(θ) = sin 3π 2 = −1. Attempting 4. If θ is coterminal with 3π to compute tan(θ) = sin(θ) 2, then cos(θ) = cos 3π cos(θ) results in −1 √ 2 0, so tan(θ) is undeﬁned. 5. We are given that csc(θ) = 1 √ 5 5 so sin(θ) = − 1√ 5. As we saw in Section 10.2, 5 we can use the Pythagorean Identity, cos2(θ) + sin2(θ) = 1, to ﬁnd cos(θ) by knowing sin(θ). Substituting, we get cos2(θ) + θ is a Quadrant IV angle, cos(θ) > 0, so cos(θ) = 2 = 1, which gives cos2(θ) = 4 √ 5 5. 5, or cos(θ) = ± 2 √ 5 5. Since sin(θ. If tan(θ) = 3, then sin(θ) cos(θ) = 3. Be careful - this does NOT mean we can take sin(θ) = 3 and cos(θ) = 1. Instead, from sin(θ) cos(θ) = 3 we get: sin(θ) = 3 cos(θ). To relate cos(θ) and sin(θ), we once again employ the Pythagorean Identity, cos2(θ) + sin2(θ) = 1. Solving sin(θ) = 3 cos(θ) for cos(θ), we ﬁnd cos(θ) = 1 3 sin(θ). Substituting this into the Pythagorean Identity, we ﬁnd sin2(θ) + 1 3 sin(θ)2 10 so sin(θ) = ± 3 10.
Since π < θ < 3π 2, θ is a Quadrant III angle. This means sin(θ) < 0, so our ﬁnal answer is sin(θ) = − 3 = 1. Solving, we get sin2(θ) = 9 10 √ √ 10 10. While the Reciprocal and Quotient Identities presented in Theorem 10.6 allow us to always reduce problems involving secant, cosecant, tangent and cotangent to problems involving cosine and sine, it is not always convenient to do so.3 It is worth taking the time to memorize the tangent and cotangent values of the common angles summarized below. 3As we shall see shortly, when solving equations involving secant and cosecant, we usually convert back to cosines and sines. However, when solving for tangent or cotangent, we usually stick with what we’re dealt. 10.3 The Six Circular Functions and Fundamental Identities 747 Tangent and Cotangent Values of Common Angles θ(degrees) 0◦ 30◦ 45◦ 60◦ 90◦ θ(radians) tan(θ) cot(θ undeﬁned undeﬁned √ 3 1 √ 3 3 0 Coupling Theorem 10.6 with the Reference Angle Theorem, Theorem 10.2, we get the following. Theorem 10.7. Generalized Reference Angle Theorem. The values of the circular functions of an angle, if they exist, are the same, up to a sign, of the corresponding circular functions of its reference angle. More speciﬁcally, if α is the reference angle for θ, then: cos(θ) = ± cos(α), sin(θ) = ± sin(α), sec(θ) = ± sec(α), csc(θ) = ± csc(α), tan(θ) = ± tan(α) and cot(θ) = ± cot(α). The choice of the (±) depends on the quadrant in which the terminal side of θ lies. We put Theorem 10.7 to good use in the following example. Example 10.3.2. Find all angles which satisfy the given equation. 1. sec(θ
) = 2 Solution. 2. tan(θ) = √ 3 3. cot(θ) = −1. 1. To solve sec(θ) = 2, we convert to cosines and get 1 cos(θ) = 2 or cos(θ) = 1 same equation we solved in Example 10.2.5, number 1, so we know the answer is: θ = π or θ = 5π 3 + 2πk for integers k. 2. This is the exact 3 + 2πk √ √ = 2. From the table of common values, we see tan π 3 3 must, therefore, have a reference angle of π 3. According to Theorem 10.7, we know 3. Our next task is the solutions to tan(θ) = to determine in which quadrants the solutions to this equation lie. Since tangent is deﬁned as the ratio y x of points (x, y) on the Unit Circle with x = 0, tangent is positive when x and y have the same sign (i.e., when they are both positive or both negative.) This happens in Quadrants I and III. In Quadrant I, we get the solutions: θ = π 3 + 2πk for integers k, and for Quadrant III, we get θ = 4π 3 + 2πk for integers k. While these descriptions of the solutions are correct, they can be combined into one list as θ = π 3 + πk for integers k. The latter form of the solution is best understood looking at the geometry of the situation in the diagram below.4 4See Example 10.2.5 number 3 in Section 10.2 for another example of this kind of simpliﬁcation of the solution. 748 Foundations of Trigonometry. From the table of common values, we see that π 4 has a cotangent of 1, which means the solutions to cot(θ) = −1 have a reference angle of π 4. To ﬁnd the quadrants in which our solutions lie, we note that cot(θ) = x y for a point (x, y) on the Unit Circle where y = 0. If cot(θ) is negative, then x and y must have diﬀerent signs (i.e., one positive and
one negative.) Hence, our solutions lie in Quadrants II and IV. Our Quadrant II solution is θ = 3π 4 + 2πk, and for Quadrant IV, we get θ = 7π 4 +2πk for integers k. Can these lists be combined? Indeed they can - one such way to capture all the solutions is: θ = 3π 4 + πk for integers k We have already seen the importance of identities in trigonometry. Our next task is to use use the Reciprocal and Quotient Identities found in Theorem 10.6 coupled with the Pythagorean Identity found in Theorem 10.1 to derive new Pythagorean-like identities for the remaining four circular functions. Assuming cos(θ) = 0, we may start with cos2(θ) + sin2(θ) = 1 and divide both sides by cos2(θ) to obtain 1 + sin2(θ) cos2(θ). Using properties of exponents along with the Reciprocal and Quotient Identities, this reduces to 1 + tan2(θ) = sec2(θ). If sin(θ) = 0, we can divide both sides of the identity cos2(θ) + sin2(θ) = 1 by sin2(θ), apply Theorem 10.6 once again, and obtain cot2(θ) + 1 = csc2(θ). These three Pythagorean Identities are worth memorizing and they, along with some of their other common forms, are summarized in the following theorem. cos2(θ) = 1 10.3 The Six Circular Functions and Fundamental Identities 749 Theorem 10.8. The Pythagorean Identities: 1. cos2(θ) + sin2(θ) = 1. Common Alternate Forms: 1 − sin2(θ) = cos2(θ) 1 − cos2(θ) = sin2(θ) 2. 1 + tan2(θ) = sec2(θ), provided cos(θ) = 0. Common Alternate Forms: sec2(θ) − tan2(θ) = 1 sec2(θ) − 1 = tan2(θ) 3. 1 + cot2(θ) = csc2(θ), provided sin
(θ) = 0. Common Alternate Forms: csc2(θ) − cot2(θ) = 1 csc2(θ) − 1 = cot2(θ) Trigonometric identities play an important role in not just Trigonometry, but in Calculus as well. We’ll use them in this book to ﬁnd the values of the circular functions of an angle and solve equations and inequalities. In Calculus, they are needed to simplify otherwise complicated expressions. In the next example, we make good use of the Theorems 10.6 and 10.8. Example 10.3.3. Verify the following identities. Assume that all quantities are deﬁned. 1. 1 csc(θ) = sin(θ) 3. (sec(θ) − tan(θ))(sec(θ) + tan(θ)) = 1 5. 6 sec(θ) tan(θ) = 3 1 − sin(θ) − 3 1 + sin(θ) 2. tan(θ) = sin(θ) sec(θ) 4. 6. sec(θ) 1 − tan(θ) = 1 cos(θ) − sin(θ) sin(θ) 1 − cos(θ) = 1 + cos(θ) sin(θ) Solution. In verifying identities, we typically start with the more complicated side of the equation and use known identities to transform it into the other side of the equation. 1. To verify 1 csc(θ) = sin(θ), we start with the left side. Using csc(θ) = 1 sin(θ), we get: 1 csc(θ) = 1 1 sin(θ) = sin(θ), which is what we were trying to prove. 750 Foundations of Trigonometry 2. Starting with the right hand side of tan(θ) = sin(θ) sec(θ), we use sec(θ) = 1 cos(θ) and ﬁnd: sin(θ) sec(θ) = sin(θ) 1 cos(θ) = sin(θ) cos(θ) = tan(θ), where the last equality is courtesy of Theorem 10.6. 3.
Expanding the left hand side of the equation gives: (sec(θ) − tan(θ))(sec(θ) + tan(θ)) = sec2(θ) − tan2(θ). According to Theorem 10.8, sec2(θ) − tan2(θ) = 1. Putting it all together, (sec(θ) − tan(θ))(sec(θ) + tan(θ)) = sec2(θ) − tan2(θ) = 1. 4. While both sides of our last identity contain fractions, the left side aﬀords us more opportu- nities to use our identities.5 Substituting sec(θ) = 1 cos(θ) and tan(θ) = sin(θ) cos(θ), we get: sec(θ) 1 − tan(θ) = 1 cos(θ) 1 − sin(θ) cos(θ) = 1 cos(θ) 1 − sin(θ) cos(θ) · cos(θ) cos(θ) = 1 cos(θ) 1 − sin(θ) cos(θ) (cos(θ)) = (cos(θ)) (1)(cos(θ)) − 1 sin(θ) cos(θ) (cos(θ)) = 1 cos(θ) − sin(θ), which is exactly what we had set out to show. 5. The right hand side of the equation seems to hold more promise. We get common denomina- tors and add: 3 1 − sin(θ) − 3 1 + sin(θ) = = = = 3(1 + sin(θ)) (1 − sin(θ))(1 + sin(θ)) − 3(1 − sin(θ)) (1 + sin(θ))(1 − sin(θ)) 3 + 3 sin(θ) 1 − sin2(θ) − 3 − 3 sin(θ) 1 − sin2(θ) (3 + 3 sin(θ)) − (3 − 3 sin(θ)) 1 − sin2(θ) 6 sin(θ) 1 − sin2(θ) 5Or, to put
to another way, earn more partial credit if this were an exam question! 10.3 The Six Circular Functions and Fundamental Identities 751 At this point, it is worth pausing to remind ourselves of our goal. We wish to transform this expression into 6 sec(θ) tan(θ). Using a reciprocal and quotient identity, we ﬁnd 6 sec(θ) tan(θ) = 6. In other words, we need to get cosines in our denominator. Theorem 10.8 tells us 1 − sin2(θ) = cos2(θ) so we get: sin(θ) cos(θ) 1 cos(θ) 3 1 − sin(θ) − 3 1 + sin(θ) = = 6 sin(θ) cos2(θ) 6 sin(θ) 1 − sin2(θ) 1 cos(θ) = 6 sin(θ) cos(θ) = 6 sec(θ) tan(θ) 6. It is debatable which side of the identity is more complicated. One thing which stands out is that the denominator on the left hand side is 1 − cos(θ), while the numerator of the right hand side is 1 + cos(θ). This suggests the strategy of starting with the left hand side and multiplying the numerator and denominator by the quantity 1 + cos(θ): sin(θ) 1 − cos(θ) = = = sin(θ) (1 − cos(θ)) · (1 + cos(θ)) (1 + cos(θ)) = sin(θ)(1 + cos(θ)) (1 − cos(θ))(1 + cos(θ)) sin(θ)(1 + cos(θ)) 1 − cos2(θ) sin(θ)(1 + cos(θ)) sin(θ) sin(θ) = = sin(θ)(1 + cos(θ)) sin2(θ) 1 + cos(θ) sin(θ) In Example 10.3.3 number 6 above, we see that multiplying 1 − cos(θ) by 1 + cos(θ) produces a diﬀerence of squares that can be simpliﬁed to one term using Theorem 10
.8. This is exactly the √ same kind of phenomenon that occurs when we multiply expressions such as 1 − 2 or 3 − 4i by 3 + 4i. (Can you recall instances from Algebra where we did such things?) For this reason, the quantities (1 − cos(θ)) and (1 + cos(θ)) are called ‘Pythagorean Conjugates.’ Below is a list of other common Pythagorean Conjugates. 2 by 1 + √ Pythagorean Conjugates 1 − cos(θ) and 1 + cos(θ): (1 − cos(θ))(1 + cos(θ)) = 1 − cos2(θ) = sin2(θ) 1 − sin(θ) and 1 + sin(θ): (1 − sin(θ))(1 + sin(θ)) = 1 − sin2(θ) = cos2(θ) sec(θ) − 1 and sec(θ) + 1: (sec(θ) − 1)(sec(θ) + 1) = sec2(θ) − 1 = tan2(θ) sec(θ)−tan(θ) and sec(θ)+tan(θ): (sec(θ)−tan(θ))(sec(θ)+tan(θ)) = sec2(θ)−tan2(θ) = 1 csc(θ) − 1 and csc(θ) + 1: (csc(θ) − 1)(csc(θ) + 1) = csc2(θ) − 1 = cot2(θ) csc(θ) − cot(θ) and csc(θ) + cot(θ): (csc(θ) − cot(θ))(csc(θ) + cot(θ)) = csc2(θ) − cot2(θ) = 1 752 Foundations of Trigonometry Verifying trigonometric identities requires a healthy mix of tenacity and inspiration. You will need to spend many hours struggling with them just to become proﬁcient in the basics. Like many things in life, there is no short-cut here – there is no complete algorithm for verifying
identities. Nevertheless, a summary of some strategies which may be helpful (depending on the situation) is provided below and ample practice is provided for you in the Exercises. Try working on the more complicated side of the identity. Strategies for Verifying Identities Use the Reciprocal and Quotient Identities in Theorem 10.6 to write functions on one side of the identity in terms of the functions on the other side of the identity. Simplify the resulting complex fractions. Add rational expressions with unlike denominators by obtaining common denominators. Use the Pythagorean Identities in Theorem 10.8 to ‘exchange’ sines and cosines, secants and tangents, cosecants and cotangents, and simplify sums or diﬀerences of squares to one term. Multiply numerator and denominator by Pythagorean Conjugates in order to take advan- tage of the Pythagorean Identities in Theorem 10.8. If you ﬁnd yourself stuck working with one side of the identity, try starting with the other side of the identity and see if you can ﬁnd a way to bridge the two parts of your work. 10.3.1 Beyond the Unit Circle In Section 10.2, we generalized the cosine and sine functions from coordinates on the Unit Circle to coordinates on circles of radius r. Using Theorem 10.3 in conjunction with Theorem 10.8, we generalize the remaining circular functions in kind. Theorem 10.9. Suppose Q(x, y) is the point on the terminal side of an angle θ (plotted in standard position) which lies on the circle of radius r, x2 + y2 = r2. Then: r x r y = = x2 + y2 x x2 + y2 y, provided x = 0., provided y = 0., provided x = 0., provided y = 0. sec(θ) = csc(θ) = tan(θ) = cot(θ) = y x x y 10.3 The Six Circular Functions and Fundamental Identities 753 Example 10.3.4. 1. Suppose the terminal side of θ, when plotted in standard position, contains the point Q(3, −4). Find the values of the six circular functions of θ. 2. Suppose θ is a Quadrant IV angle with cot
(θ) = −4. Find the values of the ﬁve remaining circular functions of θ. Solution. 1. Since x = 3 and y = −4, r = x2 + y2 = (3)2 + (−4)2 = √ cos(θ) = 3 5, sin(θ) = − 4 5, sec(θ) = 5 3, csc(θ) = − 5 4, tan(θ) = − 4 25 = 5. Theorem 10.9 tells us 3 and cot(θ) = − 3 4. 2. In order to use Theorem 10.9, we need to ﬁnd a point Q(x, y) which lies on the terminal side y, and since θ is a −1, we may choose6 x = 4 17. Applying Theorem 10.9 once √ √ 17 17 4, csc(θ) = − 17 17, sec(θ) = of θ, when θ is plotted in standard position. We have that cot(θ) = −4 = x Quadrant IV angle, we also know x > 0 and y < 0. Viewing −4 = 4 x2 + y2 = and y = −1 so that r = √ = 4 more, we ﬁnd cos(θ) = 4√ 17 and tan(θ) = − 1 4. (4)2 + (−1)2 = = − 17, sin(θ) = − 1√ √ √ 17 17 We may also specialize Theorem 10.9 to the case of acute angles θ which reside in a right triangle, as visualized below. b c a θ Theorem 10.10. Suppose θ is an acute angle residing in a right triangle. If the length of the side adjacent to θ is a, the length of the side opposite θ is b, and the length of the hypotenuse is c, then tan(θ) = sec(θ) = csc(θ) = cot(θ The following example uses Theorem 10.10 as well as the concept of an ‘angle of inclination.’ The angle of inclination (or angle of elevation) of an object refers to the angle whose initial side is some kind of base-line (say, the ground),

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