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cos π 2 − 4x + 1 11. y = − cos 2x + 3 2 π 3 − 1 2 12. y = 4 sin(−2πx + π) In Exercises 13 - 24, graph one cycle of the given function. State the period of the function. 13. y = tan 16. y = sec x − x − π 3 π 2 14. y = 2 tan 17. y = − csc x − 3 1 4 x + π 3 15. y = 1 3 18. y = − 19. y = csc(2x − π) 20. y = sec(3x − 2π) + 4 21. y = csc −x − tan(−2x − π) + 1 1 3 sec 22. y = cot x + π 6 23. y = −11 cot x 1 5 24. y = cot 2x + 3π 2 + 1 1 3 In Exercises 25 - 34, use Example 10.5.3 as a guide to show that the function is a sinusoid by rewriting it in the forms C(x) = A cos(ωx + φ) + B and S(x) = A sin(ωx + φ) + B for ω > 0 and 0 ≤ φ < 2π. 25. f (x) = √ 2 sin(x) + √ 2 cos(x) + 1 26. f (x) = 3 √ 3 sin(3x) − 3 cos(3x) 27. f (x) = − sin(x) + cos(x) − 2 28. f (x) = − 1 2 sin(2x) − √ 3 2 cos(2x) 29. f (x) = 2 √ 3 cos(x) − 2 sin(x) 31. f (x) = − 1 2 cos(5x) − √ 3 2 sin(5x) 30. f (x) = 3 2 cos(2x) − √ 3 3 2 sin(2x) + 6 32. f (x) = −6 √ 3 cos(3x) − 6 sin(3x) − 3 810 Foundations of Trigonometry 33. f (x) = √ 5 2 2 sin(x) − √ 5 2 2 cos(x)
34. f (x) = 3 sin x 6 √ − 3 3 cos x 6 35. In Exercises 25 - 34, you should have noticed a relationship between the phases φ for the S(x) and C(x). Show that if f (x) = A sin(ωx + α) + B, then f (x) = A cos(ωx + β) + B where β = α −. π 2 36. Let φ be an angle measured in radians and let P (a, b) be a point on the terminal side of φ when it is drawn in standard position. Use Theorem 10.3 and the sum identity for sine in Theorem 10.15 to show that f (x) = a sin(ωx) + b cos(ωx) + B (with ω > 0) can be rewritten as f (x) = a2 + b2 sin(ωx + φ) + B. √ 37. With the help of your classmates, express the domains of the functions in Examples 10.5.4 and 10.5.5 using extended interval notation. (We will revisit this in Section 10.7.) In Exercises 38 - 43, verify the identity by graphing the right and left hand sides on a calculator. 38. sin2(x) + cos2(x) = 1 39. sec2(x) − tan2(x) = 1 40. cos(x) = sin 41. tan(x + π) = tan(x) 42. sin(2x) = 2 sin(x) cos(x) 43. tan x 2 = − x π 2 sin(x) 1 + cos(x) In Exercises 44 - 50, graph the function with the help of your calculator and discuss the given questions with your classmates. 44. f (x) = cos(3x) + sin(x). Is this function periodic? If so, what is the period? 45. f (x) = sin(x) x. What appears to be the horizontal asymptote of the graph? 46. f (x) = x sin(x). Graph y = ±x on the same set of axes and describe the behavior of f. 47. f (x) = sin 1 x. What’s happening as x → 0? 48. f (x) = x − tan
(x). Graph y = x on the same set of axes and describe the behavior of f. 49. f (x) = e−0.1x (cos(2x) + sin(2x)). Graph y = ±e−0.1x on the same set of axes and describe the behavior of f. 50. f (x) = e−0.1x (cos(2x) + 2 sin(x)). Graph y = ±e−0.1x on the same set of axes and describe the behavior of f. 51. Show that a constant function f is periodic by showing that f (x + 117) = f (x) for all real numbers x. Then show that f has no period by showing that you cannot ﬁnd a smallest number p such that f (x + p) = f (x) for all real numbers x. Said another way, show that f (x + p) = f (x) for all real numbers x for ALL values of p > 0, so no smallest value exists to satisfy the deﬁnition of ‘period’. 10.5 Graphs of the Trigonometric Functions 811 10.5.5 Answers 1. y = 3 sin(x) Period: 2π Amplitude: 3 Phase Shift: 0 Vertical Shift: 0 2. y = sin(3x) 2π 3 Period: Amplitude: 1 Phase Shift: 0 Vertical Shift: 0 3. y = −2 cos(x) Period: 2π Amplitude: 2 Phase Shift: 0 Vertical Shift: 0 4. y = cos x − π 2 Period: 2π Amplitude: 1 Phase Shift: Vertical Shift: 0 π 2 y 3 −3 y 1 −1 y 2 −2 y 1 −1 π 2 π 3π 2 x 2π π 6 π 3 π 2 2π 3 x π 2 π 3π 2 2π x π 2 π 3π 2 2π 5π 2 x 812 5. y = − sin x + Period: 2π Amplitude: 1 π Phase Shift: − 3 Vertical Shift: 0 π 3 Foundations of Trigonometry y 1 − π 3 π 6 2π 3 7π 6 5π 3 x 6. y = sin(2x − π) Period: π Amplitude: 1 Phase Shift: Vertical Shift: 0 π
2 7. y = − 1 3 cos 1 2 x + π 3 Period: 4π Amplitude: 1 3 2π Phase Shift: − 3 Vertical Shift: 0 8. y = cos(3x − 2π) + 4 2π 3 Period: Amplitude: 1 2π Phase Shift: 3 Vertical Shift: 4 −1 y 1 −1 y 1 3 π 2 3π 4 π 5π 4 3π 2 x − 2π 3 π 3 4π 3 7π 3 x 10π 3 − 1 3 y 5 4 3 2π 3 5π 6 π 7π 6 4π 3 x 10.5 Graphs of the Trigonometric Functions 813 9. y = sin −x − π 4 − 2 Period: 2π Amplitude: 1 Phase Shift: − π 4 π x + 4 Vertical Shift: −2 y = − sin (You need to use − 2 to ﬁnd this.)15 − 4x + 1 10. y = 2 3 Period: π 2 cos π 2 Amplitude: 2 3 π 8 4x − cos y = Vertical Shift: 1 Phase Shift: 2 3 (You need to use π 2 + 1 to ﬁnd this.)16 − 9π 4 − 7π 4 − 5π 4 − 3π 4 y π 4 3π 4 5π 4 x 7π 4 − π 4 −1 −2 − 3π 8 π 2 5π 8 x − 3π 8 − π 4 − π 8 11. y = − cos 2x + 3 2 π 3 − 1 2 y 1 Period: π Amplitude: 3 2 Phase Shift: − π 6 Vertical Shift: − 1 2 12. y = 4 sin(−2πx + π) Period: 1 Amplitude: 4 1 2 (You need to use Phase Shift: y = −4 sin(2πx − π) to ﬁnd this.)17 Vertical Shift: 0 − π 6 − 1 2 π 12 π 3 7π 12 x 5π 6 −4 15Two cycles of the graph are shown to illustrate the discrepancy discussed on page 796. 16Again, we graph two cycles to illustrate the discrepancy discussed on page 796. 17This will be the last time we graph two cycles to illustrate the discrepancy discussed on page 796. 814 Foundations of Trigonometry 13. y
= tan x − Period: π π 3 14. y = 2 tan x − 3 1 4 Period: 4π y 1 − π 6 −1 π 12 π 3 7π 12 5π 6 x −2π −π y −1 −3 −5 π 2π x 15. y = tan(−2x − π) + 1 1 3 is equivalent to 1 3 tan(2x + π) + 1 y = − via the Even / Odd identity for tangent. Period − 3π 8 − π 4 x − 3π 4 − 5π 8 10.5 Graphs of the Trigonometric Functions 815 16. y = sec x − π 2 Start with y = cos x − π 2 Period: 2π 17. y = − csc x + π 3 Start with y = − sin Period: 2π x + π 3 y 1 −1 y 1 π 2 π 3π 2 2π x 5π 2 − π 3 −1 π 6 2π 3 7π 6 x 5π 3 18. y = − 1 3 sec 1 2 x + Start with y = − Period: 4π 1 3 π 3 cos 1 2 x + π 3 y 1 3 − 2π 3 − 1 3 π 3 4π 3 7π 3 x 10π 3 816 Foundations of Trigonometry 19. y = csc(2x − π) Start with y = sin(2x − π) Period: π 20. y = sec(3x − 2π) + 4 Start with y = cos(3x − 2π) + 4 Period: 2π 3 21. y = csc −x − π 4 Start with y = sin Period: 2π − 2 −x − π 4 − 2 y 1 −1 −2 −3 π 2 3π 4 π 5π 4 x 3π 2 2π 3 5π 6 π 7π 6 4π 3 x π 4 3π 4 5π 4 x 7π 4 10.5 Graphs of the Trigonometric Functions 817 22. y = cot x + Period: π π 6 23. y = −11 cot Period: 5π 1 5 x y 1 − π 6 π 12 −1 π 3 7π 12 x 5π 6 y 11 −11 5π 4 5
π 2 15π 4 5π x 24. y = 1 3 Period: cot π 2 2x + 3π − 3π 8 − π 4 x − 3π 4 − 5π 8 818 Foundations of Trigonometry 25. f (x) = √ 2 sin(x) + √ 2 cos(x) + 1 = 2 sin x + π 4 + 1 = 2 cos x + 7π 4 + 1 26. f (x) = 3 √ 3 sin(3x) − 3 cos(3x) = 6 sin 3x + 11π 6 = 6 cos 3x + 4π 3 27. f (x) = − sin(x) + cos(x) − 2 = √ 2 sin x + 3π 4 − 2 = √ x + 2 cos π 4 − 2 28. f (x) = − 1 2 sin(2x) − √ 3 2 cos(2x) = sin 2x + 4π 3 = cos 2x + 5π 6 29. f (x) = 2 √ 3 cos(x) − 2 sin(x) = 4 sin x + 30. f (x) = 3 2 cos(2x) − 31. f (x) = − 1 2 cos(5x) − 3 2 √ 3 3 2 √ 2π 3 = 4 cos x + π 6 5π 6 + 6 = 3 cos = cos 5x + 2π 3 sin(2x) + 6 = 3 sin 2x + sin(5x) = sin 5x + 7π 6 2x + π 3 + 6 32. f (x) = −6 √ 3 cos(3x) − 6 sin(3x) − 3 = 12 sin 3x + 4π 3 − 3 = 12 cos 3x + 5π 6 − 3 33. f (x) = √ 5 2 2 sin(x) − √ 5 2 2 cos(x) = 5 sin x + 34. f (x) = 3 sin x 6 √ − 3 3 cos x 6 = 6 sin x 6 + 7π 4 5π 3 = 5 cos x + 5π 4 = 6 cos x 6 + 7π 6 10.6 The Inverse Trigonometric Functions 819 10.6 The Inverse Trigonometric Functions As the title indicates, in this section we concern ourselves with ﬁ
nding inverses of the (circular) trigonometric functions. Our immediate problem is that, owing to their periodic nature, none of the six circular functions is one-to-one. To remedy this, we restrict the domains of the circular functions in the same way we restricted the domain of the quadratic function in Example 5.2.3 in Section 5.2 to obtain a one-to-one function. We ﬁrst consider f (x) = cos(x). Choosing the interval [0, π] allows us to keep the range as [−1, 1] as well as the properties of being smooth and continuous. y x Restricting the domain of f (x) = cos(x) to [0, π]. 1 Recall from Section 5.2 that the inverse of a function f is typically denoted f −1. For this reason, some textbooks use the notation f −1(x) = cos−1(x) for the inverse of f (x) = cos(x). The obvious pitfall here is our convention of writing (cos(x))2 as cos2(x), (cos(x))3 as cos3(x) and so on. It is far too easy to confuse cos−1(x) with cos(x) = sec(x) so we will not use this notation in our text.1 Instead, we use the notation f −1(x) = arccos(x), read ‘arc-cosine of x’. To understand the ‘arc’ in ‘arccosine’, recall that an inverse function, by deﬁnition, reverses the process of the original function. The function f (t) = cos(t) takes a real number input t, associates it with the angle θ = t radians, and returns the value cos(θ). Digging deeper,2 we have that cos(θ) = cos(t) is the x-coordinate of the terminal point on the Unit Circle of an oriented arc of length \|t\| whose initial point is (1, 0). Hence, we may view the inputs to f (t) = cos(t) as oriented arcs and the outputs as x-coordinates on the Unit Circle. The function f −1, then, would take x-coordinates on the Unit Circle and return oriented arcs, hence the
‘arc’ in arccosine. Below are the graphs of f (x) = cos(x) and f −1(x) = arccos(x), where we obtain the latter from the former by reﬂecting it across the line y = x, in accordance with Theorem 5.3. y 1 −x) = cos(x), 0 ≤ x ≤ π reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates −1 1 x f −1(x) = arccos(x). 1But be aware that many books do! As always, be sure to check the context! 2See page 704 if you need a review of how we associate real numbers with angles in radian measure. 820 Foundations of Trigonometry We restrict g(x) = sin(x) in a similar manner, although the interval of choice is − π 2, π 2 y. x Restricting the domain of f (x) = sin(x) to − π 2, π 2. It should be no surprise that we call g−1(x) = arcsin(x), which is read ‘arc-sine of x’. y 1 − π 2 x π 2 −1 g(x) = sin(x), − 1 1 x reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates − π 2 g−1(x) = arcsin(x). We list some important facts about the arccosine and arcsine functions in the following theorem. Theorem 10.26. Properties of the Arccosine and Arcsine Functions Properties of F (x) = arccos(x) – Domain: [−1, 1] – Range: [0, π] – arccos(x) = t if and only if 0 ≤ t ≤ π and cos(t) = x – cos(arccos(x)) = x provided −1 ≤ x ≤ 1 – arccos(cos(x)) = x provided 0 ≤ x ≤ π Properties of G(x) = arcsin(x) – Domain: [−1, 1] – Range – arcsin(x) = t if and only if − π – sin(
arcsin(x)) = x provided − – arcsin(sin(x)) = x provided − π – additionally, arcsine is odd 2 2 and sin(t) = x 10.6 The Inverse Trigonometric Functions 821 Everything in Theorem 10.26 is a direct consequence of the facts that f (x) = cos(x) for 0 ≤ x ≤ π and F (x) = arccos(x) are inverses of each other as are g(x) = sin(x) for − π 2 and G(x) = arcsin(x). It’s about time for an example. 2 ≤ x ≤ π Example 10.6.1. 1. Find the exact values of the following. (a) arccos 1 2 (c) arccos − √ 2 2 (e) arccos cos π 6 (g) cos arccos − 3 5 (b) arcsin √ 2 2 (d) arcsin − 1 2 (f) arccos cos 11π 6 (h) sin arccos − 3 5 2. Rewrite the following as algebraic expressions of x and state the domain on which the equiv- alence is valid. (a) tan (arccos (x)) (b) cos (2 arcsin(x)) Solution. 1. (a) To ﬁnd arccos 1 2, we need to ﬁnd the real number t (or, equivalently, an angle measuring 3 meets these 2. We know t = π 2 and π 2 with sin(t) = √ 2 2. The (b) The value of arcsin t radians) which lies between 0 and π with cos(t) = 1 = π criteria, so arccos 1 3. 2 √ 2 is a real number t between − π 2 √ 2 number we seek is t = π 4. Hence, arcsin 2 = π 4. √ (c) The number t = arccos √ − 2 2 lies in the interval [0, π] with cos(t) = − √ 2 2. Our answer 2 2 − is arccos (d) To ﬁnd arcsin − 1 2 answer is t = − π 6 so that arcsin − 1 = 3π 4., we seek the
number t in the interval −, we could simply invoke Theorem 10.26 to get arccos cos π 6. However, in order to make sure we understand why this is the case, we choose to work the example through using the deﬁnition of arccosine. Working from the inside out, arccos cos π is the real number t with 0 ≤ t ≤ π 6 with sin(t) = − 1 (e) Since. The 2, π 2 6 2 = arccos √ 3 2. We ﬁnd t = π. Now, arccos 6, so that arccos cos π 6 = π 6. and cos(t) = 822 Foundations of Trigonometry (f) Since 11π 6 does not fall between 0 and π, Theorem 10.26 does not apply. We are forced to. From work through from the inside out starting with arccos cos 11π 6 = arccos 6. Hence, arccos cos 11π is to use Theorem 10.26 directly. Since − the previous problem, we know arccos (g) One way to simplify cos arccos − 3 5 between −1 and 1, we have that cos arccos − 3 5 before, to really understand why this cancellation occurs, we let t = arccos − 3 5 by deﬁnition, cos(t) = − 3 in (nearly) the same amount of time. 5 is 5 and we are done. However, as. Then, 5, and we are ﬁnished 5. Hence, cos arccos − 3 = cos(t) = − 3 = − 3 5 2. (h) As in the previous example, we let t = arccos − 3 5 5 for some t where 0 ≤ t ≤ π. Since cos(t) < 0, we can narrow this down a bit and conclude that π 2 < t < π, so that t corresponds to an angle in Quadrant II. In terms of t, then, we need to ﬁnd = sin(t). Using the Pythagorean Identity cos2(t) + sin2(t) = 1, we get sin arccos − 3 5 − 3 5. Since t corresponds to a Quadrants II angle, we 5 choose sin(t) = 4 + sin2
(t) = 1 or sin(t) = ± 4 5. Hence, sin arccos − 3 so that cos(ta) We begin this problem in the same manner we began the previous two problems. To help us see the forest for the trees, we let t = arccos(x), so our goal is to ﬁnd a way to express tan (arccos (x)) = tan(t) in terms of x. Since t = arccos(x), we know cos(t) = x where 0 ≤ t ≤ π, but since we are after an expression for tan(t), we know we need to throw out t = π 2 < t ≤ π so that, geometrically, t corresponds to an angle in Quadrant I or Quadrant II. One approach3 to ﬁnding tan(t) is to use the quotient identity tan(t) = sin(t) cos(t). Substituting cos(t) = x into the Pythagorean Identity cos2(t) + sin2(t) = 1 gives x2 + sin2(t) = 1, from which we 1 − x2. Since t corresponds to angles in Quadrants I and II, sin(t) ≥ 0, get sin(t) = ± so we choose sin(t) = 2 from consideration. Hence, either 0 ≤ t < π 1 − x2. Thus, 2 or π √ √ tan(t) = sin(t) cos(t) = √ 1 − x2 x To determine the values of x for which this equivalence is valid, we consider our substitution t = arccos(x). Since the domain of arccos(x) is [−1, 1], we know we must restrict −1 ≤ x ≤ 1. Additionally, since we had to discard t = π 2, we need to discard x = cos π 2 = 0. Hence, tan (arccos (x)) = is valid for x in [−1, 0) ∪ (0, 1]. 1−x2 x √ 2, π (b) We proceed as in the previous problem by writing t = arcsin(x) so that t lies in the with sin(t) = x. We aim to express cos (2 arcsin(x)) = cos(2t) in terms interval
− π of x. Since cos(2t) is deﬁned everywhere, we get no additional restrictions on t as we did in the previous problem. We have three choices for rewriting cos(2t): cos2(t) − sin2(t), 2 cos2(t) − 1 and 1 − 2 sin2(t). Since we know x = sin(t), it is easiest to use the last form: 2 cos (2 arcsin(x)) = cos(2t) = 1 − 2 sin2(t) = 1 − 2x2 3Alternatively, we could use the identity: 1 + tan2(t) = sec2(t). Since x = cos(t), sec(t) = 1 cos(t) = 1 x. The reader is invited to work through this approach to see what, if any, diﬃculties arise. 10.6 The Inverse Trigonometric Functions 823 To ﬁnd the restrictions on x, we once again appeal to our substitution t = arcsin(x). Since arcsin(x) is deﬁned only for −1 ≤ x ≤ 1, the equivalence cos (2 arcsin(x)) = 1−2x2 is valid only on [−1, 1]. A few remarks about Example 10.6.1 are in order. Most of the common errors encountered in dealing with the inverse circular functions come from the need to restrict the domains of the original functions so that they are one-to-one. One instance of this phenomenon is the fact that arccos cos 11π 6 as opposed to 11π 6. This is the exact same phenomenon discussed in Section 6 (−2)2 = 2 as opposed to −2. Additionally, even though the expression we 5.2 when we saw arrived at in part 2b above, namely 1 − 2x2, is deﬁned for all real numbers, the equivalence cos (2 arcsin(x)) = 1 − 2x2 is valid for only −1 ≤ x ≤ 1. This is akin to the fact that while the x)2 = x is valid only for x ≥ 0. For expression x is deﬁned for all real numbers, the equivalence ( this reason, it pays to be careful when we determine the intervals where such equivalences are valid. = π √ The next pair of functions we wish to discuss are
the inverses of tangent and cotangent, which are named arctangent and arccotangent, respectively. First, we restrict f (x) = tan(x) to its fundamental cycle on − π to obtain f −1(x) = arctan(x). Among other things, note that the 2, π 2 vertical asymptotes x = − π 2 and x = π 2 of the graph of f (x) = tan(x) become the horizontal 2 and y = π asymptotes y = − π 2 of the graph of f −1(x) = arctan(x). y 1 −x) = tan(x), − π 2 < x < π 2. reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates y π 2 π 4 −1(x) = arctan(x). Next, we restrict g(x) = cot(x) to its fundamental cycle on (0, π) to obtain g−1(x) = arccot(x). Once again, the vertical asymptotes x = 0 and x = π of the graph of g(x) = cot(x) become the horizontal asymptotes y = 0 and y = π of the graph of g−1(x) = arccot(x). We show these graphs on the next page and list some of the basic properties of the arctangent and arccotangent functions. 824 Foundations of Trigonometry y 1 −1 π 4 π 2 3π 4 π x y π 3π 4 π 2 π 4 g(x) = cot(x), 0 < x < π. reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates g−1(x) = arccot(x). −1 1 x Theorem 10.27. Properties of the Arctangent and Arccotangent Functions Properties of F (x) = arctan(x) – Domain: (−∞, ∞) – Range: − π 2, π 2 – as x → −∞, arctan(x) → − π
2 – arctan(x) = t if and only if − π for x > 0 – arctan(x) = arccot 1 x – tan (arctan(x)) = x for all real numbers x 2 < x < π – arctan(tan(x)) = x provided − π – additionally, arctangent is odd 2 +; as x → ∞, arctan(x and tan(t) = x − Properties of G(x) = arccot(x) – Domain: (−∞, ∞) – Range: (0, π) – as x → −∞, arccot(x) → π−; as x → ∞, arccot(x) → 0+ – arccot(x) = t if and only if 0 < t < π and cot(t) = x – arccot(x) = arctan 1 x – cot (arccot(x)) = x for all real numbers x for x > 0 – arccot(cot(x)) = x provided 0 < x < π 10.6 The Inverse Trigonometric Functions 825 Example 10.6.2. 1. Find the exact values of the following. √ (a) arctan( 3) (c) cot(arccot(−5)) √ 3) (b) arccot(− (d) sin arctan − 3 4 2. Rewrite the following as algebraic expressions of x and state the domain on which the equiv- alence is valid. (a) tan(2 arctan(x)) (b) cos(arccot(2x)) Solution. 1. 2. (a) We know arctan( 3, so arctan( t = π 3) is the real number t between − π 3) = π 3. √ √ √ (b) The real number t = arccot(− 3) = 5π 6. arccot(− √ 2 and π 2 with tan(t) = √ 3. We ﬁnd 3) lies in the interval (0, π) with cot(t) = − 3. We get √ (c) We can apply Theorem 10.27 directly and obtain c
ot(arccot(−5)) = −5. However, working it through provides us with yet another opportunity to understand why this is the case. Letting t = arccot(−5), we have that t belongs to the interval (0, π) and cot(t) = −5. Hence, cot(arccot(−5)) = cot(t) = −5. (a) If we let t = arctan(x), then − π (d) We start simplifying sin arctan − 3 4. Then tan(t) = − 3 by letting t = arctan −, we choose csc(t) = − 5 2 3, so sin(t) = − 3 3. Substituting, we get 1 + − 4 2. Since tan(t) < 0, we know, in fact, − π 4 for some − π 2 < t < 0. One way to proceed is to use The Pythagorean Identity, 1+cot2(t) = csc2(t), since this relates the reciprocals of tan(t) and sin(t) and is valid for all t under consideration.4 From tan(t) = − 3 4, we get cot(t) = − 4 3. Since = − 3 − π 5. = csc2(t) so that csc(t) = ± 5 5. Hence, sin arctan − 3 2 and tan(t) = x. We look for a way to express tan(2 arctan(x)) = tan(2t) in terms of x. Before we get started using identities, we note that tan(2t) is undeﬁned when 2t = π 2 + πk for integers k. Dividing both sides of this equation by 2 tells us we need to exclude values of t where t = π 4 + π 2 k, where k is are t = ± π an integer. The only members of this family which lie in − π 2, π 4, which 2. Returning ∪ − π means the values of t under consideration are − to arctan(2t), we note the double angle identity tan(2t) = 2 tan(t) 1−tan2(t), is valid for all the values of t under consideration, hence we
get tan(2 arctan(x)) = tan(2t) = 2 tan(t) 1 − tan2(t) = 2x 1 − x2 4It’s always a good idea to make sure the identities used in these situations are valid for all values t under consideration. Check our work back in Example 10.6.1. Were the identities we used there valid for all t under consideration? A pedantic point, to be sure, but what else do you expect from this book? 826 Foundations of Trigonometry To ﬁnd where this equivalence is valid we check back with our substitution t = arctan(x). Since the domain of arctan(x) is all real numbers, the only exclusions come from the values of t we discarded earlier, t = ± π 4. Since x = tan(t), this means we exclude x = tan ± π 1−x2 holds for all x in 4 (−∞, −1) ∪ (−1, 1) ∪ (1, ∞). = ±1. Hence, the equivalence tan(2 arctan(x)) = 2x (b) To get started, we let t = arccot(2x) so that cot(t) = 2x where 0 < t < π. In terms of t, cos(arccot(2x)) = cos(t), and our goal is to express the latter in terms of x. Since cos(t) is always deﬁned, there are no additional restrictions on t, so we can begin using identities to relate cot(t) to cos(t). The identity cot(t) = cos(t) sin(t) is valid for t in (0, π), so our strategy is to obtain sin(t) in terms of x, then write cos(t) = cot(t) sin(t). The identity 1 + cot2(t) = csc2(t) holds for all t in (0, π) and relates cot(t) and csc(t) = 1 sin(t). Substituting cot(t) = 2x, we get 1 + (2x)2 = csc2(t), or csc(t) = ± 4x2 + 1. Since t is. Hence, between 0 and �
�, csc(t) > 0, so csc(t) = 4x2 + 1 which gives sin(t) = 1√ √ √ 4x2+1 cos(arccot(2x)) = cos(t) = cot(t) sin(t) = √ 2x 4x2 + 1 Since arccot(2x) is deﬁned for all real numbers x and we encountered no additional restrictions on t, we have cos (arccot(2x)) = 2x√ for all real numbers x. 4x2+1 The last two functions to invert are secant and cosecant. A portion of each of their graphs, which were ﬁrst discussed in Subsection 10.5.2, are given below with the fundamental cycles highlighted. y y x x The graph of y = sec(x). The graph of y = csc(x). It is clear from the graph of secant that we cannot ﬁnd one single continuous piece of its graph which covers its entire range of (−∞, −1] ∪ [1, ∞) and restricts the domain of the function so that it is one-to-one. The same is true for cosecant. Thus in order to deﬁne the arcsecant and arccosecant functions, we must settle for a piecewise approach wherein we choose one piece to cover the top of the range, namely [1, ∞), and another piece to cover the bottom, namely (−∞, −1]. There are two generally accepted ways make these choices which restrict the domains of these functions so that they are one-to-one. One approach simpliﬁes the Trigonometry associated with the inverse functions, but complicates the Calculus; the other makes the Calculus easier, but the Trigonometry less so. We present both points of view. 10.6 The Inverse Trigonometric Functions 827 10.6.1 Inverses of Secant and Cosecant: Trigonometry Friendly Approach In this subsection, we restrict the secant and cosecant functions to coincide with the restrictions on cosine and sine, respectively. For f (x) = sec(x), we restrict the domain to 0x) = sec(x) on 0, π 2 ∪ π 2,
π reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates −1 1 x f −1(x) = arcsec(x) and we restrict g(x) = csc(x) to − π 2, 0 ∪ 0, π 2. y 1 −1 − π 2 x π 2 g(x) = csc(x) on − π 2, 0 ∪ 0, π 2 y π 2 −1 1 x reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates − π 2 g−1(x) = arccsc(x) Note that for both arcsecant and arccosecant, the domain is (−∞, −1] ∪ [1, ∞). Taking a page from Section 2.2, we can rewrite this as {x : \|x\| ≥ 1}. This is often done in Calculus textbooks, so we include it here for completeness. Using these deﬁnitions, we get the following properties of the arcsecant and arccosecant functions. 828 Foundations of Trigonometry Theorem 10.28. Properties of the Arcsecant and Arccosecant Functionsa Properties of F (x) = arcsec(x) ∪ π – Domain: {x : \|x\| ≥ 1} = (−∞, −1] ∪ [1, ∞) – Range: 0, π 2, π 2 – as x → −∞, arcsec(x) → π 2 – arcsec(x) = t if and only if 0 ≤ t < π – arcsec(x) = arccos 1 x – sec (arcsec(x)) = x provided \|x\| ≥ 1 – arcsec(sec(x)) = x provided 0 ≤ x < π provided \|x\| ≥ 1 2 or π 2 or π 2 < x ≤ π +; as x → ∞, arcsec(x and sec(t) = x Properties of G(x) = arccsc(x) 2 2, 0 ∪ 0, π – Domain: {x : \|x\| ≥ 1} = (−∞, −1] ∪ [1
, ∞) – Range: − π – as x → −∞, arccsc(x) → 0−; as x → ∞, arccsc(x) → 0+ – arccsc(x) = t if and only if − π – arccsc(x) = arcsin 1 x – csc (arccsc(x)) = x provided \|x\| ≥ 1 – arccsc(csc(x)) = x provided − π – additionally, arccosecant is odd 2 ≤ x < 0 or or 0 < t ≤ π provided \|x\| ≥ 1 2 2 and csc(t) = x a... assuming the “Trigonometry Friendly” ranges are used. Example 10.6.3. 1. Find the exact values of the following. (a) arcsec(2) (b) arccsc(−2) (c) arcsec sec 5π 4 (d) cot (arccsc (−3)) 2. Rewrite the following as algebraic expressions of x and state the domain on which the equiv- alence is valid. (a) tan(arcsec(x)) (b) cos(arccsc(4x)) 10.6 The Inverse Trigonometric Functions 829 Solution. 1. 2. (a) Using Theorem 10.28, we have arcsec(2) = arccos 1 2 (b) Once again, Theorem 10.28 comes to our aid giving arccsc(−2) = arcsin − 1 2 (c) Since 5π 4 doesn’t fall between 0 and π 2 and π, we cannot use the inverse property stated in Theorem 10.28. We can, nevertheless, begin by working ‘inside out’ which yields arcsec sec 5π 4 = arcsec(− = − π 6. 2) = arccos = π 3. 2 or π = 3π 4. 2 2 √ − √ (d) One way to begin to simplify cot (arccsc (−3)) is to let t = arccsc(−3). Then, csc(t) = −3 and, since this is negative, we have that t lies in the interval − π 2, 0. We are after cot (arccsc (−3)) = cot(
t), so we use the Pythagorean Identity 1 + cot2(t) = csc2(t). Substituting, we have 1 + cot2(t) = (−3)2, or cot(t) = ± 2 ≤ t < 0, cot(t) < 0, so we get cot (arccsc (−3)) = −2 2. Since − π 8 = ±2 √ √ √ 2. ∪ π (a) We begin simplifying tan(arcsec(x)) by letting t = arcsec(x). Then, sec(t) = x for t in 0, π 2, π, and we seek a formula for tan(t). Since tan(t) is deﬁned for all t values 2 under consideration, we have no additional restrictions on t. To relate sec(t) to tan(t), we use the identity 1 + tan2(t) = sec2(t). This is valid for all values of t under consideration, √ and when we substitute sec(t) = x, we get 1 + tan2(t) = x2. Hence, tan(t) = ± x2 − 1. 2, π then then tan(t) ≥ 0; if, on the the other hand, t belongs to π If t belongs to 0, π 2 tan(t) ≤ 0. As a result, we get a piecewise deﬁned function for tan(t) tan(t) = √ √ − x2 − 1, x2 − 1, if 0 ≤ t < π 2 if π 2 < t ≤ π Now we need to determine what these conditions on t mean for x. Since x = sec(t), when, x ≤ −1. Since we encountered no further restrictions on t, the equivalence below holds for all x in (−∞, −1] ∪ [1, ∞). 2, x ≥ 1, and when π tan(arcsec(x)) = √ √ − x2 − 1, if x ≥ 1 x2 − 1, if x ≤ −1 2, 0∪0, π (b) To simplify cos(arccsc(4x)), we start by letting t = arccsc(4x). Then csc(t) = 4x
for t in, and we now set about ﬁnding an expression for cos(arccsc(4x)) = cos(t). − π Since cos(t) is deﬁned for all t, we do not encounter any additional restrictions on t. From csc(t) = 4x, we get sin(t) = 1 4x, so to ﬁnd cos(t), we can make use if the identity cos2(t) + sin2(t) = 1. Substituting sin(t) = 1 = 1. Solving, we get 2 2 cos(t) = ± 4x gives cos2(t) + 1 √ 4x 16x2 − 1 16x2 = ± 16x2 − 1 4\|x\| Since t belongs to − π. (The absolute values here are necessary, since x could be negative.) To ﬁnd the values for, we know cos(t) ≥ 0, so we choose cos(t) = 2, 0 ∪ 0, π 16−x2 4\|x\| 2 √ 830 Foundations of Trigonometry which this equivalence is valid, we look back at our original substution, t = arccsc(4x). Since the domain of arccsc(x) requires its argument x to satisfy \|x\| ≥ 1, the domain of arccsc(4x) requires \|4x\| ≥ 1. Using Theorem 2.4, we rewrite this inequality and solve to get x ≤ − 1 4. Since we had no additional restrictions on t, the equivalence ∪ 1 cos(arccsc(4x)) = holds for all x in −∞, − 1 4 4 or x ≥ 1 16x2−1 4\|x\| 4, ∞. √ Inverses of Secant and Cosecant: Calculus Friendly Approach 10.6.2 In this subsection, we restrict f (x) = sec(x) to 0, π 2 y 1 −1 π 2 π 3π 2 x ∪ π, 3π 2 y 3π 2 π π 2 f (x) = sec(x) on 0, π 2 ∪ π, 3π 2 reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates
−1 1 x f −1(x) = arcsec(x) and we restrict g(x) = csc(x) to 0, π 2 ∪ π, 3π 2. y 1 −1 π 2 π 3π 2 x y 3π 2 π π 2 g(x) = csc(x) on 0, π 2 ∪ π, 3π 2 reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates −1 1 x g−1(x) = arccsc(x) Using these deﬁnitions, we get the following result. 10.6 The Inverse Trigonometric Functions 831 Theorem 10.29. Properties of the Arcsecant and Arccosecant Functionsa Properties of F (x) = arcsec(x) ∪ π, 3π 2 – Domain: {x : \|x\| ≥ 1} = (−∞, −1] ∪ [1, ∞) – Range: 0, π 2 – as x → −∞, arcsec(x) → 3π 2 – arcsec(x) = t if and only if 0 ≤ t < π – arcsec(x) = arccos 1 for x ≥ 1 onlyb x – sec (arcsec(x)) = x provided \|x\| ≥ 1 – arcsec(sec(x)) = x provided 0 ≤ x < π − 2 or π ≤ x < 3π 2 ; as x → ∞, arcsec(x) → π 2 2 or π ≤ t < 3π − 2 and sec(t) = x Properties of G(x) = arccsc(x) ∪ π, 3π 2 – Domain: {x : \|x\| ≥ 1} = (−∞, −1] ∪ [1, ∞) – Range: 0, π 2 – as x → −∞, arccsc(x) → π+; as x → ∞, arccsc(x) → 0+ – arccsc(x) = t if and only if 0 < t ≤ π for x ≥ 1 onlyc – arccsc(x) = arcsin 1 x – csc (arccsc(x)) = x provided \|x
\| ≥ 1 – arccsc(csc(x)) = x provided 0 < x ≤ π 2 or π < t ≤ 3π 2 or π < x ≤ 3π 2 2 and csc(t) = x a... assuming the “Calculus Friendly” ranges are used. bCompare this with the similar result in Theorem 10.28. cCompare this with the similar result in Theorem 10.28. Our next example is a duplicate of Example 10.6.3. The interested reader is invited to compare and contrast the solution to each. Example 10.6.4. 1. Find the exact values of the following. (a) arcsec(2) (b) arccsc(−2) (c) arcsec sec 5π 4 (d) cot (arccsc (−3)) 2. Rewrite the following as algebraic expressions of x and state the domain on which the equiv- alence is valid. (a) tan(arcsec(x)) (b) cos(arccsc(4x)) 832 Solution. Foundations of Trigonometry 1. 2. (a) Since 2 ≥ 1, we may invoke Theorem 10.29 to get arcsec(2) = arccos 1 2 (b) Unfortunately, −2 is not greater to or equal to 1, so we cannot apply Theorem 10.29 to arccsc(−2) and convert this into an arcsine problem. Instead, we appeal to the deﬁnition. and satisﬁes csc(t) = −2. The t The real number t = arccsc(−2) lies in 0, π 2 we’re after is t = 7π ∪ π, 3π 2 6, so arccsc(−2) = 7π 6. = π 3. lies between π and 3π (c) Since 5π 4 arcsec sec 5π 4 tion as we have done in the previous examples to see how it goes. 2, we may apply Theorem 10.29 directly to simplify 4. We encourage the reader to work this through using the deﬁni- = 5π (d) To simplify cot (arccsc (−3)) we let t = arccsc (−3) so that cot (arccsc (−3)) = cot(t).
. Using the identity 2. Since √ 2. We know csc(t) = −3, and since this is negative, t lies in π, 3π 2 1 + cot2(t) = csc2(t), we ﬁnd 1 + cot2(t) = (−3)2 so that cot(t) = ± t is in the interval π, 3π 2, we know cot(t) > 0. Our answer is cot (arccsc (−3)) = 2 8 = ±2 √ √ ∪ π, 3π 2 (a) We begin simplifying tan(arcsec(x)) by letting t = arcsec(x). Then, sec(t) = x for t in 0, π, and we seek a formula for tan(t). Since tan(t) is deﬁned for all t values 2 under consideration, we have no additional restrictions on t. To relate sec(t) to tan(t), we use the identity 1 + tan2(t) = sec2(t). This is valid for all values of t under consideration, and when we substitute sec(t) = x, we get 1 + tan2(t) = x2. Hence, tan(t) = ± x2 − 1. Since t lies in 0, π, tan(t) ≥ 0, so we choose tan(t) = x2 − 1. Since we found √ 2 x2 − 1 holds for all x no additional restrictions on t, the equivalence tan(arcsec(x)) = in the domain of t = arcsec(x), namely (−∞, −1] ∪ [1, ∞). ∪ π, 3π 2 √ √ ∪ π, 3π 2 (b) To simplify cos(arccsc(4x)), we start by letting t = arccsc(4x). Then csc(t) = 4x for t in, and we now set about ﬁnding an expression for cos(arccsc(4x)) = cos(t). 0, π 2 Since cos(t) is deﬁned for all t, we do not encounter any additional restrictions on t. From csc(t) = 4x, we get sin(t) = 1 4x, so to �
��nd cos(t), we can make use if the identity cos2(t) + sin2(t) = 1. Substituting sin(t) = 1 = 1. Solving, we get 2 cos(t) = ± 4x gives cos2(t) + 1 √ 4x 16x2 − 1 16x2 = ± 16x2 − 1 4\|x\|, then cos(t) ≥ 0, and we choose cos(t) = in which case cos(t) ≤ 0, so, we choose cos(t) = − If t lies in 0, π 2 to π, 3π 2 (momentarily) piecewise deﬁned function for cos(t) √ 16x2−1 4\|x\| √. Otherwise, t belongs 16x2−1 4\|x\| This leads us to a cos(t) = √ √ −    16x2 − 1 4\|x\| 16x2 − 1 4\|x\|,, if 0 ≤ t ≤ π 2 if π < t ≤ 3π 2 10.6 The Inverse Trigonometric Functions 833 We now see what these restrictions mean in terms of x. Since 4x = csc(t), we get that for 0 ≤ t ≤ π 2, 4x ≥ 1, or x ≥ 1 4. In this case, we can simplify \|x\| = x so √ √ cos(t) = 16x2 − 1 4\|x\| = 16x2 − 1 4x Similarly, for π < t ≤ 3π get 2, we get 4x ≤ −1, or x ≤ − 1 4. In this case, \|x\| = −x, so we also cos(t) = − √ 16x2 − 1 4\|x\| = − √ 16x2 − 1 4(−x) = √ 16x2 − 1 4x 16x2−1 Hence, in all cases, cos(arccsc(4x)) = 4x the domain of t = arccsc(4x), namely −∞, − 1 4, and this equivalence is valid for all x in ∪ 1 4, ∞ √ 10.6.3 Calculators and the Inverse Circular Functions
. In the sections to come, we will have need to approximate the values of the inverse circular functions. On most calculators, only the arcsine, arccosine and arctangent functions are available and they are usually labeled as sin−1, cos−1 and tan−1, respectively. If we are asked to approximate these values, it is a simple matter to punch up the appropriate decimal on the calculator. If we are asked for an arccotangent, arcsecant or arccosecant, however, we often need to employ some ingenuity, as our next example illustrates. Example 10.6.5. 1. Use a calculator to approximate the following values to four decimal places. (a) arccot(2) (b) arcsec(5) (c) arccot(−2) (d) arccsc − 3 2 2. Find the domain and range of the following functions. Check your answers using a calculator. (a) f (x) = π 2 − arccos x 5 (b) f (x) = 3 arctan (4x). (c) f (x) = arccot x 2 + π Solution. 1. (a) Since 2 > 0, we can use the property listed in Theorem 10.27 to rewrite arccot(2) as arccot(2) = arctan 1 2. In ‘radian’ mode, we ﬁnd arccot(2) = arctan 1 2 ≈ 0.4636. (b) Since 5 ≥ 1, we can use the property from either Theorem 10.28 or Theorem 10.29 to write arcsec(5) = arccos 1 5 ≈ 1.3694. 834 Foundations of Trigonometry (c) Since the argument −2 is negative, we cannot directly apply Theorem 10.27 to help us ﬁnd arccot(−2). Let t = arccot(−2). Then t is a real number such that 0 < t < π and cot(t) = −2. Moreover, since cot(t) < 0, we know π 2 < t < π. Geometrically, this means t corresponds to a Quadrant II angle θ = t radians. This allows us to proceed using a ‘reference angle’ approach.
Consider α, the reference angle for θ, as pictured below. By deﬁnition, α is an acute angle so 0 < α < π 2, and the Reference Angle Theorem, Theorem 10.2, tells us that cot(α) = 2. This means α = arccot(2) radians. Since the argument of arccotangent is now a positive 2, we can use Theorem 10.27 to get α = arccot(2) = arctan 1 ≈ 2.6779 radians, 2 we get arccot(−2) ≈ 2.6779. radians. Since θ = π − α = π − arctan 1 2 y 1 θ = arccot(−2) radians α 1 x. By deﬁnition, the real number Another way to attack the problem is to use arctan − 1 2 t = arctan − 1 satisﬁes tan(t) = − 1 2. Since tan(t) < 0, we know 2 more speciﬁcally that − π 2 < t < 0, so t corresponds to an angle β in Quadrant IV. To ﬁnd the value of arccot(−2), we once again visualize the angle θ = arccot(−2) radians and note that it is a Quadrant II angle with tan(θ) = − 1 2. This means it is exactly π units away from β, and we get θ = π + β = π + arctan − 1 ≈ 2.6779 radians. Hence, 2 as before, arccot(−2) ≈ 2.6779. 2 < t < π 2 with − π 10.6 The Inverse Trigonometric Functions 835 y 1 θ = arccot(−2) radians, π (d) If the range of arccosecant is taken to be − π = arcsin − 2 3 ∪ π, 3π 2, we can use Theorem 10.28 to ≈ −0.7297. If, on the other hand, the range of arccosecant get arccsc − 3 2 is taken to be 0, π, then we proceed as in the previous problem by
letting 2. Then t is a real number with csc(t) = − 3 t = arccsc − 3 2. Since csc(t) < 0, we have 2 that π < θ ≤ 3π 2, so t corresponds to a Quadrant III angle, θ. As above, we let α be 2, which means α = arccsc 3 the reference angle for θ. Then 0 < α < π 2 radians. Since the argument of arccosecant is now positive, we may use Theorem 10.29 to get α = arccsc 3 ≈ 3.8713 = arcsin 2 2 3 ≈ 3.8713. radians, arccsc − 3 2 radians. Since θ = π + α = π + arcsin 2 3 2 and csc(α) = 3 y 1 θ = arccsc − 3 2 radians α 1 x 836 2. Foundations of Trigonometry 5 by setting the argument of the arccosine, in this case x (a) Since the domain of F (x) = arccos(x) is −1 ≤ x ≤ 1, we can ﬁnd the domain of 2 − arccos x f (x) = π 5, between −1 and 1. Solving −1 ≤ x 5 ≤ 1 gives −5 ≤ x ≤ 5, so the domain is [−5, 5]. To determine the range of f, we take a cue from Section 1.7. Three ‘key’ points on the graph of and (1, 0). Following the procedure outlined in F (x) = arccos(x) are (−1, π), 0, π 2 Theorem 1.7, we track these points to −5, − π. Plotting these values 2 tells us that the range5 of f is − π, (0, 0) and 5, π 2. Our graph conﬁrms our results. 2, π 2 (b) To ﬁnd the domain and range of f (x) = 3 arctan (4x), we note that since the domain of F (x) = arctan(x) is all real numbers, the only restrictions, if any, on the domain of f (x) = 3 arctan (4x) come from the
argument of the arctangent, in this case, 4x. Since 4x is deﬁned for all real numbers, we have established that the domain of f is all real numbers. To determine the range of f, we can, once again, appeal to Theorem 1.7. Choosing our ‘key’ point to be (0, 0) and tracking the horizontal asymptotes y = − π 2 and y = π 2, we ﬁnd that the graph of y = f (x) = 3 arctan (4x) diﬀers from the graph of y = F (x) = arctan(x) by a horizontal compression by a factor of 4 and a vertical stretch by a factor of 3. It is the latter which aﬀects the range, producing a range of − 3π. We conﬁrm our ﬁndings on the calculator below. 2, 3π 2 y = f (x) = − arccos y = f (x) = 3 arctan (4x) π 2 x 5 (c) To ﬁnd the domain of g(x) = arccot x + π, we proceed as above. Since the domain of 2 G(x) = arccot(x) is (−∞, ∞), and x 2 is deﬁned for all x, we get that the domain of g is (−∞, ∞) as well. As for the range, we note that the range of G(x) = arccot(x), like that of F (x) = arctan(x), is limited by a pair of horizontal asymptotes, in this case y = 0 and y = π. Following Theorem 1.7, we graph y = g(x) = arccot x + π starting with 2 y = G(x) = arccot(x) and ﬁrst performing a horizontal expansion by a factor of 2 and following that with a vertical shift upwards by π. This latter transformation is the one which aﬀects the range, making it now (π, 2π). To check this graphically, we encounter a bit of a problem, since on many calculators, there is no shortcut button corresponding to the arccotangent function. Taking
a cue from number 1c, we attempt to rewrite +π in terms of the arctangent function. Using Theorem 10.27, we have g(x) = arccot x 2 when x = arctan 2 that arccot x 2 > 0, or, in this case, when x > 0. Hence, for x > 0, x 2 + π. When x we have g(x) = arctan 2 2 < 0, we can use the same argument in number x. 1c that gave us arccot(−2) = π + arctan − 1 2 to give us arccot x 2 = π + arctan 2 x 5It also conﬁrms our domain! 10.6 The Inverse Trigonometric Functions 837 Hence, for x < 0, g(x) = π + arctan 2 + 2π. What about x = 0? We x know g(0) = arccot(0) + π = π, and neither of the formulas for g involving arctangent will produce this result.6 Hence, in order to graph y = g(x) on our calculators, we need to write it as a piecewise deﬁned function: + π = arctan 2 x g(x) = arccot x 2 + π =    We show the input and the result below. arctan 2 x arctan 2 x + 2π, π, + π, if x < 0 if x = 0 if x > 0 y = g(x) in terms of arctangent y = g(x) = arccot x 2 + π The inverse trigonometric functions are typically found in applications whenever the measure of an angle is required. One such scenario is presented in the following example. Example 10.6.6. 7 The roof on the house below has a ‘6/12 pitch’. This means that when viewed from the side, the roof line has a rise of 6 feet over a run of 12 feet. Find the angle of inclination from the bottom of the roof to the top of the roof. Express your answer in decimal degrees, rounded to the nearest hundredth of a degree. Front View Side View Solution. If we divide the side view of the house
down the middle, we ﬁnd that the roof line forms the hypotenuse of a right triangle with legs of length 6 feet and 12 feet. Using Theorem 10.10, we 6Without Calculus, of course... 7The authors would like to thank Dan Stitz for this problem and associated graphics. 838 Foundations of Trigonometry ﬁnd the angle of inclination, labeled θ below, satisﬁes tan(θ) = 6 we can use the arctangent function and we ﬁnd θ = arctan 1 2 12 = 1 radians ≈ 26.56◦. 2. Since θ is an acute angle, 6 feet θ 12 feet 10.6.4 Solving Equations Using the Inverse Trigonometric Functions. In Sections 10.2 and 10.3, we learned how to solve equations like sin(θ) = 1 2 for angles θ and tan(t) = −1 for real numbers t. In each case, we ultimately appealed to the Unit Circle and relied on the fact that the answers corresponded to a set of ‘common angles’ listed on page 724. If, on the other hand, we had been asked to ﬁnd all angles with sin(θ) = 1 3 or solve tan(t) = −2 for real numbers t, we would have been hard-pressed to do so. With the introduction of the inverse trigonometric functions, however, we are now in a position to solve these equations. A good parallel to keep in mind is how the square root function can be used to solve certain quadratic equations. The equation x2 = 4 is a lot like sin(θ) = 1 2 in that it has friendly, ‘common value’ answers x = ±2. The equation x2 = 7, on the other hand, is a lot like sin(θ) = 1 3. We know8 there are answers, but we can’t express them using ‘friendly’ numbers.9 To solve x2 = 7, we make use of the square root function and write x = ± 7. We can certainly approximate these answers using a calculator, but as far as exact answers go, we leave them as x = ± 7. In the same way, we will use the arcsine function to solve sin(θ) = 1 3, as seen in
the following example. √ √ Example 10.6.7. Solve the following equations. 1. Find all angles θ for which sin(θ) = 1 3. 2. Find all real numbers t for which tan(t) = −2 3. Solve sec(x) = − 5 3 for x. Solution. 1. If sin(θ) = 1 3, then the terminal side of θ, when plotted in standard position, intersects the Unit Circle at y = 1 3. Geometrically, we see that this happens at two places: in Quadrant I and Quadrant II. If we let α denote the acute solution to the equation, then all the solutions 8How do we know this again? 9This is all, of course, a matter of opinion. For the record, the authors ﬁnd ± √ 7 just as ‘nice’ as ±2. 10.6 The Inverse Trigonometric Functions 839 to this equation in Quadrant I are coterminal with α, and α serves as the reference angle for all of the solutions to this equation in Quadrant II = arcsin 1 3 radians α 1 x Since 1 3 isn’t the sine of any of the ‘common angles’ discussed earlier, we use the arcsine functions to express our answers. The real number t = arcsin 1 is deﬁned so it satisﬁes 3 radians. Since the solutions in Quadrant I 0 < t < π + 2πk are all coterminal with α, we get part of our solution to be θ = α + 2πk = arcsin 1 3 for integers k. Turning our attention to Quadrant II, we get one solution to be π − α. Hence, the Quadrant II solutions are θ = π − α + 2πk = π − arcsin 1 3 3. Hence, α = arcsin 1 + 2πk, for integers k. 2 with sin(t) = 1 3 2. We may visualize the solutions to tan(t) = −2 as angles θ with tan(θ) = −2. Since tangent is negative only in Quadrants II and IV, we focus our eﬀorts there = arctan(−2) radians π β Since −2 isn’t the tangent of any of
the ‘common angles’, we need to use the arctangent function to express our answers. The real number t = arctan(−2) satisﬁes tan(t) = −2 and − π 2 < t < 0. If we let β = arctan(−2) radians, we see that all of the Quadrant IV solutions 840 Foundations of Trigonometry to tan(θ) = −2 are coterminal with β. Moreover, the solutions from Quadrant II diﬀer by exactly π units from the solutions in Quadrant IV, so all the solutions to tan(θ) = −2 are of the form θ = β + πk = arctan(−2) + πk for some integer k. Switching back to the variable t, we record our ﬁnal answer to tan(t) = −2 as t = arctan(−2) + πk for integers k. 3. The last equation we are asked to solve, sec(x) = − 5 3, poses two immediate problems. First, we are not told whether or not x represents an angle or a real number. We assume the latter, but note that we will use angles and the Unit Circle to solve the equation regardless. Second, as we have mentioned, there is no universally accepted range of the arcsecant function. For that reason, we adopt the advice given in Section 10.3 and convert this to the cosine problem cos(x) = − 3 5. Adopting an angle approach, we consider the equation cos(θ) = − 3 5 and note that our solutions lie in Quadrants II and III. Since − 3 5 isn’t the cosine of any of the ‘common angles’, we’ll need to express our solutions in terms of the arccosine function. The real number 5. If we let β = arccos − 3 t = arccos − 3 5 radians, we see that β is a Quadrant II angle. To obtain a Quadrant III angle solution, we may simply use −β = − arccos − 3. Since all angle solutions are coterminal with β 5 + 2πk or 5 to be θ = β + 2πk = arccos − 3 or −β, we get our solutions to
cos(θ) = − 3 + 2πk for integers k. Switching back to the variable x, we θ = −β + 2πk = − arccos − 3 5 + 2πk record our ﬁnal answer to sec(x) = − 5 for integers k. + 2πk or x = − arccos − 3 5 2 < t < π with cos(t) = − 3 3 as x = arccos − 3 is deﬁned so that π 5 5 5 y 1 y 1 β = arccos − 3 5 radians β = arccos − 3 5 radians 1 x 1 −β = − arccos − 3 5 x radians The reader is encouraged to check the answers found in Example 10.6.7 - both analytically and with the calculator (see Section 10.6.3). With practice, the inverse trigonometric functions will become as familiar to you as the square root function. Speaking of practice... 10.6 The Inverse Trigonometric Functions 841 10.6.5 Exercises In Exercises 1 - 40, ﬁnd the exact value. 1. arcsin (−1) 2. arcsin − √ 3 2 5. arcsin (0) 6. arcsin 1 2 3. arcsin − √ 2 2 7. arcsin √ 2 2 4. arcsin − 1 2 8. arcsin √ 3 2 9. arcsin (1) 10. arccos (−1) 11. arccos − √ 3 2 12. arccos − √ 2 2 13. arccos − 1 2 17. arccos √ 3 2 21. arctan − √ 3 3 14. arccos (0) 15. arccos 1 2 16. arccos √ 2 2 18. arccos (1) 19. arctan − √ 3 20. arctan (−1) 22. arctan (0) 23. arctan 25. arctan √ 3 26. arccot − √ 3 29. arccot (0) 30. arccot √ 3 3 33. arcsec (2) 34. arccsc (2) 37. arcsec √ 2 3 3 38. arccsc √ 2 3 3 √ 3 3 24
. arctan (1) 27. arccot (−1) 28. arccot − √ 3 3 31. arccot (1) 32. arccot √ 3 35. arcsec √ 2 36. arccsc √ 2 39. arcsec (1) 40. arccsc (1) In Exercises 41 - 48, assume that the range of arcsecant is 0, π 2 arccosecant is 0, π 2 when ﬁnding the exact value. ∪ π, 3π 2 ∪ π, 3π 2 and that the range of 41. arcsec (−2) 45. arccsc (−2) 42. arcsec − √ 46. arccsc − √ 2 2 43. arcsec − 47. arccsc − √ 2 3 3 √ 2 3 3 44. arcsec (−1) 48. arccsc (−1) 842 Foundations of Trigonometry In Exercises 49 - 56, assume that the range of arcsecant is 0, π 2 arccosecant is − π when ﬁnding the exact value. 2, 0 ∪ 0, π 2 ∪ π 2, π and that the range of 49. arcsec (−2) 53. arccsc (−2) 50. arcsec − √ 54. arccsc − √ 2 2 51. arcsec − 55. arccsc − √ 2 3 3 √ 2 3 3 52. arcsec (−1) 56. arccsc (−1) In Exercises 57 - 86, ﬁnd the exact value or state that it is undeﬁned. 57. sin arcsin 1 2 58. sin arcsin − √ 2 2 60. sin (arcsin (−0.42)) 61. sin arcsin 5 4 63. cos arccos − 1 2 64. cos arccos 5 13 66. cos (arccos (π)) 67. tan (arctan (−1)) 59. sin arcsin 3 5 √ 62. cos arccos 2 2 65. cos (arccos (−0.998)) 68. tan arctan √ 3 70. tan (arctan (0.965)) 71. tan (arctan (3π)) 69. tan arctan 5 12
72. cot (arccot (1)) 73. cot arccot − √ 3 75. cot (arccot (−0.001)) 76. cot arccot 78. sec (arcsec (−1)) 79. sec arcsec 17π 4 1 2 82. csc arccsc √ 2 81. sec (arcsec (117π)) 84. csc arccsc √ 2 2 74. cot arccot − 7 24 77. sec (arcsec (2)) 80. sec (arcsec (0.75)) 83. csc arccsc − √ 2 3 3 85. csc (arccsc (1.0001)) 86. csc arccsc π 4 In Exercises 87 - 106, ﬁnd the exact value or state that it is undeﬁned. 87. arcsin sin π 6 88. arcsin − sin π 3 89. arcsin sin 3π 4 10.6 The Inverse Trigonometric Functions 843 90. arcsin sin 93. arccos 96. arccos cos cos 11π 6 2π 3 5π 4 91. arcsin sin 94. arccos 97. arctan 4π 3 3π 2 cos 92. arccos 95. arccos cos π 4 cos − π 6 tan tan π 3 π 2 98. arctan tan 101. arctan tan − π 4 2π 3 99. arctan (tan (π)) 100. arctan 102. arccot 105. arccot cot cot π 3 π 2 103. arccot cot 106. arccot cot − π 4 2π 3 104. arccot (cot (π)) In Exercises 107 - 118, assume that the range of arcsecant is 0, π 2 arccosecant is 0, π 2 when ﬁnding the exact value. ∪ π, 3π 2 117. arcsec sec 118. arccsc csc In Exercises 119 - 130, assume that the range of arcsecant is 0, π 2 arccosecant is − π when ﬁnding the exact value. 2, 0 ∪ 0, π 2 sec π 4 107. arcsec 110. arcsec sec 113. ar
ccsc csc 116. arccsc csc − π 2 5π 4 11π 6 sec π 4 119. arcsec 122. arcsec sec 125. arccsc csc 128. arccsc csc − π 2 5π 4 11π 6 108. arcsec sec 111. arcsec sec 114. arccsc csc 120. arcsec sec 123. arcsec sec 126. arccsc csc 4π 3 5π 3 2π 3 11π 12 4π 3 5π 3 2π 3 11π 12 ∪ π, 3π 2 and that the range of 109. arcsec sec 5π 6 csc π 6 112. arccsc 115. arccsc csc − π 2 9π 8 ∪ π 2, π and that the range of 121. arcsec sec 5π 6 csc π 6 124. arccsc 127. arccsc csc − π 2 9π 8 129. arcsec sec 130. arccsc csc 844 Foundations of Trigonometry In Exercises 131 - 154, ﬁnd the exact value or state that it is undeﬁned. 131. sin arccos − 1 2 132. sin arccos 3 5 134. sin arccot √ 5 137. cos arctan √ 7 135. sin (arccsc (−3)) 140. tan arcsin − √ 2 5 5 141. tan arccos − 1 2 143. tan (arccot (12)) 144. cot arcsin 12 13 138. cos (arccot (3)) 139. cos (arcsec (5)) 133. sin (arctan (−2)) 136. cos arcsin − 5 13 142. tan arcsec 145. cot arccos 5 3 √ 3 2 √ 3 2 146. cot arccsc √ 5 147. cot (arctan (0.25)) 148. sec arccos 149. sec arcsin − 12 13 150. sec (arctan (10)) 151. sec arccot − √ 10 10 152. csc (arccot (9)) 153. csc arcsin 3 5 154. csc arctan − 2 3 In Exercises 155 - 164, ﬁnd the exact value or state that it is undeﬁned. 155. sin arcs
in 5 13 + π 4 156. cos (arcsec(3) + arctan(2)) 157. tan arctan(3) + arccos − 3 5 158. sin 2 arcsin − 4 5 159. sin 2arccsc 13 5 161. cos 2 arcsin 3 5 163. cos 2arccot − √ 5 160. sin (2 arctan (2)) 162. cos 2arcsec 25 7 164. sin arctan(2) 2 10.6 The Inverse Trigonometric Functions 845 In Exercises 165 - 184, rewrite the quantity as algebraic expressions of x and state the domain on which the equivalence is valid. 165. sin (arccos (x)) 166. cos (arctan (x)) 167. tan (arcsin (x)) 168. sec (arctan (x)) 169. csc (arccos (x)) 170. sin (2 arctan (x)) 171. sin (2 arccos (x)) 172. cos (2 arctan (x)) 173. sin(arccos(2x)) 174. sin arccos x 5 175. cos arcsin x 2 176. cos (arctan (3x)) 177. sin(2 arcsin(7x)) 178. sin 2 arcsin √ x 3 3 179. cos(2 arcsin(4x)) 180. sec(arctan(2x)) tan(arctan(2x)) 181. sin (arcsin(x) + arccos(x)) 182. cos (arcsin(x) + arctan(x)) 183. tan (2 arcsin(x)) 184. sin arctan(x) 1 2 185. If sin(θ) = 186. If tan(θ) = 187. If sec(θ) = x 2 x 7 x 4 for − π 2 for − π 2 < θ < π 2, ﬁnd an expression for θ + sin(2θ) in terms of x. < θ < π 2, ﬁnd an expression for 1 2 θ − 1 2 sin(2θ) in terms of x. for 0 < θ < π 2, ﬁnd an expression for 4 tan(θ) − 4θ in terms of x. In Ex
ercises 188 - 207, solve the equation using the techniques discussed in Example 10.6.7 then approximate the solutions which lie in the interval [0, 2π) to four decimal places. 188. sin(x) = 7 11 189. cos(x) = − 2 9 190. sin(x) = −0.569 191. cos(x) = 0.117 192. sin(x) = 0.008 193. cos(x) = 194. tan(x) = 117 195. cot(x) = −12 196. sec(x) = 197. csc(x) = − 200. cos(x) = − 90 17 7 16 198. tan(x) = − √ 10 199. sin(x) = 201. tan(x) = 0.03 202. sin(x) = 0.3502 359 360 3 2 3 8 846 Foundations of Trigonometry 203. sin(x) = −0.721 204. cos(x) = 0.9824 205. cos(x) = −0.5637 206. cot(x) = 1 117 207. tan(x) = −0.6109 In Exercises 208 - 210, ﬁnd the two acute angles in the right triangle whose sides have the given lengths. Express your answers using degree measure rounded to two decimal places. 208. 3, 4 and 5 209. 5, 12 and 13 210. 336, 527 and 625 211. A guy wire 1000 feet long is attached to the top of a tower. When pulled taut it touches level ground 360 feet from the base of the tower. What angle does the wire make with the ground? Express your answer using degree measure rounded to one decimal place. 212. At Cliﬀs of Insanity Point, The Great Sasquatch Canyon is 7117 feet deep. From that point, a ﬁre is seen at a location known to be 10 miles away from the base of the sheer canyon wall. What angle of depression is made by the line of sight from the canyon edge to the ﬁre? Express your answer using degree measure rounded to one decimal place. 213. Shelving is being built at the Utility Muﬃn Research Library which is to be 14 inches deep. An 18-inch rod will be attached to the wall and the underside of the shelf at its edge away from the wall, forming
a right triangle under the shelf to support it. What angle, to the nearest degree, will the rod make with the wall? 214. A parasailor is being pulled by a boat on Lake Ippizuti. The cable is 300 feet long and the parasailor is 100 feet above the surface of the water. What is the angle of elevation from the boat to the parasailor? Express your answer using degree measure rounded to one decimal place. 215. A tag-and-release program to study the Sasquatch population of the eponymous Sasquatch National Park is begun. From a 200 foot tall tower, a ranger spots a Sasquatch lumbering through the wilderness directly towards the tower. Let θ denote the angle of depression from the top of the tower to a point on the ground. If the range of the riﬂe with a tranquilizer dart is 300 feet, ﬁnd the smallest value of θ for which the corresponding point on the ground is in range of the riﬂe. Round your answer to the nearest hundreth of a degree. In Exercises 216 - 221, rewrite the given function as a sinusoid of the form S(x) = A sin(ωx + φ) using Exercises 35 and 36 in Section 10.5 for reference. Approximate the value of φ (which is in radians, of course) to four decimal places. 216. f (x) = 5 sin(3x) + 12 cos(3x) 217. f (x) = 3 cos(2x) + 4 sin(2x) 218. f (x) = cos(x) − 3 sin(x) 219. f (x) = 7 sin(10x) − 24 cos(10x) 10.6 The Inverse Trigonometric Functions 847 220. f (x) = − cos(x) − 2 √ 2 sin(x) 221. f (x) = 2 sin(x) − cos(x) In Exercises 222 - 233, ﬁnd the domain of the given function. Write your answers in interval notation. 222. f (x) = arcsin(5x) 223. f (x) = arccos 3x − 1 2 224. f (x) = arcsin 2x2 225. f (x) = arccos 1 x2 − 4 226.
f (x) = arctan(4x) 227. f (x) = arccot 2x x2 − 9 228. f (x) = arctan(ln(2x − 1)) 229. f (x) = arccot( √ 2x − 1) 230. f (x) = arcsec(12x) 231. f (x) = arccsc(x + 5) 232. f (x) = arcsec x3 8 233. f (x) = arccsc e2x 234. Show that arcsec(x) = arccos of f (x) = arcsec(x). 1 x for \|x\| ≥ 1 as long as we use 0, ∪ π 2, π π 2 as the range 235. Show that arccsc(x) = arcsin of f (x) = arccsc(x). 1 x for \|x\| ≥ 1 as long as we use − π 2, 0 0, ∪ π 2 as the range 236. Show that arcsin(x) + arccos(x) = π 2 for −1 ≤ x ≤ 1. 237. Discuss with your classmates why arcsin 1 2 = 30◦. 238. Use the following picture and the series of exercises on the next page to show that arctan(1) + arctan(2) + arctan(3) = π y D(2, 3) A(0, 1) α β γ B(1, 0) x C(2, 0) O(0, 0) 848 Foundations of Trigonometry (a) Clearly AOB and BCD are right triangles because the line through O and A and the line through C and D are perpendicular to the x-axis. Use the distance formula to show that BAD is also a right triangle (with ∠BAD being the right angle) by showing that the sides of the triangle satisfy the Pythagorean Theorem. (b) Use AOB to show that α = arctan(1) (c) Use BAD to show that β = arctan(2) (d) Use BCD to show that γ = arctan(3) (e) Use the fact that O, B and C all lie on the x-axis to conclude that α + β + γ = π. Thus
arctan(1) + arctan(2) + arctan(3) = π. 10.6 The Inverse Trigonometric Functions 849 10.6.6 Answers 1. arcsin (−1) = − π 2 2. arcsin − √ 3 2 = − π 3 3. arcsin − √ 2 2 = − π 4 4. arcsin 7. arcsin −. arcsin (0) = 0 8. arcsin √ 3 2 10. arccos (−1) = π 11. arccos − √ 3 2 = π 3 = 5π 6 13. arccos 16. arccos = 2π 14. arccos (0) = 17. arccos √ 3 2 π 2 = π 6 19. arctan − √ 3 = − π 3 20. arctan (−1) = − π 4 22. arctan (0) = 0 23. arctan √ 3 3 = π 6 25. arctan √ 3 = √ π 3 28. arccot − = 3 3 26. arccot − √ 3 = 5π 6 2π 3 29. arccot (0) = π 2 31. arccot (1) = 34. arccsc (2) = 37. arcsec √ 2 3 3 π 4 π 6 32. arccot √ 35. arcsec √ 38. arccsc. arcsin 1 2 = π 6 9. arcsin (1) = π 2 12. arccos − √ 2 2 = 3π 4 15. arccos 1 2 = π 3 18. arccos (1) = 0 21. arctan − √ 3 3 = − π 6 24. arctan (1) = π 4 27. arccot (−1) = 3π 4 30. arccot √ 3 3 = π 3 33. arcsec (2) = 36. arccsc √ π 3 2 = π 4 39. arcsec (1) = 0 40. arccsc (1) = π 2 41. arcsec (−2) = 4π 3 42. arcsec − √ 2 = 5π 4 850 Foundations of Trigonometry 44. arcsec
(−1) = π 45. arccsc (−2) = 47. arccsc − √ 3 2 3 = 4π 3 48. arccsc (−1) = 7π 6 3π 2 50. arcsec − 2 = √ 51. arcsec 2 − √ 3 3 = 5π 6 54. arccsc − √ 2 = − π 4 3π 4 π 6 π 2 52. arcsec (−1) = π 53. arccsc (−2) = − 55. arccsc − √ 3 2 = − π 3 56. arccsc (−1) = − 43. arcsec − √ 3 2 3 = 7π 6 46. arccsc − √ 2 = 5π 4 49. arcsec (−2) = 2π 3 57. sin arcsin 59. sin arcsin 61. sin arcsin 58. sin arcsin − √ 2 2 √ 2 2 = − 60. sin (arcsin (−0.42)) = −0.42 √ 2 2 √ 2 2 = is undeﬁned. 62. cos arccos 63. cos arccos − 1 2 = − 1 2 65. cos (arccos (−0.998)) = −0.998 67. tan (arctan (−1)) = −1 69. tan arctan 5 12 = 5 12 64. cos arccos 5 13 = 5 13 66. cos (arccos (π)) is undeﬁned. 68. tan arctan √ 3 = √ 3 70. tan (arctan (0.965)) = 0.965 71. tan (arctan (3π)) = 3π 72. cot (arccot (1)) = 1 73. cot arccot − √ √ 3 = − 3 75. cot (arccot (−0.001)) = −0.001 74. cot arccot 76. cot arccot − 7 24 17π 4 = − 7 24 = 17π 4 77. sec (arcsec (2)) = 2 78. sec (arcsec (−1)) = −1 10.6 The Inverse Trigonometric Functions 851 79. sec arcsec 1 2 is undeﬁned. 81. sec (arcsec (117π)) = 117π 83. csc
arccsc 2 − √ 3 3 √ 3 2 3 = − 85. csc (arccsc (1.0001)) = 1.0001 87. arcsin sin 89. arcsin sin 91. arcsin sin 93. arccos cos π 6 3π 4 4π 3 2π 3 95. arccos − cos π 6 = π 4 = − π 3 = 2π 3 = π 6 = π 6 97. arctan tan π 3 = π 3 99. arctan (tan (π)) = 0 101. arctan tan 2π 3 = − π 3 80. sec (arcsec (0.75)) is undeﬁned. 82. csc arccsc √ √ 2 = √ 2 84. csc arccsc is undeﬁned. 2 2 86. csc arccsc π 4 is undeﬁned. 88. arcsin sin 90. arcsin sin − π 3 11π 6 = − π 3 = − π 6 92. arccos cos 94. arccos cos 96. arccos cos π 4 3π 2 5π 4 = − tan 98. arctan 100. arctan tan π 4 π 4 = = π 2 3π 4 = − π 4 102. arccot cot π 2 π 3 is undeﬁned = π 3 103. arccot cot 105. arccot cot − π 4 3π 2 = 107. arcsec sec 109. arcsec sec 111. arcsec sec π 4 5π 6 5π 3 = 3π 4 π 2 7π 6 π 3 104. arccot (cot (π)) is undeﬁned 106. arccot cot 108. arcsec sec 2π 3 4π 3 = = 2π 3 4π 3 110. arcsec 112. arccsc sec − π 2 is undeﬁned. csc π 6 = π 6 = π 4 = = 852 Foundations of Trigonometry 113. arccsc csc 5π 4 = 5π 4 115. arccsc csc 117. arcsec sec 119. arcsec sec 121. arcsec sec 123. arcsec sec 125. arccsc csc 127. arccsc
csc 129. arcsec sec − π 2 11π 12 = 3π 2 = 13π 12 π 4 = = 5π 5π 6 5π 3 5π 4 − π 2 11π 12 = 11π 12 131. sin arccos − = 1 2 133. sin (arctan (−2)) = − 135. sin (arccsc (−3)) = − 137. cos arctan √ 7 = 139. cos (arcsec (5)) = 1 5 114. arccsc csc 116. arccsc csc 118. arccsc csc 120. arcsec sec 2π 3 11π 6 9π 8 4π 3 = π 3 = 7π 6 = = 9π 8 2π 3 122. arcsec sec − π 2 is undeﬁned csc csc csc 126. arccsc 128. arccsc 124. arccsc π 6 2π 3 11π 6 9π 8 3 5 134. sin arccot √ 130. arccsc 132. sin arccos csc 136. cos arcsin − 5 = = 4 5 √ 6 6 5 13 = 12 13 √ 3 10 138. cos (arccot (3)) = 140. tan arcsin − √ 2 5 = −2 10 5 = 4 3 141. tan arccos − √ 3 = − 1 2 143. tan (arccot (12)) = 1 12 142. tan arcsec 144. cot arcsin 5 3 12 13 = 5 12 10.6 The Inverse Trigonometric Functions 853 145. cot arccos √ 3 2 √ 3 = 147. cot (arctan (0.25)) = 4 = 13 5 149. sec arcsin − 12 13 √ 151. sec arccot − 10 10 153. csc arcsin 155. sin arcsin 3 5 = 5 3 5 13 + π 4 = 157. tan arctan(3) + arccos − 146. cot arccsc √ 5 = 2 148. sec arccos √ 3 2 √ 3 2 3 = 150. sec (arctan (10)) = √ 101 √ = − 11 152. csc (arccot (9)) = √ 82 154. csc arctan − 2 3 = − √ 13 2 156. cos (
arcsec(3) + arctan(2)) = √ 10 √ 5 − 4 15 √ 17 2 26 3 5 = 1 3 158. sin 2 arcsin − 4 5 = − 24 25 159. sin 2arccsc 161. cos 2 arcsin 13 5 3 5 = 120 169 = 7 25 160. sin (2 arctan (2)) = 4 5 162. cos 2arcsec 25 7 = − 163. cos 2arccot − √ 5 = 2 3 165. sin (arccos (x)) = √ 1 − x2 for −1 ≤ x ≤ 1 164. sin arctan(2) 2 = 5 − 10 527 625 √ 5 166. cos (arctan (x)) = √ 167. tan (arcsin (x)) = √ 1 1 + x2 x 1 − x2 for all x for −1 < x < 1 168. sec (arctan (x)) = √ 1 + x2 for all x 169. csc (arccos (x)) = √ 1 1 − x2 for −1 < x < 1 170. sin (2 arctan (x)) = for all x 2x x2 + 1 √ 171. sin (2 arccos (x)) = 2x 1 − x2 for −1 ≤ x ≤ 1 854 Foundations of Trigonometry 172. cos (2 arctan (x)) = 173. sin(arccos(2x)) = 174. sin arccos 175. cos arcsin x 5 x 2 = = √ 1 − x2 1 + x2 for all x √ 1 − 4x2 for − 1 √ 25 − x2 5 2 ≤ x ≤ 1 2 for −5 ≤ x ≤ 5 4 − x2 2 for −2 ≤ x ≤ 2 176. cos (arctan (3x)) = √ for all x 1 1 + 9x2 √ 177. sin(2 arcsin(7x)) = 14x 1 − 49x2 for − 1 7 ≤ x ≤ 1 7 178. sin 2 arcsin √ x 3 3 √ 2x = 3 − x2 3 √ for − 3 ≤ x ≤ √ 3 179. cos(2 arcsin(4x)) = 1 − 32x2 for − 1 4 1 4 ≤ x ≤ √ 180. sec(ar
ctan(2x)) tan(arctan(2x)) = 2x 1 + 4x2 for all x 181. sin (arcsin(x) + arccos(x)) = 1 for −1 ≤ x ≤ 1 182. cos (arcsin(x) + arctan(x)) = √ 1 − x2 − x2 √ 1 + x2 for −1 ≤ x ≤ 1 183. 10 tan (2 arcsin(x)) = √ 2x 1 − x2 1 − 2x2 for x in −1 184. sin arctan(x) = 1 2    √ x2 + 1 − 1 √ x2 + 1 2 √ x2 + 1 − 1 √ x2 + 1 2 − for x ≥ 0 for x < 0 185. If sin(θ) = 186. If tan(θ) = x 2 x 7 for − π 2 for − π 2 < θ < π 2, then θ + sin(2θ) = arcsin x 2 x + √ 4 − x2 2 < θ < π 2, then 1 2 θ − 1 2 sin(2θ) = 1 2 arctan x 7 − 7x x2 + 49 10The equivalence for x = ±1 can be veriﬁed independently of the derivation of the formula, but Calculus is required to fully understand what is happening at those x values. You’ll see what we mean when you work through the details of the identity for tan(2t). For now, we exclude x = ±1 from our answer. 10.6 The Inverse Trigonometric Functions 855 187. If sec(θ) = 188. x = arcsin x 4 7 11 for 0 < θ < π 2, then 4 tan(θ) − 4θ = √ x2 − 16 − 4arcsec x 4 + 2πk or x = π − arcsin 7 11 + 2πk, in [0, 2π), x ≈ 0.6898, 2.4518 189. x = arccos − 2 9 + 2πk or x = − arccos − 2 9 +
2πk, in [0, 2π), x ≈ 1.7949, 4.4883 190. x = π + arcsin(0.569) + 2πk or x = 2π − arcsin(0.569) + 2πk, in [0, 2π), x ≈ 3.7469, 5.6779 191. x = arccos(0.117) + 2πk or x = 2π − arccos(0.117) + 2πk, in [0, 2π), x ≈ 1.4535, 4.8297 192. x = arcsin(0.008) + 2πk or x = π − arcsin(0.008) + 2πk, in [0, 2π), x ≈ 0.0080, 3.1336 193. x = arccos 359 360 + 2πk or x = 2π − arccos 359 360 + 2πk, in [0, 2π), x ≈ 0.0746, 6.2086 194. x = arctan(117) + πk, in [0, 2π), x ≈ 1.56225, 4.70384 195. x = arctan − 1 12 + πk, in [0, 2π), x ≈ 3.0585, 6.2000 196. x = arccos 2 3 + 2πk or x = 2π − arccos 2 3 + 2πk, in [0, 2π), x ≈ 0.8411, 5.4422 197. x = π + arcsin 198. x = arctan − 3 8 199. x = arcsin 200. x = arccos − 7 16 + 2πk or x = 2π − arcsin 17 90 10 + πk, in [0, 2π), x ≈ 1.8771, 5.0187 17 90 √ + 2πk or x = π − arcsin 3 8 + 2πk, in [0, 2π), x ≈ 0.3844, 2.7572 7 16 + 2πk, in [0, 2π), x ≈ 2.0236, 4.2596 + 2πk or x = − arccos − + 2πk, in [0
, 2π), x ≈ 3.3316, 6.0932 201. x = arctan(0.03) + πk, in [0, 2π), x ≈ 0.0300, 3.1716 202. x = arcsin(0.3502) + 2πk or x = π − arcsin(0.3502) + 2πk, in [0, 2π), x ≈ 0.3578, 2.784 203. x = π + arcsin(0.721) + 2πk or x = 2π − arcsin(0.721) + 2πk, in [0, 2π), x ≈ 3.9468, 5.4780 204. x = arccos(0.9824) + 2πk or x = 2π − arccos(0.9824) + 2πk, in [0, 2π), x ≈ 0.1879, 6.0953 205. x = arccos(−0.5637) + 2πk or x = − arccos(−0.5637) + 2πk, in [0, 2π), x ≈ 2.1697, 4.1135 206. x = arctan(117) + πk, in [0, 2π), x ≈ 1.5622, 4.7038 207. x = arctan(−0.6109) + πk, in [0, 2π), x ≈ 2.5932, 5.7348 856 Foundations of Trigonometry 208. 36.87◦ and 53.13◦ 209. 22.62◦ and 67.38◦ 210. 32.52◦ and 57.48◦ 211. 68.9◦ 212. 7.7◦ 213. 51◦ 216. f (x) = 5 sin(3x) + 12 cos(3x) = 13 sin 3x + arcsin 214. 19.5◦ 12 13 ≈ 13 sin(3x + 1.1760) 215. 41.81◦ 217. f (x) = 3 cos(2x) + 4 sin(2x) = 5 sin 2x + arcsin 3 5 ≈ 5 sin(2x + 0.6435
) 218. f (x) = cos(x) − 3 sin(x) = √ 10 sin x + arccos − 10 219. f (x) = 7 sin(10x) − 24 cos(10x) = 25 sin 10x + arcsin − √ 3 10 √ ≈ 10 sin(x + 2.8198) 24 25 ≈ 25 sin(10x − 1.2870) 220. f (x) = − cos(x) − 2 √ 2 sin(x) = 3 sin x + π + arcsin 221. f (x) = 2 sin(x) − cos(x) = √ 5 sin x + arcsin − 1 3 ≈ 3 sin(x + 3.4814) √ ≈ 5 sin(x − 0.4636) √ 5 5 223. −, 1 1 3 1 5 222. − 224. −, 1 5 √ √ 2 2, 225. (−∞, − √ √ 5] ∪ [− √ 3, √ 3] ∪ [ 5, ∞) 2 2 226. (−∞, ∞) 227. (−∞, −3) ∪ (−3, 3) ∪ (3, ∞) 228., ∞ 1 2 230. −∞, − 1 12 ∪ 1 12, ∞ 229., ∞ 1 2 231. (−∞, −6] ∪ [−4, ∞) 232. (−∞, −2] ∪ [2, ∞) 233. [0, ∞) 10.7 Trigonometric Equations and Inequalities 857 10.7 Trigonometric Equations and Inequalities In Sections 10.2, 10.3 and most recently 10.6, we solved some basic equations involving the trigonometric functions. Below we summarize the techniques we’ve employed thus far. Note that we use the neutral letter ‘u’ as the argument1 of each circular function for generality. Strategies for Solving Basic Equations Involving Trigonometric Functions To solve cos(u) = c or sin(u) = c for −1 ≤ c ≤ 1, ﬁrst solve for u in the interval [0, 2π) and add integer multiples of the period 2π. If c < −1 or of c
> 1, there are no real solutions. To solve sec(u) = c or csc(u) = c for c ≤ −1 or c ≥ 1, convert to cosine or sine, respectively, and solve as above. If −1 < c < 1, there are no real solutions. To solve tan(u) = c for any real number c, ﬁrst solve for u in the interval − π 2, π 2 and add integer multiples of the period π. To solve cot(u) = c for c = 0, convert to tangent and solve as above. If c = 0, the solution to cot(u) = 0 is u = π 2 + πk for integers k. 2 and ﬁnd the solution x = π 2, we know the solutions take the form u = π 6 + 2πk for integers k. How do we solve something like sin(3x) = 1 Using the above guidelines, we can comfortably solve sin(x) = 1 or x = 5π has the form sin(u) = 1 integers k. Since the argument of sine here is 3x, we have 3x = π integers k. To solve for x, we divide both sides2 of these equations by 3, and obtain x = π or x = 5π 6 + 2πk 2? Since this equation 6 + 2πk for 6 + 2πk for 18 + 2π 3 k 3 k for integers k. This is the technique employed in the example below. Example 10.7.1. Solve the following equations and check your answers analytically. List the solutions which lie in the interval [0, 2π) and verify them using a graphing utility. 2. csc 1 6 + 2πk or u = 5π 6 + 2πk or 3x = 5π 3. cot (3x) = 0 18 + 2π 1. cos(2x3 5. tan x 2 6. sin(2x) = 0.87 4. sec2(x) = 4 Solution. 1. The solutions to cos(u) = − √ the argument of cosine here is 2x, this means 2x = 5π k. Solving for x gives x = 5π analytically, we substitute them into the original equation. For any integer k we have 6 +
2πk for integers k. Since 6 + 2πk for integers 12 + πk for integers k. To check these answers 6 + 2πk or 2x = 7π 12 + πk or x = 7π 6 + 2πk or u = 7π 2 are u = 5π 3 cos 2 5π 12 + πk = cos 5π = cos 5π 6 √ = − 3 2 6 + 2πk (the period of cosine is 2π) 1See the comments at the beginning of Section 10.5 for a review of this concept. 2Don’t forget to divide the 2πk by 3 as well! 858 Foundations of Trigonometry 12 + πk = cos 7π √ Similarly, we ﬁnd cos 2 7π 3 2. To determine which of our solutions lie in [0, 2π), we substitute integer values for k. The solutions we keep come from the values of k = 0 and k = 1 and are x = 5π 12, 17π 12. Using a √ 3 calculator, we graph y = cos(2x) and y = − 2 over [0, 2π) and examine where these two graphs intersect. We see that the x-coordinates of the intersection points correspond to the decimal representations of our exact answers. 6 + 2πk = cos 7π 12 and 19π = − 12, 7π 6 2. Since this equation has the form csc(u) = 2, we rewrite this as sin(u) = u = π 4 + 2πk or u = 3π 4 + 2πk for integers k. Since the argument of cosecant here is 1 √ 2 2 and ﬁnd 3 x − π + 2πk or 1 3 x − π = 3π 4 + 2πk To solve 1 3 x − π = π 4 + 2πk, we ﬁrst add π to both sides 1 3 x = π 4 + 2πk + π A common error is to treat the ‘2πk’ and ‘π’ terms as ‘like’ terms and try to combine them when they are not.3 We can, however, combine the ‘π’ and ‘ π 4 ’ terms to get We now ﬁnish by multiplying both sides by 3
to get 1 3 x = 5π 4 + 2πk x = 3 5π 4 + 2πk = 15π 4 + 6πk Solving the other equation, 1 check the ﬁrst family of answers, we substitute, combine line terms, and simplify. 4 + 2πk produces x = 21π 3 x − π = 3π 4 + 6πk for integers k. To csc 1 3 15π 4 + 6πk − π = csc 5π = csc π = csc π √ 4 2 = 4 + 2πk − π 4 + 2πk (the period of cosecant is 2π) The family x = 21π 4 + 6πk checks similarly. Despite having inﬁnitely many solutions, we ﬁnd that none of them lie in [0, 2π). To verify this graphically, we use a reciprocal identity to 2 do not intersect at rewrite the cosecant as a sine and we ﬁnd that y = and y = √ 1 sin( 1 3 x−π) all over the interval [0, 2π). 3Do you see why? 10.7 Trigonometric Equations and Inequalities 859 y = cos(2x) and y = − √ 3 2 y = 1 sin( 1 3 x−π) and y = √ 2 3. Since cot(3x) = 0 has the form cot(u) = 0, we know u = π 2 + πk, so, in this case, 3x = π 2 + πk for integers k. Solving for x yields x = π 3 k. Checking our answers, we get 6 + π 2 + πk cot 3 π 6 + π 3 k = cot π = cot π 2 = 0 (the period of cotangent is π) As k runs through the integers, we obtain six answers, corresponding to k = 0 through k = 5, which lie in [0, 2π): x = π 2 and 11π 6. To conﬁrm these graphically, we must be careful. On many calculators, there is no function button for cotangent. We choose4 to use sin(3x). Graphing y = cos(3x) the
quotient identity cot(3x) = cos(3x) sin(3x) and y = 0 (the x-axis), we see that the x-coordinates of the intersection points approximately match our solutions. 6, 3π 6, 7π 2, 5π 6, π 4. The complication in solving an equation like sec2(x) = 4 comes not from the argument of secant, which is just x, but rather, the fact the secant is being squared. To get this equation to look like one of the forms listed on page 857, we extract square roots to get sec(x) = ±2. Converting to cosines, we have cos(x) = ± 1 3 + 2πk or x = 5π 3 + 2πk for integers k. For cos(x) = − 1 3 + 2πk for integers k. If we take a step back and think of these families of solutions geometrically, we see we are ﬁnding the measures of all angles with a reference angle of π 3. As a result, these solutions can be combined and we may write our solutions as x = π 3 + πk for integers k. To check the ﬁrst family of solutions, we note that, depending on the integer. However, it is true that for all integers k, k, sec π sec π 3 = ±2. (Can you show this?) As a result, 3 + πk doesn’t always equal sec π 2. For cos(x) = 1 2, we get x = 2π 2, we get x = π 3 + 2πk or x = 4π 3 + πk and x = 2π 3 + πk = ± sec π 3 sec2 π 3 + πk = ± sec π 3 2 = (±2)2 = 4 The same holds for the family x = 2π the values k = 0 and k = 1, namely x = π 3 + πk. The solutions which lie in [0, 2π) come from 3. To conﬁrm graphically, we use 3 and 5π 3, 4π 3, 2π 4The reader is encouraged to see what happens if we had chosen the reciprocal identity cot(3x) = 1 tan(3x) instead. The graph on the calculator
appears identical, but what happens when you try to ﬁnd the intersection points? 860 Foundations of Trigonometry a reciprocal identity to rewrite the secant as cosine. The x-coordinates of the intersection points of y = (cos(x))2 and y = 4 verify our answers. 1 y = cos(3x) sin(3x) and y = 0 y = 1 cos2(x) and y = 4 5. The equation tan x 2 Hence, x = −3 has the form tan(u) = −3, whose solution is u = arctan(−3)+πk. 2 = arctan(−3) + πk, so x = 2 arctan(−3) + 2πk for integers k. To check, we note tan 2 arctan(−3)+2πk 2 = tan (arctan(−3) + πk) = tan (arctan(−3)) = −3 (the period of tangent is π) (See Theorem 10.27) To determine which of our answers lie in the interval [0, 2π), we ﬁrst need to get an idea of the value of 2 arctan(−3). While we could easily ﬁnd an approximation using a calculator,5 we proceed analytically. Since −3 < 0, it follows that − π 2 < arctan(−3) < 0. Multiplying through by 2 gives −π < 2 arctan(−3) < 0. We are now in a position to argue which of the solutions x = 2 arctan(−3) + 2πk lie in [0, 2π). For k = 0, we get x = 2 arctan(−3) < 0, so we discard this answer and all answers x = 2 arctan(−3) + 2πk where k < 0. Next, we turn our attention to k = 1 and get x = 2 arctan(−3) + 2π. Starting with the inequality −π < 2 arctan(−3) < 0, we add 2π and get π < 2 arctan(−3) + 2π < 2π. This means x = 2 arctan(−3) + 2π lies in [0, 2π). Advancing k to 2
produces x = 2 arctan(−3) + 4π. Once again, we get from −π < 2 arctan(−3) < 0 that 3π < 2 arctan(−3) + 4π < 4π. Since this is outside the interval [0, 2π), we discard x = 2 arctan(−3) + 4π and all solutions of the form x = 2 arctan(−3) + 2πk for k > 2. Graphically, we see y = tan x and y = −3 intersect only 2 once on [0, 2π) at x = 2 arctan(−3) + 2π ≈ 3.7851. 6. To solve sin(2x) = 0.87, we ﬁrst note that it has the form sin(u) = 0.87, which has the family of solutions u = arcsin(0.87) + 2πk or u = π − arcsin(0.87) + 2πk for integers k. Since the argument of sine here is 2x, we get 2x = arcsin(0.87) + 2πk or 2x = π − arcsin(0.87) + 2πk 2 − 1 which gives x = 1 2 arcsin(0.87) + πk for integers k. To check, 2 arcsin(0.87) + πk or x = π 5Your instructor will let you know if you should abandon the analytic route at this point and use your calculator. But seriously, what fun would that be? 10.7 Trigonometric Equations and Inequalities 861 sin 2 1 2 arcsin(0.87) + πk = sin (arcsin(0.87) + 2πk) = sin (arcsin(0.87)) (the period of sine is 2π) = 0.87 (See Theorem 10.26) For the family x = π 2 − 1 2 arcsin(0.87) + πk, we get sin 2 π 2 − 1 2 arcsin(0.87) + πk = sin (π − arcsin(0.87) + 2πk) = sin (π − arcsin(0.87)) = sin (arcsin(0.87)) = 0.87 (the
period of sine is 2π) (sin(π − t) = sin(t)) (See Theorem 10.26) 2 arcsin(0.87) < π 2 so that multiplying through by 1 4. Starting with the family of solutions x = 1 To determine which of these solutions lie in [0, 2π), we ﬁrst need to get an idea of the value of x = 1 2 arcsin(0.87). Once again, we could use the calculator, but we adopt an analytic route here. By deﬁnition, 0 < arcsin(0.87) < π 2 gives us 0 < 1 2 arcsin(0.87) + πk, we use the same kind of arguments as in our solution to number 5 above and ﬁnd only the solutions corresponding to k = 0 and k = 1 lie in [0, 2π): x = 1 2 arcsin(0.87) + π. Next, we move to the family x = π 2 arcsin(0.87) + πk for integers k. Here, we need to get a better estimate of π 2 arcsin(0.87) < π 4, we ﬁrst multiply through by −1 and then add π 4, or 4 < π π 2. Proceeding with the usual arguments, we ﬁnd the only solutions which lie in [0, 2π) correspond to k = 0 and k = 1, namely x = π 2 arcsin(0.87) and x = 3π 2 arcsin(0.87). All told, we have found four solutions to sin(2x) = 0.87 in [0, 2π): x = 1 2 arcsin(0.87). By graphing y = sin(2x) and y = 0.87, we conﬁrm our results. 2 arcsin(0.87). From the inequality 0 < 1 2 − 1 2 arcsin(0.87), x = 1 2 arcsin(0.87) and x = 3π 2 arcsin(0.87) + π, x = π 2 arcsin(0.87) and x = 1 2 arcsin(0.87) > π 2 arcsin(0.87) < π 2 to get = tan x 2 and y = −
3 y = sin(2x) and y = 0.87 862 Foundations of Trigonometry Each of the problems in Example 10.7.1 featured one trigonometric function. If an equation involves two diﬀerent trigonometric functions or if the equation contains the same trigonometric function but with diﬀerent arguments, we will need to use identities and Algebra to reduce the equation to the same form as those given on page 857. Example 10.7.2. Solve the following equations and list the solutions which lie in the interval [0, 2π). Verify your solutions on [0, 2π) graphically. 1. 3 sin3(x) = sin2(x) 3. cos(2x) = 3 cos(x) − 2 5. cos(3x) = cos(5x) 7. sin(x) cos x 2 + cos(x) sin x 2 = 1 Solution. 2. sec2(x) = tan(x) + 3 4. cos(3x) = 2 − cos(x) 6. sin(2x) = √ 3 cos(x) 8. cos(x) − √ 3 sin(x) = 2 1. We resist the temptation to divide both sides of 3 sin3(x) = sin2(x) by sin2(x) (What goes wrong if you do?) and instead gather all of the terms to one side of the equation and factor. 3 sin3(x) = sin2(x) 3 sin3(x) − sin2(x) = 0 sin2(x)(3 sin(x) − 1) = 0 Factor out sin2(x) from both terms. We get sin2(x) = 0 or 3 sin(x) − 1 = 0. Solving for sin(x), we ﬁnd sin(x) = 0 or sin(x) = 1 3. The solution to the ﬁrst equation is x = πk, with x = 0 and x = π being the two solutions +2πk which lie in [0, 2π). To solve sin(x) = 1 + 2πk for integers k. We ﬁnd the two solutions here which lie in [0, 2π) or x = π − arcsin 1 3. To check graphically, we plot y = 3
(sin(x))3 and and x = π − arcsin 1 to be x = arcsin 1 3 3 y = (sin(x))2 and ﬁnd the x-coordinates of the intersection points of these two curves. Some extra zooming is required near x = 0 and x = π to verify that these two curves do in fact intersect four times.6 3, we use the arcsine function to get x = arcsin 1 3 2. Analysis of sec2(x) = tan(x) + 3 reveals two diﬀerent trigonometric functions, so an identity is in order. Since sec2(x) = 1 + tan2(x), we get sec2(x) = tan(x) + 3 1 + tan2(x) = tan(x) + 3 (Since sec2(x) = 1 + tan2(x).) tan2(x) − tan(x) − 2 = 0 u2 − u − 2 = 0 (u + 1)(u − 2) = 0 Let u = tan(x). 6Note that we are not counting the point (2π, 0) in our solution set since x = 2π is not in the interval [0, 2π). In the forthcoming solutions, remember that while x = 2π may be a solution to the equation, it isn’t counted among the solutions in [0, 2π). 10.7 Trigonometric Equations and Inequalities 863 This gives u = −1 or u = 2. Since u = tan(x), we have tan(x) = −1 or tan(x) = 2. From tan(x) = −1, we get x = − π 4 + πk for integers k. To solve tan(x) = 2, we employ the arctangent function and get x = arctan(2) + πk for integers k. From the ﬁrst set of solutions, we get x = 3π 4 as our answers which lie in [0, 2π). Using the same sort of argument we saw in Example 10.7.1, we get x = arctan(2) and x = π + arctan(2) as answers from our second set of solutions which lie in [0, 2π). Using a reciprocal identity, we rewrite the secant as a cosine and graph
y = (cos(x))2 and y = tan(x) + 3 to ﬁnd the x-values of the points where they intersect. 4 and x = 7π 1 y = 3(sin(x))3 and y = (sin(x))2 y = 1 (cos(x))2 and y = tan(x) + 3 3. In the equation cos(2x) = 3 cos(x) − 2, we have the same circular function, namely cosine, on both sides but the arguments diﬀer. Using the identity cos(2x) = 2 cos2(x) − 1, we obtain a ‘quadratic in disguise’ and proceed as we have done in the past. cos(2x) = 3 cos(x) − 2 2 cos2(x) − 1 = 3 cos(x) − 2 (Since cos(2x) = 2 cos2(x) − 1.) 2 cos2(x) − 3 cos(x) + 1 = 0 2u2 − 3u + 1 = 0 (2u − 1)(u − 1) = 0 Let u = cos(x). 2, we get x = π 2 or u = 1. Since u = cos(x), we get cos(x) = 1 This gives u = 1 2 or cos(x) = 1. Solving cos(x) = 1 3 + 2πk or x = 5π 3 + 2πk for integers k. From cos(x) = 1, we get x = 2πk for integers k. The answers which lie in [0, 2π) are x = 0, π 3. Graphing y = cos(2x) and y = 3 cos(x) − 2, we ﬁnd, after a little extra eﬀort, that the curves intersect in three places on [0, 2π), and the x-coordinates of these points conﬁrm our results. 3, and 5π 4. To solve cos(3x) = 2 − cos(x), we use the same technique as in the previous problem. From Example 10.4.3, number 4, we know that cos(3x) = 4 cos3(x) − 3 cos(x). This transforms the equation into a polynomial in terms of cos(x). cos(3x)
= 2 − cos(x) 4 cos3(x) − 3 cos(x) = 2 − cos(x) 2 cos3(x) − 2 cos(x) − 2 = 0 4u3 − 2u − 2 = 0 Let u = cos(x). 864 Foundations of Trigonometry To solve 4u3 − 2u − 2 = 0, we need the techniques in Chapter 3 to factor 4u3 − 2u − 2 into (u − 1) 4u2 + 4u + 2. We get either u − 1 = 0 or 4u2 + 2u + 2 = 0, and since the discriminant of the latter is negative, the only real solution to 4u3 − 2u − 2 = 0 is u = 1. Since u = cos(x), we get cos(x) = 1, so x = 2πk for integers k. The only solution which lies in [0, 2π) is x = 0. Graphing y = cos(3x) and y = 2 − cos(x) on the same set of axes over [0, 2π) shows that the graphs intersect at what appears to be (0, 1), as required. y = cos(2x) and y = 3 cos(x) − 2 y = cos(3x) and y = 2 − cos(x) 5. While we could approach cos(3x) = cos(5x) in the same manner as we did the previous two problems, we choose instead to showcase the utility of the Sum to Product Identities. From cos(3x) = cos(5x), we get cos(5x) − cos(3x) = 0, and it is the presence of 0 on the right hand side that indicates a switch to a product would be a good move.7 Using Theorem 10.21, we have that cos(5x) − cos(3x) = −2 sin 5x+3x = −2 sin(4x) sin(x). Hence, 2 the equation cos(5x) = cos(3x) is equivalent to −2 sin(4x) sin(x) = 0. From this, we get sin(4x) = 0 or sin(x) = 0. Solving sin(4x) = 0 gives x = π 4 k for integers k, and the solution to sin(x) =
0 is x = πk for integers k. The second set of solutions is contained in the ﬁrst set of solutions,8 so our ﬁnal solution to cos(5x) = cos(3x) is x = π 4 k for integers k. There are eight of these answers which lie in [0, 2π): x = 0, π 4, 3π 4, π, 5π 4. Our plot of the graphs of y = cos(3x) and y = cos(5x) below (after some careful zooming) bears this out. sin 5x−3x 2 and 7π 2, 3π 4, π 2 √ 6. In examining the equation sin(2x) = 3 cos(x), not only do we have diﬀerent circular functions involved, namely sine and cosine, we also have diﬀerent arguments to contend with, namely 2x and x. Using the identity sin(2x) = 2 sin(x) cos(x) makes all of the arguments the same and we proceed as we would solving any nonlinear equation – gather all of the nonzero terms on one side of the equation and factor. sin(2x) = 2 sin(x) cos(x) = √ 2 sin(x) cos(x) − cos(x)(2 sin(x) − 3 cos(x) = 0 3) = 0 √ √ √ 3 cos(x) 3 cos(x) (Since sin(2x) = 2 sin(x) cos(x).) from which we get cos(x) = 0 or sin(x) = 2, we get x = π integers k. From sin(x) = √ 3 3 2. From cos(x) = 0, we obtain x = π 2 + πk for 3 +2πk for integers k. The answers 3 +2πk or x = 2π √ 7As always, experience is the greatest teacher here! 8As always, when in doubt, write it out! 10.7 Trigonometric Equations and Inequalities which lie in [0, 2π) are x = π after some careful zooming, verify our answers. 3 and 2π 2, 3π 2, π 3. We graph y = sin(2x) and y = 865
√ 3 cos(x) and, y = cos(3x) and y = cos(5x) y = sin(2x) and y = √ 3 cos(x) 7. Unlike the previous problem, there seems to be no quick way to get the circular functions or their arguments to match in the equation sin(x) cos x = 1. If we stare at 2 it long enough, however, we realize that the left hand side is the expanded form of the sum formula for sin x + x 2 x = 1. Solving, 2 we ﬁnd x = π 3 and x = 5π 3. Graphing y = sin(x) cos x 2. Hence, our original equation is equivalent to sin 3 3 k for integers k. Two of these solutions lie in [0, 2π): x = π and y = 1 validates our solutions. + cos(x) sin x 2 + cos(x) sin x 2 3 + 4π 8. With the absence of double angles or squares, there doesn’t seem to be much we can do. However, since the arguments of the cosine and sine are the same, we can rewrite the left hand side of this equation as a sinusoid.9 To ﬁt f (x) = cos(x) − 3 sin(x) to the form A sin(ωt + φ) + B, we use what we learned in Example 10.5.3 and ﬁnd A = 2, B = 0, ω = 1 = 2, 3 sin(x) = 2 as 2 sin x + 5π and φ = 5π 6 or sin x + 5π 3 + 2πk for integers k. Only one of 6 these solutions, x = 5π 3, which corresponds to k = 1, lies in [0, 2π). Geometrically, we see that y = cos(x) − 6. Hence, we can rewrite the equation cos(x) − = 1. Solving the latter, we get x = − π 3 sin(x) and y = 2 intersect just once, supporting our answer. √ √ √ y = sin(x) cos x 2 + cos(x) sin x 2 and y = 1 y = cos(x) − √ 3 sin(x) and y = 2 We repeat here the advice given when solving systems
of nonlinear equations in section 8.7 – when it comes to solving equations involving the trigonometric functions, it helps to just try something. 9We are essentially ‘undoing’ the sum / diﬀerence formula for cosine or sine, depending on which form we use, so this problem is actually closely related to the previous one! 866 Foundations of Trigonometry Next, we focus on solving inequalities involving the trigonometric functions. Since these functions are continuous on their domains, we may use the sign diagram technique we’ve used in the past to solve the inequalities.10 Example 10.7.3. Solve the following inequalities on [0, 2π). Express your answers using interval notation and verify your answers graphically. 1. 2 sin(x) ≤ 1 2. sin(2x) > cos(x) 3. tan(x) ≥ 3 Solution. 1. We begin solving 2 sin(x) ≤ 1 by collecting all of the terms on one side of the equation and zero on the other to get 2 sin(x) − 1 ≤ 0. Next, we let f (x) = 2 sin(x) − 1 and note that our original inequality is equivalent to solving f (x) ≤ 0. We now look to see where, if ever, f is undeﬁned and where f (x) = 0. Since the domain of f is all real numbers, we can immediately set about ﬁnding the zeros of f. Solving f (x) = 0, we have 2 sin(x) − 1 = 0 or sin(x) = 1 2. The solutions here are x = π 6 + 2πk for integers k. Since we are restricting our attention to [0, 2π), only x = π 6 are of concern to us. Next, we choose test values in [0, 2π) other than the zeros and determine if f is positive or negative there. For = 1 and for x = π we get f (π) = −1. 2 we get f π x = 0 we have f (0) = −1, for x = π Since our original inequality is equivalent to f (x) ≤ 0, we are looking for where the function is negative (−) or 0, and we get the intervals 0, π 6, 2π. We can conﬁrm our answer 6 graph
ically by seeing where the graph of y = 2 sin(x) crosses or is below the graph of y = 1. 6 + 2πk and x = 5π 6 and x = 5π ∪ 5π 2 (−) 0 (+) π 6 0 (−) 5π 6 2π 0 y = 2 sin(x) and y = 1 2. We ﬁrst rewrite sin(2x) > cos(x) as sin(2x) − cos(x) > 0 and let f (x) = sin(2x) − cos(x). Our original inequality is thus equivalent to f (x) > 0. The domain of f is all real numbers, so we can advance to ﬁnding the zeros of f. Setting f (x) = 0 yields sin(2x) − cos(x) = 0, which, by way of the double angle identity for sine, becomes 2 sin(x) cos(x) − cos(x) = 0 or 2 +πk for integers k of which only x = π cos(x)(2 sin(x)−1) = 0. From cos(x) = 0, we get x = π 2 and x = 3π 2 which gives x = π 6 + 2πk or x = 5π 6 lie in [0, 2π). Next, we choose 2 lie in [0, 2π). For 2 sin(x) − 1 = 0, we get sin(x) = 1 6 + 2πk for integers k. Of those, only x = π 6 and x = 5π 10See page 214, Example 3.1.5, page 321, page 399, Example 6.3.2 and Example 6.4.2 for discussion of this technique. 10.7 Trigonometric Equations and Inequalities 867 4 we get f 3π our test values. For x = 0 we ﬁnd f (0) = −1; when x = π = −1 + for x = 3π ; when x = π we have f (π) = 1, and lastly, for √, so this is 2 x = 7π 6, π our answer. We can use the calculator to check that the graph of y = sin(2x) is indeed above the graph of y = cos(x) on those intervals. 2−
2 2. We see f (x) > 0 on π 2 √ 2 2 = 2 = −2− 4 we get f 7π = −1 − 2 = 2− ∪ 5π 6, 3π 4 we get ; √ √ (−) 0 (+) π 6 0 (−) π 2 0 (+) 5π 6 0 (−) 3π 2 2π 0 y = sin(2x) and y = cos(x) 2 and 3π 3. Proceeding as in the last two problems, we rewrite tan(x) ≥ 3 as tan(x) − 3 ≥ 0 and let f (x) = tan(x) − 3. We note that on [0, 2π), f is undeﬁned at x = π 2, so those values will need the usual disclaimer on the sign diagram.11 Moving along to zeros, solving f (x) = tan(x) − 3 = 0 requires the arctangent function. We ﬁnd x = arctan(3) + πk for integers k and of these, only x = arctan(3) and x = arctan(3) + π lie in [0, 2π). Since 3 > 0, we know 0 < arctan(3) < π 2 which allows us to position these zeros correctly on the sign diagram. To choose test values, we begin with x = 0 and ﬁnd f (0) = −3. Finding a is a bit more challenging. Keep in mind convenient test value in the interval arctan(3), π 2 that the arctangent function is increasing and is bounded above by π 2. This means that the number x = arctan(117) is guaranteed12 to lie between arctan(3) and π 2. We see that f (arctan(117)) = tan(arctan(117)) − 3 = 114. For our next test value, we take x = π and ﬁnd f (π) = −3. To ﬁnd our next test value, we note that since arctan(3) < arctan(117) < π 2, it follows13 that arctan(3) + π < arctan(117) + π < 3π 2. Evaluating f at x = ar
ctan(117) + π yields f (arctan(117) + π) = tan(arctan(117) + π) − 3 = tan(arctan(117)) − 3 = 114. We = −4. Since we want f (x) ≥ 0, we 4 and ﬁnd f 7π choose our last test value to be x = 7π. Using the graphs of y = tan(x) ∪ arctan(3) + π, 3π see that our answer is arctan(3), π 2 2 and y = 3, we see when the graph of the former is above (or meets) the graph of the latter. 4 11See page 321 for a discussion of the non-standard character known as the interrobang. 12We could have chosen any value arctan(t) where t > 3. 13... by adding π through the inequality... 868 Foundations of Trigonometry (−) 0 (+) 0 arctan(3) (−) π 2 (arctan(3) + π) 0 (+) (−) 3π 2 2π y = tan(x) and y = 3 Our next example puts solving equations and inequalities to good use – ﬁnding domains of functions. Example 10.7.4. Express the domain of the following functions using extended interval notation.14 1. f (x) = csc 2x + π 3 2. f (x) = sin(x) 2 cos(x) − 1 3. f (x) = 1 − cot(x) Solution. 1. To ﬁnd the domain of f (x) = csc 2x + π 3, we rewrite f in terms of sine as f (x) = 1 sin(2x+ π 3 ) Since the sine function is deﬁned everywhere, our only concern comes from zeros in the denominator. Solving sin 2x + π 2 k for integers k. In set-builder notation, 3 our domain is x : x = − π 2 k for integers k. To help visualize the domain, we follow the 6,..., old mantra ‘When in doubt, write it out!’ We get x : x = − π where we have kept the denominators 6 throughout to help see the
pattern. Graphing the situation on a numberline, we have = 0, we get x = − π 6 + π 6, − 7π 6, − 4π 6 + π 6, 8π 6, 5π 6, 2π. − 7π 6 − 4π 6 − π 6 2π 6 5π 6 8π 6 Proceeding as we did on page 756 in Section 10.3.1, we let xk denote the kth number excluded from the domain and we have xk = − π for integers k. The intervals which comprise the domain are of the form (xk, xk + 1) = integers. Using extended interval notation, we have that the domain is as k runs through the 2 k = (3k−1)π 6 (3k−1)π 6, (3k+2)π 6 6 + π ∞ k=−∞ (3k − 1)π 6, (3k + 2)π 6 We can check our answer by substituting in values of k to see that it matches our diagram. 14See page 756 for details about this notation. 10.7 Trigonometric Equations and Inequalities 869 2. Since the domains of sin(x) and cos(x) are all real numbers, the only concern when ﬁnding 2 cos(x)−1 is division by zero so we set the denominator equal to zero and 3 +2πk for integers 3 + 2πk for integers k, the domain of f (x) = sin(x) solve. From 2 cos(x)−1 = 0 we get cos(x) = 1 k. Using set-builder notation, the domain is x : x = π or x : x = ± π 3, ± 11π 3,..., so we have 3 + 2πk and x = 5π 3 +2πk or x = 5π 2 so that x = π 3, ± 7π 3, ± 5π − 11π 3 − 7π 3 − 5π 3 − π 3 π 3 5π 3 7π 3 11π 3 Unlike the previous example, we have two diﬀerent families of points to consider, and we present two ways of dealing with this kind of situation. One way is to generalize what we did in the previous example and use the formulas we found
in our domain work to describe the intervals. To that end, we let ak = π for integers k. The goal now is to write the domain in terms of the a’s an b’s. We ﬁnd a0 = π 3, a1 = 7π 3 and b−2 = − 7π 3. Hence, in terms of the a’s and b’s, our domain is 3 + 2πk = (6k+5)π 3 + 2πk = (6k+1)π 3, a−2 = − 11π 3, a−1 = − 5π 3, b−1 = − π 3, a2 = 13π 3, b1 = 11π 3, b0 = 5π 3, b2 = 17π and bk = 5π 3 3... (a−2, b−2) ∪ (b−2, a−1) ∪ (a−1, b−1) ∪ (b−1, a0) ∪ (a0, b0) ∪ (b0, a1) ∪ (a1, b1) ∪... If we group these intervals in pairs, (a−2, b−2)∪(b−2, a−1), (a−1, b−1)∪(b−1, a0), (a0, b0)∪(b0, a1) and so forth, we see a pattern emerge of the form (ak, bk) ∪ (bk, ak + 1) for integers k so that our domain can be written as ∞ k=−∞ (ak, bk) ∪ (bk, ak + 1) = ∞ k=−∞ (6k + 1)π 3, (6k + 5)π 3 ∪ (6k + 5)π 3, (6k + 7)π 3 A second approach to the problem exploits the periodic nature of f. Since cos(x) and sin(x) have period 2π, it’s not too diﬃcult to show the function f repeats itself every 2π units.15 This means if we can ﬁnd a formula for the domain on an interval of length 2π, we can express the entire domain by translating our answer left
and right on the x-axis by adding integer multiples of 2π. One such interval that arises from our domain work is π. The portion. Adding integer multiples of 2π, we get the family of of the domain here is π 3, 7π 3 + 2πk for integers k. We leave it to the reader intervals π 3 + 2πk, 7π to show that getting common denominators leads to our previous answer. ∪ 5π 3 + 2πk ∪ 5π 3 + 2πk, 5π 3, 5π 3, 7π 3 3 3 15This doesn’t necessarily mean the period of f is 2π. The tangent function is comprised of cos(x) and sin(x), but its period is half theirs. The reader is invited to investigate the period of f. 870 Foundations of Trigonometry 3. To ﬁnd the domain of f (x) = 1 − cot(x), we ﬁrst note that, due to the presence of the cot(x) term, x = πk for integers k. Next, we recall that for the square root to be deﬁned, we need 1−cot(x) ≥ 0. Unlike the inequalities we solved in Example 10.7.3, we are not restricted here to a given interval. Our strategy is to solve this inequality over (0, π) (the same interval which generates a fundamental cycle of cotangent) and then add integer multiples of the period, in this case, π. We let g(x) = 1 − cot(x) and set about making a sign diagram for g over the interval (0, π) to ﬁnd where g(x) ≥ 0. We note that g is undeﬁned for x = πk for integers k, in particular, at the endpoints of our interval x = 0 and x = π. Next, we look for the zeros of g. Solving g(x) = 0, we get cot(x) = 1 or x = π 4 + πk for integers k and 6 and x = π only one of these, x = π 2, we get √ 3, and g π g π 2 6 4, lies in (0, π). Choosing the test values x =
π = 1. = 1 − (−) 0 0 (+) π 4 π We ﬁnd g(x) ≥ 0 on π the intervals π express our ﬁnal answer as 4 + πk, π + πk = 4, π. Adding multiples of the period we get our solution to consist of. Using extended interval notation, we, (k + 1)π (4k+1)π 4 ∞ k=−∞ (4k + 1)π 4, (k + 1)π We close this section with an example which demonstrates how to solve equations and inequalities involving the inverse trigonometric functions. Example 10.7.5. Solve the following equations and inequalities analytically. Check your answers using a graphing utility. 1. arcsin(2x) = π 3 2. 4 arccos(x) − 3π = 0 3. 3 arcsec(2x − 1) + π = 2π 4. 4 arctan2(x) − 3π arctan(x) − π2 = 0 5. π2 − 4 arccos2(x) < 0 6. 4 arccot(3x) > π Solution. 1. To solve arcsin(2x) = π 3 is in the range of the arcsine function (so a solution exists!) Next, we exploit the inverse property of sine and arcsine from Theorem 10.26 3, we ﬁrst note that π 10.7 Trigonometric Equations and Inequalities 871 arcsin(2x) = π 3 sin (arcsin(2x)) = sin π 3 2x = x = √ 3 2 √ 3 4 Since sin(arcsin(u)) = u Graphing y = arcsin(2x) and the horizontal line y = π √ 3 4 ≈ 0.4430. 2. Our ﬁrst step in solving 4 arccos(x) − 3π = 0 is to isolate the arccosine. Doing so, we get 3, we see they intersect at arccos(x) = 3π 4. Since 3π 4 is in the range of arccosine, we may apply Theorem 10.26 arccos(x) = 3π 4 cos (arcc
os(x)) = cos 3π 4 x = − √ 2 2 Since cos(arccos(u)) = u The calculator conﬁrms y = 4 arccos(x) − 3π crosses y = 0 (the x-axis) at − √ 2 2 ≈ −0.7071. y = arcsin(2x) and y = π 3 y = 4 arccos(x) − 3π 3. From 3 arcsec(2x − 1) + π = 2π, we get arcsec(2x − 1) = π 3. As we saw in Section 10.6, there are two possible ranges for the arcsecant function. Fortunately, both ranges contain π 3. Applying Theorem 10.28 / 10.29, we get arcsec(2x − 1) = π 3 sec(arcsec(2x − 1)) = sec π 3 2x − 1 = 2 x = 3 2 Since sec(arcsec(u)) = u To check using our calculator, we need to graph y = 3 arcsec(2x − 1) + π. To do so, we make from Theorems 10.28 and 10.29.16 We see the graph use of the identity arcsec(u) = arccos 1 u of y = 3 arccos + π and the horizontal line y = 2π intersect at 3 1 2 = 1.5. 2x−1 16Since we are checking for solutions where arcsecant is positive, we know u = 2x − 1 ≥ 1, and so the identity applies in both cases. 872 Foundations of Trigonometry 4. With the presence of both arctan2(x) ( = (arctan(x))2) and arctan(x), we substitute u = arctan(x). The equation 4 arctan2(x) − 3π arctan(x) − π2 = 0 becomes 4u2 − 3πu − π2 = 0. Factoring,17 we get (4u + π)(u − π) = 0, so u = arctan(x) = − π 4 or u = arctan(x) = π. Since − π 4 is in the range of arctangent, but π is not,
we only get solutions from the ﬁrst equation. Using Theorem 10.27, we get arctan(x) = − π 4 tan(arctan(x)) = tan − π 4 x = −1 Since tan(arctan(u)) = u. The calculator veriﬁes our result. y = 3 arcsec(2x − 1) + π and y = 2π y = 4 arctan2(x) − 3π arctan(x) − π2 5. Since the inverse trigonometric functions are continuous on their domains, we can solve inequalities featuring these functions using sign diagrams. Since all of the nonzero terms of π2 − 4 arccos2(x) < 0 are on one side of the inequality, we let f (x) = π2 − 4 arccos2(x) and note the domain of f is limited by the arccos(x) to [−1, 1]. Next, we ﬁnd the zeros of f by setting f (x) = π2 − 4 arccos2(x) = 0. We get arccos(x) = ± π 2, and since the range of arccosine is = 0 [0, π], we focus our attention on arccos(x) = π as our only zero. Hence, we have two test intervals, [−1, 0) and (0, 1]. Choosing test values x = ±1, we get f (−1) = −3π2 < 0 and f (1) = π2 > 0. Since we are looking for where f (x) = π2 − 4 arccos2(x) < 0, our answer is [−1, 0). The calculator conﬁrms that for these values of x, the graph of y = π2 − 4 arccos2(x) is below y = 0 (the x-axis.) 2. Using Theorem 10.26, we get x = cos π 2 17It’s not as bad as it looks... don’t let the π throw you! 10.7 Trigonometric Equations and Inequalities 873 (−) (+) 0 0 1 −1 y = π2 − 4 arccos2(x) 6. To
begin, we rewrite 4 arccot(3x) > π as 4 arccot(3x) − π > 0. We let f (x) = 4 arccot(3x) − π, and note the domain of f is all real numbers, (−∞, ∞). To ﬁnd the zeros of f, we set f (x) = 4 arccot(3x) − π = 0 and solve. We get arccot(3x) = π 4 is in the range of arccotangent, we may apply Theorem 10.27 and solve 4, and since π arccot(3x) = π 4 cot(arccot(3x)) = cot π 4 3x = 1 x = 1 3 Since cot(arccot(u)) = u. 3, we have two test intervals, −∞, 1 Next, we make a sign diagram for f. Since the domain of f is all real numbers, and there is 3, ∞. Ideally, we wish only one zero of f, x = 1 to ﬁnd test values x in these intervals so that arccot(4x) corresponds to one of our oft-used ‘common’ angles. After a bit of computation,18 we choose x = 0 for x < 1 3, we √ 3 = − π choose x = 3 < 0. Since we are looking for where 3. We ﬁnd f (0) = π > 0 and f. To check graphically, we use the f (x) = 4 arccot(3x) − π > 0, we get our answer −∞, 1 3 technique in number 2c of Example 10.6.5 in Section 10.6 to graph y = 4 arccot(3x) and we see it is above the horizontal line y = π on −∞, 1 3 3 and for x > 1 = −∞, 0.3. and 1 √ 3 3 3 (−) (+) 0 1 3 y = 4 arccot(3x) and y = π 18Set 3x equal to the cotangents of the ‘common angles’ and choose accordingly. 874 Foundations of Trigonometry 10.7.1 Exercises In Exercises 1 -
18, ﬁnd all of the exact solutions of the equation and then list those solutions which are in the interval [0, 2π). 1. sin (5x) = 0 2. cos (3x) = 1 2 4. tan (6x) = 1 5. csc (4x) = −1 √ 3 3 = 0 7. cot (2x) = − 10. cos x + 5π 6 13. csc(x) = 0 16. sec2 (x) = 4 3 8. cos (9x) = 9 11. sin 2x − π 3 = − 1 2 14. tan (2x − π) = 1 17. cos2 (x. sin (−2x) = 6. sec (3x) = 9. sin = x 3 12. 2 cos x + √ 3 = 7π 4 15. tan2 (x) = 3 18. sin2 (x) = 3 4 In Exercises 19 - 42, solve the equation, giving the exact solutions which lie in [0, 2π) 19. sin (x) = cos (x) 21. sin (2x) = cos (x) 23. cos (2x) = cos (x) 20. sin (2x) = sin (x) 22. cos (2x) = sin (x) 24. cos(2x) = 2 − 5 cos(x) 25. 3 cos(2x) + cos(x) + 2 = 0 26. cos(2x) = 5 sin(x) − 2 27. 3 cos(2x) = sin(x) + 2 29. tan2(x) = 1 − sec(x) 31. sec(x) = 2 csc(x) 33. sin(2x) = tan(x) 35. cos(2x) + csc2(x) = 0 37. tan2 (x) = 3 2 sec (x) 39. tan(2x) − 2 cos(x) = 0 28. 2 sec2(x) = 3 − tan(x) 30. cot2(x) = 3 csc(x) − 3 32. cos(x) csc(x) cot(x) = 6 − cot2(x) 34. cot4(x) = 4 csc2(x) − 7 36
. tan3 (x) = 3 tan (x) 38. cos3 (x) = − cos (x) 40. csc3(x) + csc2(x) = 4 csc(x) + 4 41. 2 tan(x) = 1 − tan2(x) 42. tan (x) = sec (x) 10.7 Trigonometric Equations and Inequalities 875 In Exercises 43 - 58, solve the equation, giving the exact solutions which lie in [0, 2π) 43. sin(6x) cos(x) = − cos(6x) sin(x) 44. sin(3x) cos(x) = cos(3x) sin(x) 45. cos(2x) cos(x) + sin(2x) sin(x) = 1 46. cos(5x) cos(3x) − sin(5x) sin(3x) = √ 3 2 47. sin(x) + cos(x) = 1 √ 49. 2 cos(x) − √ 2 sin(x) = 1 51. cos(2x) − √ 3 sin(2x) = √ 2 48. sin(x) + √ 3 cos(x) = 1 √ 50. 3 sin(2x) + cos(2x) = 1 √ 52. 3 3 sin(3x) − 3 cos(3x) = 3 √ 3 53. cos(3x) = cos(5x) 54. cos(4x) = cos(2x) 55. sin(5x) = sin(3x) 56. cos(5x) = − cos(2x) 57. sin(6x) + sin(x) = 0 58. tan(x) = cos(x) In Exercises 59 - 68, solve the equation. 59. arccos(2x) = π 60. π − 2 arcsin(x) = 2π 61. 4 arctan(3x − 1) − π = 0 62. 6 arccot(2x) − 5π = 0 63. 4 arcsec x 2 = π 64. 12 arccsc x 3 = 2π 65. 9 arcsin2(x) − π2 = 0 66. 9 arccos2(x)
− π2 = 0 67. 8 arccot2(x) + 3π2 = 10π arccot(x) 68. 6 arctan(x)2 = π arctan(x) + π2 In Exercises 69 - 80, solve the inequality. Express the exact answer in interval notation, restricting your attention to 0 ≤ x ≤ 2π. 69. sin (x) ≤ 0 72. cos2 (x) > 75. cot2 (x) ≥ 1 2 1 3 78. cos(3x) ≤ 1 70. tan (x) ≥ √ 3 73. cos (2x) ≤ 0 76. 2 cos(x) ≥ 1 79. sec(x) ≤ √ 2 71. sec2 (x) ≤ 4 74. sin x + π 3 > 1 2 77. sin(5x) ≥ 5 80. cot(x) ≤ 4 876 Foundations of Trigonometry In Exercises 81 - 86, solve the inequality. Express the exact answer in interval notation, restricting your attention to −π ≤ x ≤ π. 81. cos (x) > 84. sin2 (x) < √ 3 2 3 4 82. sin(x) > 1 3 83. sec (x) ≤ 2 85. cot (x) ≥ −1 86. cos(x) ≥ sin(x) In Exercises 87 - 92, solve the inequality. Express the exact answer in interval notation, restricting your attention to −2π ≤ x ≤ 2π. 87. csc (x) > 1 90. tan2 (x) ≥ 1 88. cos(x) ≤ 5 3 89. cot(x) ≥ 5 91. sin(2x) ≥ sin(x) 92. cos(2x) ≤ sin(x) In Exercises 93 - 98, solve the given inequality. 93. arcsin(2x) > 0 94. 3 arccos(x) ≤ π 95. 6 arccot(7x) ≥ π 96. π > 2 arctan(x) 97. 2 arcsin(x)2 > π arcsin(x) 98. 12 arccos(x)2 + 2π2 > 11π arccos(x) In Exercises 99 - 107, express the domain of the function using the extended interval notation. (See
page 756 in Section 10.3.1 for details.) 99. f (x) = 1 cos(x) − 1 100. f (x) = cos(x) sin(x) + 1 101. f (x) = tan2(x) − 1 102. f (x) = 2 − sec(x) 103. f (x) = csc(2x) 104. f (x) = sin(x) 2 + cos(x) 105. f (x) = 3 csc(x) + 4 sec(x) 106. f (x) = ln (\| cos(x)\|) 107. f (x) = arcsin(tan(x)) 108. With the help of your classmates, determine the number of solutions to sin(x) = 1 2 and sin(4x) = 1 2 in [0, 2π). Then ﬁnd the number of solutions to sin(2x) = 1 2 in [0, 2π). A pattern should emerge. Explain how this pattern would help you solve equations like sin(11x) = 1 2. What do you ﬁnd? 2, sin(3x) = 1 2 and sin 5x = 1 = 1 2 2 2 = 1 2. Now consider sin x with −1 and repeat the whole exploration. 2, sin 3x Replace 1 2 10.7 Trigonometric Equations and Inequalities 877 10.7.2 Answers 1. x = πk 5 ; x = 0, π 5, 2π 5, 3π 5, 4π 5, π, 6π 5, 7π 5, 8π 5, 9π 5 2. x = π 9 + 2πk 3 or x = + πk or x = 5π 9 5π 6 + 2πk 3 ; x = π 9, 5π 9, 7π 9, 11π 9, 13π 9, 17π 9 + πk; x = 2π 3, 5π 6, 5π 3, 11π 6 + + πk 6 πk 2 ; x = π 24, 5π 24, 3π 8, 13π 24, 17π 24, 7π 8, 25π 24, 29π 24, 11π 8, 37π 24, 41π 24, 15π 8 ; x = 3π 8, 7π 8, 11
π 8, 15π 8 + 2πk 3 or x = 7π 12 + 2πk 3 ; x = π 12, 7π 12, 3π 4, 5π 4, 17π 12, 23π 12 3. x = 4. x = 5. x = 6. x = 2π 3 π 24 3π 8 π 12 7. x = π 3 + πk 2 ; x = π 3, 5π 6, 4π 3, 11π 6 8. No solution 9. x = 3π 4 10. x = − π 3 11. x = 3π 4 + 6πk or x = 9π 4 + 6πk; x = 3π 4 + πk; x = 2π 3, 5π 3 + πk or x = 13π 12 + πk; x = π 12, 3π 4, 13π 12, 7π 4 12. x = − 19π 12 + 2πk or x = π 12 + 2πk; x = π 12, 5π 12 13. No solution 14. x = 5π 8 + πk 2 ; x = π 8, 5π 8, 9π 8, 13π 8 15. x = 16. x = 17. x = 18 + πk or x = + πk or x = 2π 3 5π 6 + πk; x = + πk; x = π 3 π 6,, 2π 3 5π 6,, 4π 3 7π 6,, 5π 3 11π 6 + πk 2 ; x = π 4, 3π 4, 5π 4, 7π 4 + πk or x = 2π 3 + πk; x = π 3, 2π 3, 4π 3, 5π 3 878 19. x = 21. x = π 4 π 6 23. x = 0,,, 5π 4 π 2 2π 3,, 3π 2, 5π 6 4π 3 25. x = 27. x = 2π 3 7π 6,, 4π 3, arccos 11π 6, arcsin 1 3 1 3, 2π − arccos, π − arcsin 1 3 1 3 29. x = 0, 2π 3, 4π 3 31. x = arctan
(2), π + arctan(2) Foundations of Trigonometry 20. x = 0, 22. x = 24. x = 26, 5π 3 3π 2, π 3 5π 6 5π 3 5π 6,,, 28. x = 3π 4, 7π 4, arctan 1 2, π + arctan 1 2 30. x = 32. x = 34. x = π 6 π 6 π 6 36. x = 0, 38. x = 40. x = π 2 π 6,,,,,, 5π 6 7π 6 π 4 π 3 3π 2 5π 6,,, π 2 5π 6 3π 4 2π 3,, 11π 6 5π 6 4π 3,, 7π 6 5π 3,, π,, 7π 6, 3π 2, 11π 6 5π 4, 7π 4, 11π 6 33. x = 0, π,, 3π 4, 5π 4, 7π 4 π 4 3π 2 5π 3 π 2 5π 8, 35. x = 37. x = 39. x = 41,,,,, 5π 6 9π 8, 3π 2 13π 8 4π 7,,, 3π 7 3π 2 13π 48, 43. x = 0, 44. x = 0, π 7 π 2, 2π 7, π, 46. x = π 48, 11π 48 47. x = 0, π 2 49. x = 51. x =, π 12 17π 24 17π 12 41π 24,, 23π 24, 47π 24 42. No solution, 5π 7, 6π 7, π, 8π 7, 9π 7, 10π 7, 11π 7, 12π 7, 13π 7 45. x = 0, 23π 48, 25π 48, 35π 48, 37π 48, 47π 48, 49π 48, 59π 48, 48. x =, 71π 48, 73π 48, 83π 48, 85π 48, 95π 48 61π 48 π 2, 11π 6 π 3 5π 18,, 50. x = 0, π, 52. x = π 6, 4π 3 5π 6, 17π 18
, 3π 2, 29π 18 10.7 Trigonometric Equations and Inequalities 879 53. x = 0, 55. x = 0 3π 8 3π 4 5π 8,, π,, 5π 4 7π 8, 3π 2 9π 8 7π 4 11π 8,, 13π 8, 15π 8 54. x = 0, π 3, 2π 3, π, 4π 3, 5π 3,, π, 56. x = π 7, π 3, 3π 7, 5π 7, π, 9π 7, 11π 7, 5π 3, 13π 7 57. x = 0, 2π 7, 58. x = arcsin 59. x = − 1 2 61. x = 2 3 63. x = 2 √ 2 65. x = ± √ 3 2 67. x = −1, 0, 6π 7, 4π 7 −1 + 2 8π 7 √ 5, 10π 7, 12π 7, π 5, 3π 5, π,, 7π 5 ≈ 0.6662, π − arcsin 9π 5 −1 + 2 √ 5 ≈ 2.4754 60. x = −1 62. x = − √ 3 2 64. x = 6 66. x = 1 2 √ 68. x = − 3 69. [π, 2π] 71. 73. 0, π 3 ∪ π 4, 3π 4, 2π 3 5π 4 ∪, 7π 4 4π 3 ∪ 5π 3, 2π 0, 75. π 3 ∪ 2π 3, π ∪ π, 4π 3 ∪ 5π 3, 2π 70. 72. 74. 76. π 3, π 2 ∪ 0, 0, 0,, 4π 3 3π 4 11π 6 5π 3, 2π 3π 2 5π 4 ∪ 7π 4, 2π, 2π 77. No solution 78. [0, 2π] 0, 79. π 4 ∪ π 2, 3π 2 ∪ 7π 4, 2π 81. − π 6, π 6 −π, − 83 80. [arccot(4), π) ∪ [π + arccot(
4), 2π) 82. arcsin 1 3, π − arcsin 1 3 84. − 2π 3, − π 3 ∪ π 3, 2π 3 880 Foundations of Trigonometry −π, − 85. π 4 ∪ 0, 3π 4 −2π, − 87. 3π 2 ∪ − 3π 2, −π ∪ 0, π 2 ∪ π 2, π 86. − 3π 4, π 4 88. [−2π, 2π] 89. (−2π, arccot(5) − 2π] ∪ (−π, arccot(5) − π] ∪ (0, arccot(5)] ∪ (π, π + arccot(5)] − 3π 2, − 5π 4 − ∪ 3π 4, − −π, − 0, 3π 4 ∪ 5π 4, 3π 2 ∪ 3π 2, 7π 4 ∪ π, 5π 3 90. − 7π 4, − 3π 2 −2π, − 91. 5π 3 ∪ 92. − 93. 0, 1 2, − 7π 6 11π 6 ∪ ∪ π 6, 5π 6 ∪,, − π 3π 2 2 2, 1 94. 1 96. (−∞, ∞) 97. [−1, 0) 99. ∞ k=−∞ (2kπ, (2k + 2)π) 100. ∞ k=−∞ (4k − 1)π 2, (4k + 3)π 2 −∞, √ 3 7 95. 98. −14k + 1)π k=−∞ 4 ∞ (6k − 1)π 3 (2k + 1)π 2 ∪ (2k + 1)π 2, (4k + 3)π 4 (6k + 1)π 3 ∪ (4k + 1)π 2, (4k + 3)π 2,, kπ 2 kπ 2,, (k + 1)π 2 (k + 1)π 2 (4k − 1)π 4, (4k + 1)π 4 104. (−∞, ∞) 106. ∞ k=−∞ (2k − 1
)π 2, (2k + 1)π 2 101. 102. 103. 105. 107. k=−∞ ∞ k=−∞ ∞ k=−∞ ∞ k=−∞ Chapter 11 Applications of Trigonometry 11.1 Applications of Sinusoids In the same way exponential functions can be used to model a wide variety of phenomena in nature,1 the cosine and sine functions can be used to model their fair share of natural behaviors. In section 10.5, we introduced the concept of a sinusoid as a function which can be written either in the form C(x) = A cos(ωx+φ)+B for ω > 0 or equivalently, in the form S(x) = A sin(ωx+φ)+B for ω > 0. At the time, we remained undecided as to which form we preferred, but the time for such indecision is over. For clarity of exposition we focus on the sine function2 in this section and switch to the independent variable t, since the applications in this section are time-dependent. We reintroduce and summarize all of the important facts and deﬁnitions about this form of the sinusoid below. Properties of the Sinusoid S(t) = A sin(ωt + φ) + B The amplitude is \|A\| The angular frequency is ω and the ordinary frequency is f = ω 2π The period is T = 1 f = 2π ω The phase is φ and the phase shift is − φ ω The vertical shift or baseline is B Along with knowing these formulas, it is helpful to remember what these quantities mean in context. The amplitude measures the maximum displacement of the sine wave from its baseline (determined by the vertical shift), the period is the length of time it takes to complete one cycle of the sinusoid, the angular frequency tells how many cycles are completed over an interval of length 2π, and the ordinary frequency measures how many cycles occur per unit of time. The phase indicates what 1See Section 6.5. 2Sine haters can use the co-function identity cos π 2 − θ = sin(θ) to turn all of the sines into cosines. 882 Applications of Trigonometry angle φ corresponds to t = 0, and the phase shift represents how much of a ‘head start’ the sinus
oid has over the un-shifted sine function. The ﬁgure below is repeated from Section 10.5. amplitude baseline period In Section 10.1.1, we introduced the concept of circular motion and in Section 10.2.1, we developed formulas for circular motion. Our ﬁrst foray into sinusoidal motion puts these notions to good use. Example 11.1.1. Recall from Exercise 55 in Section 10.1 that 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 136 feet. It completes two revolutions in 2 minutes and 7 seconds. Assuming that the riders are at the edge of the circle, ﬁnd a sinusoid which describes the height of the passengers above the ground t seconds after they pass the point on the wheel closest to the ground. Solution. We sketch the problem situation below and assume a counter-clockwise rotation.3 θ Q h P O 3Otherwise, we could just observe the motion of the wheel from the other side. 11.1 Applications of Sinusoids 883 We know from the equations given on page 732 in Section 10.2.1 that the y-coordinate for counterclockwise motion on a circle of radius r centered at the origin with constant angular velocity (frequency) ω is given by y = r sin(ωt). Here, t = 0 corresponds to the point (r, 0) so that θ, the angle measuring the amount of rotation, is in standard position. In our case, the diameter of the wheel is 128 feet, so the radius is r = 64 feet. Since the wheel completes two revolutions in 2 minutes and 7 seconds (which is 127 seconds) the period T = 1 seconds. Hence, the angular frequency is ω = 2π T = 4π 127 radians per second. Putting these two pieces of information together, we have that y = 64 sin 4π 127 t describes the y-coordinate on the Giant Wheel after t seconds, assuming it is centered at (0, 0) with t = 0 corresponding to the point Q. In order to ﬁnd an expression for h, we take the point O in the ﬁgure as the origin. Since the base of the Giant Wheel ride is 8 feet above the ground and the Giant Wheel itself has a radius of 64 feet, its center is 72 feet above the ground. To account for this vertical
shift upward,4 we add 72 to our formula for y to obtain the new formula h = y + 72 = 64 sin 4π 127 t + 72. Next, we need to adjust things so that t = 0 corresponds to the point P instead of the point Q. This is where the phase comes into play. Geometrically, we need to shift the angle θ in the ﬁgure back π 2 radians. From Section 10.2.1, we know θ = ωt = 4π 127 t, so we (temporarily) write the height in terms of θ as h = 64 sin (θ) + 72. + 72. We 127 t − π Subtracting π 2. can check the reasonableness of our answer by graphing y = h(t) over the interval 0, 127 2 2 from θ gives the ﬁnal answer h(t) = 64 sin θ − π + 72 = 64 sin 4π 2 (127) = 127 2 2 y 136 72 8 t 127 2 A few remarks about Example 11.1.1 are in order. First, note that the amplitude of 64 in our answer corresponds to the radius of the Giant Wheel. This means that passengers on the Giant Wheel never stray more than 64 feet vertically from the center of the Wheel, which makes sense. 8 = 15.875. This represents the Second, the phase shift of our answer works out to be ‘time delay’ (in seconds) we introduce by starting the motion at the point P as opposed to the point Q. Said diﬀerently, passengers which ‘start’ at P take 15.875 seconds to ‘catch up’ to the point Q. 4π/127 = 127 π/2 Our next example revisits the daylight data ﬁrst introduced in Section 2.5, Exercise 6b. 4We are readjusting our ‘baseline’ from y = 0 to y = 72. 884 Applications of Trigonometry Example 11.1.2. According to the U.S. Naval Observatory website, the number of hours H of daylight that Fairbanks, Alaska received on the 21st day of the nth month of 2009 is given below. Here t = 1 represents January 21, 2009, t = 2 represents February 21, 2009, and so on. Month Number Hours of Daylight 1 2 3
4 5 6 7 8 9 10 11 12 5.8 9.3 12.4 15.9 19.4 21.8 19.4 15.6 12.4 9.1 5.6 3.3 1. Find a sinusoid which models these data and use a graphing utility to graph your answer along with the data. 2. Compare your answer to part 1 to one obtained using the regression feature of a calculator. Solution. 1. To get a feel for the data, we plot it below. H 22 20 18 16 14 12 10 10 11 12 6 The data certainly appear sinusoidal,5 but when it comes down to it, ﬁtting a sinusoid to data manually is not an exact science. We do our best to ﬁnd the constants A, ω, φ and B so that the function H(t) = A sin(ωt + φ) + B closely matches the data. We ﬁrst go after the vertical shift B whose value determines the baseline. In a typical sinusoid, the value of B is the average of the maximum and minimum values. So here we take B = 3.3+21.8 = 12.55. Next is the amplitude A which is the displacement from the baseline to the maximum (and minimum) values. We ﬁnd A = 21.8 − 12.55 = 12.55 − 3.3 = 9.25. At this point, we have H(t) = 9.25 sin(ωt + φ) + 12.55. Next, we go after the angular frequency ω. Since the data collected is over the span of a year (12 months), we take the period T = 12 months.6 This 2 5Okay, it appears to be the ‘∧’ shape we saw in some of the graphs in Section 2.2. Just humor us. 6Even though the data collected lies in the interval [1, 12], which has a length of 11, we need to think of the data point at t = 1 as a representative sample of the amount of daylight for every day in January. That is, it represents H(t) over the interval [0, 1]. Similarly, t = 2 is a sample of H(t) over [1, 2], and so forth. 11.1 Applications of Sinusoids 885 12 = π T = 2π means ω
= 2π 6. The last quantity to ﬁnd is the phase φ. Unlike the previous example, it is easier in this case to ﬁnd the phase shift − φ ω. Since we picked A > 0, the phase shift corresponds to the ﬁrst value of t with H(t) = 12.55 (the baseline value).7 Here, we choose t = 3, since its corresponding H value of 12.4 is closer to 12.55 than the next value, 15.9, which corresponds to t = 4. Hence, − φ 2. We have H(t) = 9.25 sin π + 12.55. Below is a graph of our data with the curve y = H(t). ω = 3, so φ = −3ω = −. Using the ‘SinReg’ command, we graph the calculator’s regression below. While both models seem to be reasonable ﬁts to the data, the calculator model is possibly the better ﬁt. The calculator does not give us an r2 value like it did for linear regressions in Section 2.5, nor does it give us an R2 value like it did for quadratic, cubic and quartic regressions as in Section 3.1. The reason for this, much like the reason for the absence of R2 for the logistic model in Section 6.5, is beyond the scope of this course. We’ll just have to use our own good judgment when choosing the best sinusoid model. 11.1.1 Harmonic Motion One of the major applications of sinusoids in Science and Engineering is the study of harmonic motion. The equations for harmonic motion can be used to describe a wide range of phenomena, from the motion of an object on a spring, to the response of an electronic circuit. In this subsection, we restrict our attention to modeling a simple spring system. Before we jump into the Mathematics, there are some Physics terms and concepts we need to discuss. In Physics, ‘mass’ is deﬁned as a measure of an object’s resistance to straight-line motion whereas ‘weight’ is the amount of force (pull) gravity exerts on an object. An object’s mass cannot change,8 while its weight could change. 7See the ﬁgure on page 882. 8Well, assuming the object
isn’t subjected to relativistic speeds... 886 Applications of Trigonometry An object which weighs 6 pounds on the surface of the Earth would weigh 1 pound on the surface of the Moon, but its mass is the same in both places. In the English system of units, ‘pounds’ (lbs.) is a measure of force (weight), and the corresponding unit of mass is the ‘slug’. In the SI system, the unit of force is ‘Newtons’ (N) and the associated unit of mass is the ‘kilogram’ (kg). We convert between mass and weight using the formula9 w = mg. Here, w is the weight of the object, m is the mass and g is the acceleration due to gravity. In the English system, g = 32 feet second2, and in the SI system, g = 9.8 meters second2. Hence, on Earth a mass of 1 slug weighs 32 lbs. and a mass of 1 kg weighs 9.8 N.10 Suppose we attach an object with mass m to a spring as depicted below. The weight of the object will stretch the spring. The system is said to be in ‘equilibrium’ when the weight of the object is perfectly balanced with the restorative force of the spring. How far the spring stretches to reach equilibrium depends on the spring’s ‘spring constant’. Usually denoted by the letter k, the spring constant relates the force F applied to the spring to the amount d the spring stretches in accordance with Hooke’s Law11 F = kd. If the object is released above or below the equilibrium position, or if the object is released with an upward or downward velocity, the object will bounce up and down on the end of the spring until some external force stops it. If we let x(t) denote the object’s displacement from the equilibrium position at time t, then x(t) = 0 means the object is at the equilibrium position, x(t) < 0 means the object is above the equilibrium position, and x(t) > 0 means the object is below the equilibrium position. The function x(t) is called the ‘equation of motion’ of the object.12 x(t) = 0 at the equilibrium position x(t) < 0 above the equilibrium position x(t) > 0 below the equilibrium position If we ignore all other in�
�uences on the system except gravity and the spring force, then Physics tells us that gravity and the spring force will battle each other forever and the object will oscillate indeﬁnitely. In this case, we describe the motion as ‘free’ (meaning there is no external force causing the motion) and ‘undamped’ (meaning we ignore friction caused by surrounding medium, which in our case is air). The following theorem, which comes from Diﬀerential Equations, gives x(t) as a function of the mass m of the object, the spring constant k, the initial displacement x0 of the 9This is a consequence of Newton’s Second Law of Motion F = ma where F is force, m is mass and a is acceleration. In our present setting, the force involved is weight which is caused by the acceleration due to gravity. 10Note that 1 pound = 1 slug foot 11Look familiar? We saw Hooke’s Law in Section 4.3.1. 12To keep units compatible, if we are using the English system, we use feet (ft.) to measure displacement. If we second2 and 1 Newton = 1 kg meter second2. are in the SI system, we measure displacement in meters (m). Time is always measured in seconds (s). 11.1 Applications of Sinusoids 887 object and initial velocity v0 of the object. As with x(t), x0 = 0 means the object is released from the equilibrium position, x0 < 0 means the object is released above the equilibrium position and x0 > 0 means the object is released below the equilibrium position. As far as the initial velocity v0 is concerned, v0 = 0 means the object is released ‘from rest,’ v0 < 0 means the object is heading upwards and v0 > 0 means the object is heading downwards.13 Theorem 11.1. Equation for Free Undamped Harmonic Motion: Suppose an object of mass m is suspended from a spring with spring constant k. If the initial displacement from the equilibrium position is x0 and the initial velocity of the object is v0, then the displacement x from the equilibrium position at time t is given by x(t) = A sin(ωt + φ) where ω = k m and A = x2 0 + 2 v0 ω A sin(φ) = x0 and Aω cos(φ) = v
0. It is a great exercise in ‘dimensional analysis’ to verify that the formulas given in Theorem 11.1 work out so that ω has units 1 s and A has units ft. or m, depending on which system we choose. Example 11.1.3. Suppose an object weighing 64 pounds stretches a spring 8 feet. 1. If the object is attached to the spring and released 3 feet below the equilibrium position from rest, ﬁnd the equation of motion of the object, x(t). When does the object ﬁrst pass through the equilibrium position? Is the object heading upwards or downwards at this instant? 2. If the object is attached to the spring and released 3 feet below the equilibrium position with an upward velocity of 8 feet per second, ﬁnd the equation of motion of the object, x(t). What is the longest distance the object travels above the equilibrium position? When does this ﬁrst happen? Conﬁrm your result using a graphing utility. Solution. In order to use the formulas in Theorem 11.1, we ﬁrst need to determine the spring constant k and the mass of the object m. To ﬁnd k, we use Hooke’s Law F = kd. We know the object weighs 64 lbs. and stretches the spring 8 ft.. Using F = 64 and d = 8, we get 64 = k · 8, or k = 8 lbs. s2. We get m = 2 slugs. We can now proceed to apply Theorem 11.1. ft.. To ﬁnd m, we use w = mg with w = 64 lbs. and g = 32 ft. k 8 1. With k = 8 and m = 2, we get ω = 2 = 2. We are told that the object is released 3 feet below the equilibrium position ‘from rest.’ This means x0 = 3 and v0 = 0. Therefore, 32 + 02 = 3. To determine the phase φ, we have A sin(φ) = x0, 2 and angles coterminal to it A = which in this case gives 3 sin(φ) = 3 so sin(φ) = 1. Only φ = π 0 + v0 x2 m = 2 √ = ω 13The sign conventions here are carried over from Physics. If not for the spring, the
object would fall towards the ground, which is the ‘natural’ or ‘positive’ direction. Since the spring force acts in direct opposition to gravity, any movement upwards is considered ‘negative’. 888 Applications of Trigonometry = 0. Going through the usual analysis we ﬁnd t = − π satisfy this condition, so we pick14 the phase to be φ = π 2. Hence, the equation of motion is x(t) = 3 sin 2t + π. To ﬁnd when the object passes through the equilibrium position we 2 solve x(t) = 3 sin 2t + π 2 k for 2 integers k. Since we are interested in the ﬁrst time the object passes through the equilibrium position, we look for the smallest positive t value which in this case is t = π 4 ≈ 0.78 seconds after the start of the motion. Common sense suggests that if we release the object below the equilibrium position, the object should be traveling upwards when it ﬁrst passes through it. To check this answer, we graph one cycle of x(t). Since our applied domain in this situation is t ≥ 0, and the period of x(t) is T = 2π 2 = π, we graph x(t) over the interval [0, π]. Remembering that x(t) > 0 means the object is below the equilibrium position and x(t) < 0 means the object is above the equilibrium position, the fact our graph is crossing through the t-axis from positive x to negative x at t = π ω = 2π 4 + π 4 conﬁrms our answer. ω = 2 0 + v0 x2 5. From Aω cos(φ) = v0, we get 10 cos(φ) = −8, or cos(φ) = − 4 2. The only diﬀerence between this problem and the previous problem is that we now release the object with an upward velocity of 8 ft s. We still have ω = 2 and x0 = 3, but now we have v0 = −8, the negative indicating the velocity is directed upwards. Here, we get 32 + (−4)2 = 5. From A sin(φ) = x0, we get 5 sin(φ) = 3 which gives A = sin(φ)
= 3 5. This means that φ is a Quadrant II angle which we can describe in terms of either arcsine or arccosine. Since x(t) is expressed in terms of sine, we choose to express φ = π − arcsin 3. Hence, 5. Since the amplitude of x(t) is 5, the object will travel x(t) = 5 sin 2t + π − arcsin 3 5 at most 5 feet above the equilibrium position. To ﬁnd when this happens, we solve the equation x(t) = 5 sin 2t + π − arcsin 3 = −5, the negative once again signifying that 5 the object is above the equilibrium position. Going through the usual machinations, we get 2 arcsin 3 t = 1 4 + πk for integers k. The smallest of these values occurs when k = 0, 5 that is, t = 1 4 ≈ 1.107 seconds after the start of the motion. To check our answer using the calculator, we graph y = 5 sin 2x + π − arcsin 3 on a graphing utility 5 and conﬁrm the coordinates of the ﬁrst relative minimum to be approximately (1.107, −5). + π 2 arcsin 3 + π 5 x π 4 π 2 3π 4 π t 3 2 1 −1 −2 −3 x(t) = 3 sin 2t + π 2 y = 5 sin 2x + π − arcsin 3 5 It is possible, though beyond the scope of this course, to model the eﬀects of friction and other external forces acting on the system.15 While we may not have the Physics and Calculus background 14For conﬁrmation, we note that Aω cos(φ) = v0, which in this case reduces to 6 cos(φ) = 0. 15Take a good Diﬀerential Equations class to see this! 11.1 Applications of Sinusoids 889 to derive equations of motion for these scenarios, we can certainly analyze them. We examine three cases in the following example. Example 11.1.4. 1. Write x(t) = 5e−t/5 cos(t) + 5e−t/5 using a graphing utility. √ 3 sin(t) in the form x(t) =
A(t) sin(ωt + φ). Graph x(t) 2. Write x(t) = (t + 3) √ 2 cos(2t) + (t + 3) √ x(t) using a graphing utility. 2 sin(2t) in the form x(t) = A(t) sin(ωt + φ). Graph 3. Find the period of x(t) = 5 sin(6t) − 5 sin (8t). Graph x(t) using a graphing utility. Solution. √ √ √ 1. We start rewriting x(t) = 5e−t/5 cos(t) + 5e−t/5 3 sin(t) by factoring out 5e−t/5 from both terms to get x(t) = 5e−t/5 cos(t) + 3 sin(t). We convert what’s left in parentheses to the required form using the formulas introduced in Exercise 36 from Section 10.5. We ﬁnd so that x(t) = 10e−t/5 sin t + π cos(t) +. Graphing this on the 3 sin(t) = 2 sin t + π 3 3 calculator as y = 10e−x/5 sin x + π reveals some interesting behavior. The sinusoidal nature 3 continues indeﬁnitely, but it is being attenuated. In the sinusoid A sin(ωx + φ), the coeﬃcient, we can think A of the sine function is the amplitude. In the case of y = 10e−x/5 sin x + π 3 of the function A(x) = 10e−x/5 as the amplitude. As x → ∞, 10e−x/5 → 0 which means the amplitude continues to shrink towards zero. Indeed, if we graph y = ±10e−x/5 along with y = 10e−x/5 sin x + π, we see this attenuation taking place. This equation corresponds to 3 the motion of an object on a spring where there is a slight force which acts to ‘damp’, or slow the motion. An example of this kind of force would be the friction of the object against the air. In this model, the object oscillates forever, but with smaller and
smaller amplitude. y = 10e−x/5 sin x + π 3 y = 10e−x/5 sin x + π 3, y = ±10e−x/5 √ 2. Proceeding as in the ﬁrst example, we factor out (t + 3) 2 from each term in the function 2(cos(2t) + sin(2t)). We ﬁnd x(t) = (t + 3). Graphing this on the (cos(2t) + sin(2t)) = calculator as y = 2(x + 3) sin 2x + π, we ﬁnd the sinusoid’s amplitude growing. Since our 4 amplitude function here is A(x) = 2(x + 3) = 2x + 6, which continues to grow without bound, so x(t) = 2(t + 3) sin 2t + π 4 √ 2 sin 2t + π 4 2 sin(2t) to get x(t) = (t + 3) 2 cos(2t) + (t + 3) √ √ √ 890 Applications of Trigonometry as x → ∞, this is hardly surprising. The phenomenon illustrated here is ‘forced’ motion. That is, we imagine that the entire apparatus on which the spring is attached is oscillating as well. In this case, we are witnessing a ‘resonance’ eﬀect – the frequency of the external oscillation matches the frequency of the motion of the object on the spring.16 y = 2(x + 3) sin 2x + π 4 y = 2(x + 3) sin 2x + π 4 y = ±2(x + 3) 8 = 3 3. Last, but not least, we come to x(t) = 5 sin(6t) − 5 sin(8t). To ﬁnd the period of this function, we need to determine the length of the smallest interval on which both f (t) = 5 sin(6t) and g(t) = 5 sin(8t) complete a whole number of cycles. To do this, we take the ratio of their frequencies and reduce to lowest terms: 6 4. This tells us that for every 3 cycles f makes, In other words, the period of x(t) is three times
the period of f (t) (which is g makes 4. four times the period of g(t)), or π. We graph y = 5 sin(6x) − 5 sin(8x) over [0, π] on the calculator to check this. This equation of motion also results from ‘forced’ motion, but here the frequency of the external oscillation is diﬀerent than that of the object on the spring. Since the sinusoids here have diﬀerent frequencies, they are ‘out of sync’ and do not amplify each other as in the previous example. Taking things a step further, we can use a sum to product identity to rewrite x(t) = 5 sin(6t) − 5 sin(8t) as x(t) = −10 sin(t) cos(7t). The lower frequency factor in this expression, −10 sin(t), plays an interesting role in the graph of x(t). Below we graph y = 5 sin(6x) − 5 sin(8x) and y = ±10 sin(x) over [0, 2π]. This is an example of the ‘beat’ phenomena, and the curious reader is invited to explore this concept as well.17 y = 5 sin(6x) − 5 sin(8x) over [0, π] y = 5 sin(6x) − 5 sin(8x) and y = ±10 sin(x) over [0, 2π] 16The reader is invited to investigate the destructive implications of resonance. 17A good place to start is this article on beats. 11.1 Applications of Sinusoids 891 11.1.2 Exercises 1. The sounds we hear are made up of mechanical waves. The note ‘A’ above the note ‘middle second. Find a sinusoid which C’ is a sound wave with ordinary frequency f = 440 Hertz = 440 cycles models this note, assuming that the amplitude is 1 and the phase shift is 0. 2. The voltage V in an alternating current source has amplitude 220 √ 2 and ordinary frequency f = 60 Hertz. Find a sinusoid which models this voltage. Assume that the phase is 0. 3. The London Eye is a popular tourist attraction in London, England and is one of the largest Ferris Wheels in the world. It has a diameter of 135 meters
and makes one revolution (counterclockwise) every 30 minutes. It is constructed so that the lowest part of the Eye reaches ground level, enabling passengers to simply walk on to, and oﬀ of, the ride. Find a sinsuoid which models the height h of the passenger above the ground in meters t minutes after they board the Eye at ground level. 4. On page 732 in Section 10.2.1, we found the x-coordinate of counter-clockwise motion on a circle of radius r with angular frequency ω to be x = r cos(ωt), where t = 0 corresponds to the point (r, 0). Suppose we are in the situation of Exercise 3 above. Find a sinsusoid which models the horizontal displacement x of the passenger from the center of the Eye in meters t minutes after they board the Eye. Here we take x(t) > 0 to mean the passenger is to the right of the center, while x(t) < 0 means the passenger is to the left of the center. 5. In Exercise 52 in Section 10.1, we introduced the yo-yo trick ‘Around the World’ in which a yo-yo is thrown so it sweeps out a vertical circle. As in that exercise, suppose the yo-yo string is 28 inches and it completes one revolution in 3 seconds. If the closest the yo-yo ever gets to the ground is 2 inches, ﬁnd a sinsuoid which models the height h of the yo-yo above the ground in inches t seconds after it leaves its lowest point. 6. Suppose an object weighing 10 pounds is suspended from the ceiling by a spring which stretches 2 feet to its equilibrium position when the object is attached. (a) Find the spring constant k in lbs. (b) Find the equation of motion of the object if it is released from 1 foot below the equilibrium position from rest. When is the ﬁrst time the object passes through the equilibrium position? In which direction is it heading? ft. and the mass of the object in slugs. (c) Find the equation of motion of the object if it is released from 6 inches above the equilibrium position with a downward velocity of 2 feet per second. Find when the object passes through the equilibrium position heading downwards for the third time. 892 Applications of Trigonometry 7. Consider the pendulum below. Ignoring air resistance, the angular displacement of the pen- dulum from the vertical position
, θ, can be modeled as a sinusoid.18 θ The amplitude of the sinusoid is the same as the initial angular displacement, θ0, of the pendulum and the period of the motion is given by T = 2π l g where l is the length of the pendulum and g is the acceleration due to gravity. (a) Find a sinusoid which gives the angular displacement θ as a function of time, t. Arrange things so θ(0) = θ0. (b) In Exercise 40 section 5.3, you found the length of the pendulum needed in Jeﬀ’s antique Seth-Thomas clock to ensure the period of the pendulum is 1 2 of a second. Assuming the initial displacement of the pendulum is 15◦, ﬁnd a sinusoid which models the displacement of the pendulum θ as a function of time, t, in seconds. 8. The table below lists the average temperature of Lake Erie as measured in Cleveland, Ohio on the ﬁrst of the month for each month during the years 1971 – 2000.19 For example, t = 3 represents the average of the temperatures recorded for Lake Erie on every March 1 for the years 1971 through 2000. Month Number, t Temperature (◦ F), 10 11 12 36 33 34 38 47 57 67 74 73 67 56 46 (a) Using the techniques discussed in Example 11.1.2, ﬁt a sinusoid to these data. (b) Using a graphing utility, graph your model along with the data set to judge the reason- ableness of the ﬁt. 18Provided θ is kept ‘small.’ Carl remembers the ‘Rule of Thumb’ as being 20◦ or less. Check with your friendly neighborhood physicist to make sure. 19See this website: http://www.erh.noaa.gov/cle/climate/cle/normals/laketempcle.html. 11.1 Applications of Sinusoids 893 (c) Use the model you found in part 8a to predict the average temperature recorded for Lake Erie on April 15th and September 15th during the years 1971–2000.20 (d) Compare your results to those obtained using a graphing utility. 9. The fraction of the moon illuminated at midnight Eastern Standard Time on the tth day of June, 2009 is given
in the table below.21 Day of June, t Fraction Illuminated, F 3 6 9 12 15 18 21 24 27 30 0.81 0.98 0.98 0.83 0.57 0.27 0.04 0.03 0.26 0.58 (a) Using the techniques discussed in Example 11.1.2, ﬁt a sinusoid to these data.22 (b) Using a graphing utility, graph your model along with the data set to judge the reason- ableness of the ﬁt. (c) Use the model you found in part 9a to predict the fraction of the moon illuminated on June 1, 2009. 23 (d) Compare your results to those obtained using a graphing utility. 10. With the help of your classmates, research the phenomena mentioned in Example 11.1.4, namely resonance and beats. 11. With the help of your classmates, research Amplitude Modulation and Frequency Modulation. 12. What other things in the world might be roughly sinusoidal? Look to see what models you can ﬁnd for them and share your results with your class. 20The computed average is 41◦F for April 15th and 71◦F for September 15th. 21See this website: http://www.usno.navy.mil/USNO/astronomical-applications/data-services/frac-moon-ill. 22You may want to plot the data before you ﬁnd the phase shift. 23The listed fraction is 0.62. 894 Applications of Trigonometry 11.1.3 Answers 1. S(t) = sin (880πt) 2. V (t) = 220 √ 2 sin (120πt) 3. h(t) = 67.5 sin π 15 t − π 2 + 67.5 4. x(t) = 67.5 cos π 15 t − π 2 = 67.5 sin π 15 t 5. h(t) = 28 sin 2π + 30 2 3 t − π ft. and m = 5 16 slugs 6. (a) k = 5 lbs. (b) x(t) = sin 4t + π 2. The object ﬁrst passes through the equilibrium point when t = π 8 ≈ 0.39 seconds after the motion starts. At this time, the object
is heading upwards. (c) x(t) = √ 2 2 sin 4t + 7π 4. The object passes through the equilibrium point heading down- wards for the third time when t = 17π 16 ≈ 3.34 seconds. 7. (a) θ(t) = θ0 sin g l t + π 2 (b) θ(t) = π 12 sin 4πt + π 2 8. (a) T (t) = 20.5 sin π (b) Our function and the data set are graphed below. The sinusoid seems to be shifted to 6 t − π + 53.5 the right of our data. (c) The average temperature on April 15th is approximately T (4.5) ≈ 39.00◦F and the average temperature on September 15th is approximately T (9.5) ≈ 73.38◦F. (d) Using a graphing calculator, we get the following This model predicts the average temperature for April 15th to be approximately 42.43◦F and the average temperature on September 15th to be approximately 70.05◦F. This model appears to be more accurate. 11.1 Applications of Sinusoids 895 9. (a) Based on the shape of the data, we either choose A < 0 or we ﬁnd the second value of t which closely approximates the ‘baseline’ value, F = 0.505. We choose the latter to obtain F (t) = 0.475 sin π 15 t − 2π + 0.505 = 0.475 sin π 15 t + 0.505 (b) Our function and the data set are graphed below. It’s a pretty good ﬁt. (c) The fraction of the moon illuminated on June 1st, 2009 is approximately F (1) ≈ 0.60 (d) Using a graphing calculator, we get the following. This model predicts that the fraction of the moon illuminated on June 1st, 2009 is approximately 0.59. This appears to be a better ﬁt to the data than our ﬁrst model. 896 Applications of Trigonometry 11.2 The Law of Sines Trigonometry literally means ‘measuring triangles’ and with Chapter 10 under our belts, we are more than prepared to do just that. The
main goal of this section and the next is to develop theorems which allow us to ‘solve’ triangles – that is, ﬁnd the length of each side of a triangle and the measure of each of its angles. In Sections 10.2, 10.3 and 10.6, we’ve had some experience solving right triangles. The following example reviews what we know. Example 11.2.1. Given a right triangle with a hypotenuse of length 7 units and one leg of length 4 units, ﬁnd the length of the remaining side and the measures of the remaining angles. Express the angles in decimal degrees, rounded to the nearest hundreth of a degree. Solution. For deﬁnitiveness, we label the triangle below √ To ﬁnd the length of the missing side a, we use the Pythagorean Theorem to get a2 + 42 = 72 which then yields a = 33 units. Now that all three sides of the triangle are known, there are several ways we can ﬁnd α using the inverse trigonometric functions. To decrease the chances of propagating error, however, we stick to using the data given to us in the problem. In this case, the lengths 4 and 7 were given, so we want to relate these to α. According to Theorem 10.4, cos(α) = 4 7. Since α is an acute angle, α = arccos 4 radians. Converting to degrees, we ﬁnd α ≈ 55.15◦. Now 7 that we have the measure of angle α, we could ﬁnd the measure of angle β using the fact that α and β are complements so α + β = 90◦. Once again, we opt to use the data given to us in the radians and we problem. According to Theorem 10.4, we have that sin(β) = 4 have β ≈ 34.85◦. 7 so β = arcsin 4 7 A few remarks about Example 11.2.1 are in order. First, we adhere to the convention that a lower case Greek letter denotes an angle1 and the corresponding lowercase English letter represents the side2 opposite that angle. Thus, a is the side opposite α, b is the side opposite β and c is the side opposite γ. Taken together, the pairs (α, a), (β, b) and (γ

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