Section 8 Inverse Trigonometric Functions

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1 Section 8 Inverse Trigonometric Functions Inverse Sine Function Recall that for every function y = f (x), one may de ne its INVERSE FUNCTION y = f 1 (x) as the unique solution of x = f (y). In other words, the inverse function y = f 1 (x) "undo" the function y = f (x) : f f 1 (x) = x, f 1 (f (x)) = x. We also know that even if a function y = f (x) is de ned every, its inverse function y = f 1 (x) may not be de ned everywhere. The su cient condition that guarantees existence of inverse functions is called "horizontal line test": a horizontal line can intersect the graph of y = f (x) at most once. For trigonometric functions, for instance the graph of y = sin x intersects horizontal y = 0. in nite many times: Therefore, to de ne inverse function of y = sin x, we consider the restricted sine function 1

2 h i.e., the restriction of y = sin x on π, π i. This function has the domain h π, π i and the range [ 1, 1] ; it captures all information of the sine function over all real numbers. Notice that the only di erence between the restricted Sine function and the (unrestricted) Sine function is their domains: h restricted y = sin x de ned only for x on π, π i y = sin x de ned for all real numbers x h π, π i is invert- De nition: The restricted Sine function de ned on ible, and its inverse function is denoted as y = sin 1 x or y = arcsin x. h The domain of y = arcsin x is [ 1, 1], and the range y = arcsin x is π, π i. The graph of y = arcsin x is the re ection of the graph for the restricted Sine function about y = x : Example 1 Find the following inverse Sine function values: (a) arcsin ³ 1 (b) arcsin p

3 Solution: (a) By the general de nition of inverse functions, y = arcsin 1 is the solution of the restricted Sine function for y : 1 = sin y The words "restricted Sine" means that Therefore π y π. y = π, or arcsin 1 = π (Notice that the notation "arcsin" comes from the fact that the length of the arc whose sine is 1/ is π/.) (b) Similarly, we need to solve p = sin y for The solution is π y π. y = π p! Ã, or arcsin = π Recall that by the de nition of inverse function, Therefore, Example Find following values: f f 1 (x) = x for x in D f 1 f 1 (f (x)) = x for x in D (f) sin (arcsin x) = x for x in [ 1, 1] h arcsin (sin x) = x for x in π, π i

4 ³ (a) arcsin sin (b) arcsin (c) arcsin ³ π ³ ³ π = sin 1 sin = 4 4 π π sin = sin 1 sin = 4 4 7π 7π = sin 1 sin = sin Solution: (a) According to the formula above, since x = π 4 is in π, π ³ ³ π arcsin sin = π 4 4 (b) This time, since x = π 4 is not in π, π, the above formula that would have lead to arcsin π sin = π 4 4 doesn t apply. We thus solve this problem, we need to nd an angle x in π, π such that π sin x = sin. 4 This can be easily done as follows: π ³ sin = sin π π = sin π Thus π ³ ³ π arcsin sin = arcsin sin = π (c) This time again 7π is outside of the domain of the restricted sine function π, π. according to the method in (b), 7π ³ sin = sin π + π = sin π ³ = sin π. Now since π is in π, π, arcsin sin 7π ³ ³ = arcsin sin π = π. 4

5 Example Find the following values (a) sin arcsin = sin sin 1 = (b) sin (arcsin ) = sin sin 1 () = Solution: (a) Obviously, x = is in the domain of arcsin x, [ 1, 1]. sin arcsin = (b) x = is not in the [ 1, 1], arcsin is unde ned. So sin (arcsin ) is not de ned. Example 4 (a) What is cos sin 1 x? (b) What is sin sin 1 x? Solution: (a) Set θ = sin 1 x which is de ned for any 1 x 1. This means sin θ = x and π θ π Thus cos θ 0, and p cos θ = 1 sin θ = p 1 x. (b) Inverse Cosine function For y = cos x sin θ = sin θ cos θ = x p 1 x 5

6 we de ne the restricted Cosine function as so that its graph is the piece in [0, π] y = cos x for x in [0, π] Apparently, the restricted Cosine function passes the horizontal line test and thus is invertible. We call the inverse function of the restricted Cosine function inverse Cosine and is denoted by y = cos 1 x or y = arccos x. Analogous to the inverse sine function, there are some basic facts for y = cos 1 x : y = cos 1 x has domain [ 1, 1] and range [0, π]

7 cos cos 1 x = x for x in [ 1, 1] cos 1 (cos x) = x for x in [0, π] Example 5 Find the values: (a) cos 1 (0), (b) cos 1 1 Solution: (a) θ = cos 1 (0) is the angle in [0, π] satisfying cos θ = 0 =) θ = π (b) θ = cos 1 1 is the solution of cos θ = 1, θ in [0, π] Since cos π = 1, cos π ³ = cos π π = cos π = 1. So θ = cos 1 1 = π Example compute (a) cos cos 1, (b) arccos (cos 4) Solution: (a) Since x = is in the domain of arccos x, cos cos 1 = (b) Note that x = 4 > π '.14, which is beyond the domain of the restricted cosine. So the formula arccos (cos x) = x doesn t apply. 7

8 we need to nd an angle θ in [0, π] such that cos θ = cos 4. To this end cos 4 = cos (π 4), θ = π 4 is in [0, π] and arccos (cos 4) = arccos (cos θ) = θ = π 4 Example 7 (a) What is sin (cos 1 x)?(b) What is sin 1 x + cos 1 x? Solution: (a) Set θ = cos 1 x so θ is in [0, π] where sin θ > 0. Thus sin cos 1 x = sin θ = p 1 cos θ = p 1 x. (b) Set α = sin 1 x. Then x = sin α. Draw a right triangle whose one angle is α, hypotenuse = 1, and the opposite side is a = x. Then this side x is adjacent to the third angle in the triangle, i.e., β = π α, and cos β = x/1, or β = cos 1 x. We conclude sin 1 x + cos 1 x = α+ β = π Inverse function for Tangent Function Since the graph of y = tan x is and the period is π. So one entire period 8

9 passes the horizontal ³ test and thus invertible. The inverse function of this restricted piece in π, π is denoted as y = tan 1 x or y = arctan x. Some basic facts: y = tan 1 x has the domain ( 1, 1) and range ³ π, π tan 1 (tan x) = x only for x in ³ π, π tan tan 1 x = x for all x. Example 8 Find (a) tan 1 ( 1), (b) tan tan 1 p 5 9

10 ³ Solution: (a) Since tan π = 1, tan 1 ( 1) = π 4 4 (b) tan tan p 1 5 = p 5 Example 9 (a) Given tan θ = x/, 0 < θ < π/. Express tan θ as a function of x. (b) Simplify sec (tan 1 x) Solution: (a) Using double-angle formula, ³ x tan θ = tan θ 1 tan θ = ³ x = 4x 4 x 1 (b) Set θ = tan 1 x. x = tan θ. sec tan 1 x = sec x = p 1 + tan θ = p 1 + x Homework: In Exercise 1-10, evaluate each of the quantities that is de ned. quantity is unde ned, explain it. Ãp! 1. sin 1 = If a. cos 1 ( 1) =. tan p 1 = 4. sin sin 1 5 = 4 4π 5. cos cos 1 = 5 ³. tan ³tan 1 π = 5 7. sin 1 sin 4π = 10

11 8. cos cos 1 = 7 π 9. sin sin 1 = 10. cos (cos 1 ) = In Exercise 11-15, nd and simplify the exact value of each quantity tan sin cos arcsin 9 1. sin (tan 1 ) sin cos cos tan 1 4 Simplify the following expressions. 1. tan (cos 1 x) 17. sin (tan 1 x) 18. cos sin 1 x + cos 1 x 11