2 2 [ 1, 1]. Thus there exists a function sin 1 or arcsin. sin x arcsin x(= sin 1 x) arcsin(x) = y sin(y) = x. arcsin(sin x) = x if π 2 x π 2

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1 7.4 Inverse Trigonometric Functions From looking at the graphs of the trig functions, we see that they fail the horizontal line test spectacularly. However, if you restrict their domain, you can find an inverse for these functions on this domain only. Inverse Sine The function sin is one-to-one when restricted to sin : [ π, ] π [ 1, 1]. Thus there exists a function sin 1 or arcsin [ sin 1 : [ 1, 1] π, π ] The arcsin function satisfies sin x arcsin x(= sin 1 x) arcsin(x) = y sin(y) = x arcsin(sin x) = x if π x π sin(arcsin x) = x if 1 x 1 Examples: Find (a)sin 1 ( 1 ), (b)arcsin( 1 ), (c)sin 1 ( ) and (d)cos(sin 1 ( 5 )). (a) We know that sin( π 6 ) = 1, so sin 1 ( 1 ) = π 6. (b) Likewise, sin( π 6 ) = 1, so arcsin( 1 ) = π 6.

2 (c) We know sin x is never (it is never greater than 1), so sin 1 ( ) is undefined. (d) For u [ π, ] π, we have cos(u) = 1 sin u, so ( ) ( ( )) cos(sin 1 5 ) = 1 sin sin 1 5 ( ) = 1 5 = = 16 5 = 4 5 Inverse Cosine The function cos is one-to-one when restricted to cos : [0, π] [ 1, 1]. Thus there exists a function cos 1 or arccos cos 1 : [ 1, 1] [0, π] cos x arccos x(= cos 1 x) The arccos function satisfies arccos(x) = y cos(y) = x arccos(cos x) = x if 0 x π

3 cos(arccos x) = x if 1 x 1 ( ) Examples: Find (a) cos 1 and (b) cos 1 (0). (a) cos( π) = ( ), thus 6 cos 1 = π. 6 (b) cos( π ) = 0, thus cos 1 (0) = π Examples: 1. Show that sin(cos 1 (x)) = 1 x. sin u = 1 cos (u) for u [0, π]. Thus. Show that tan(cos 1 (x)) = 1 x x. tan(u) = sin u cos u, so sin(cos 1 (x)) = 1 cos (cos 1 (x)) = 1 x tan(cos 1 (x)) = sin(cos 1 (x)) cos(cos 1 (x)) =. Show that sin( cos 1 (x)) = x 1 x sin u = sin u cos u, so 1 x sin( cos 1 (x) = sin(cos 1 (x)) cos(cos 1 (x)) = 1 x x = x 1 x x Inverse Tangent The function tan is one-to-one when restricted to tan : [ π, ] π R. Thus there exists a function tan 1 or arctan [ tan 1 : R π, π ]

4 The arctan function satisfies tan x arctan x(= tan 1 x) arctan(x) = y tan(y) = x arctan(tan x) = x if π x π tan(arctan x) = x if x R Examples: Find (a) tan 1 (1), (b) tan 1 and (c)arctan( 0) (a) tan( π 4 ) = 1, thus tan 1 (1) = π 4. (b) tan( π ) =, thus tan 1 ( ) = π (c) Using a calculator, we find arctan( 0) Other Inverse Trig Functions The other trig functions cot, csc and sec also have inverses when restricted to suitable domains, namely cot 1, csc 1 and sec 1. You don t need to worry about graphing these. Just keep in mind: cot 1 1 tan 1 csc 1 1 sin 1 sec 1 1 cos 1

5 7.5 Trigonometric Equations One frequently has to solve equations involving trig functions. Sometimes the values of x you look at are restricted, while others you are asked to find all the values of x that make a given equation true. In the latter case, there are usually an infinite number of solutions whose form depends on the period of the trig functions involved. Examples: 1. Solve (a) tan (x) = 0 (b) sin(x) = cos(x) (c) 1 + sin x = cos (x) (d) sin x cos x = 0 (e) cos(x) + 1 = sin x with t [0, π] (a) tan (x) = 0 tan(x) = ±. tan(x) = when x = π. But tan has period π, so tan(x) = x = π + kπ k Z Likewise, tan(x) = x = π + kπ k Z Thus the full set of solutions is S = { π + kπ, x = π } + kπ k Z (b) If sin(x) = cos(x), cos(x) 0 because sin and cos are not 0 in the same places. Thus we can divide both sides by cos and get sin(x) cos(x) = 1 tan(x) = 1 This happens when x = π. Once again, tan has a period of π so 4 x = π 4 + kπ k Z (c) We can get the equation 1 + sin x = cos (x) into an easier form to deal with by subbing in 1 sin x for cos x: 1 + sin x = cos (x) 1 + sin x = sin x sin x + sin x 1 = 0

6 This is a quadratic in sin x that factors as ( sin x 1)(sin x + 1) = 0 Which means either sin x 1 = 0 or sin x + 1 = 0. sin x = 1, and the second gives sin x = 1. sin x = 1 x = π + kπ, k Z sin x = 1 x = π 6 + kπ, k Z or x = 5π 6 + kπ, k Z (d) We use a double angle identity on sin x cos x = 0 to get Hence either sin x cos x cos x = 0 cos x( sin x 1) = 0 cos x = 0 x = π + kπ, k Z or x = π sin x = 1 x = π 6 + kπ, k Z or x = 5π 6 + kπ, k Z. + kπ, k Z. (e) Here we have to be a little trickier and square both sides cos x + 1 = sin x (cos x + 1) = 1 cos x cos x + cos x + 1 = 1 cos x cos x + cos x = 0 cos x(cos x 1) = 0 The first gives us that Hence either cos x = 0 or cos x = 1. Between 0 and π, the first only happens at π and π and the second happens at π. Hence the three solutions are. Consider the equation sin(x) 1 = 0. (a) Find all the solutions to the equation. x = π, π, π (b) Find all the solutions to the equation in the interval [0, π). (a) The equation rearranges to sin(x) = 1. As we ve seen before, this means that x = π 6 + kπ, k Z or x = 5π 6 + kπ, k Z. x = π 18 + kπ 5π, k Z or x = 18 + kπ, k Z.

7 (b) We solve the inequalities 0 π 18 + kπ < π k < 1 18 k < 1 18 = k < k < The ks in this range are k = 0, k = 1 and k =. Hence For the other solutions we have x = π 18, 1π 18, 5π π 18 + kπ < π k < 5 18 k < 5 18 = k < k < The ks in this range are k = 0, k = 1 and k =. Hence. Consider the equation tan( x ) 1 = 0. (a) Find all the solutions of the equation. x = 5π 18, 17π 18, 9π 18 (b) Find all the solutions in the interval [0, 4π). (a) This rearranges to tan( x ) = 1 This gives us x = π 6 + kπ, k Z x = π + kπ, k Z

8 (b) As before, we find the ks we need: 0 π + kπ < 4π k < 4 1 k < 4 1 = k < The ks in this range are k = 0 and k = 1. Hence x = π or x = 7π. 4. Solve the equation tan x tan x = 0. This is a quadratic in tan x, so tan x tan x = 0 (tan x )(tan x + 1) = 0 So tan x = or tan x = 1. There isn t a convenient angle with tan x =, but we can use tan 1 to write x = tan 1 () + kπ, k Z The second one of course gives us 5. Solve the equation sin θ = 0. x = π 4 + kπ, k Z This rearranges to sin θ =. Once again there is no easy angle that gives us sin θ =. We know that sin is positive in the first and second quadrants, and thus the two solutions in [0, π) are ) ) sin 1 ( and π sin 1 ( Hence the full solution is ( ) ( ) θ = sin 1 + kπ or θ = π sin 1 + kπ