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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 : − π2 , π2 → [−1, 1]. Thus there exists a function sin−1 or arcsin h π πi sin−1 : [−1, 1] → − , 2 2 arcsin x(= sin−1 x) sin x The arcsin function satisfies arcsin(x) = y ⇔ sin(y) = x π π ≤x≤ 2 2 sin(arcsin x) = x if − 1 ≤ x ≤ 1 arcsin(sin x) = x if − Examples: Find (a)sin−1 ( 12 ), (b)arcsin(− 21 ), (c)sin−1 ( 32 ) and (d)cos(sin−1 ( 35 )). solution: (a) We know that sin( π6 ) = 12 , so sin−1 ( 12 ) = π6 . (b) Likewise, sin(− π6 ) = − 21 , so arcsin(− 12 ) = − π6 . (c) We know sin x is never 32 (it is never greater than 1), so sin−1 ( 32 ) is undefined. p (d) For u ∈ − π2 , π2 , we have cos(u) = 1 − sin2 u, so s 3 3 −1 2 −1 cos(sin ) = 1 − sin sin 5 5 s 2 3 = 1− 5 r 9 1− = 25 r 16 = 25 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 cos(arccos x) = x if − 1 ≤ x ≤ 1 √ 3 −1 and (b) cos−1 (0). Examples: Find (a) cos 2 solution: (a) cos( π6 ) = √ 3 , 2 thus cos−1 √ 3 2 (b) cos( π2 ) = 0, thus cos−1 (0) = = π6 . π 2 Examples: √ 1. Show that sin(cos−1 (x)) = 1 − x2 . p solution: sin u = 1 − cos2 (u) for u ∈ [0, π]. Thus p √ sin(cos−1 (x)) = 1 − cos2 (cos−1 (x)) = 1 − x2 2. Show that tan(cos−1 (x)) = solution: tan(u) = sin u , cos u √ 1−x2 . x so sin(cos−1 (x)) tan(cos (x)) = = cos(cos−1 (x)) −1 √ 1 − x2 x √ 3. Show that sin(2 cos−1 (x)) = 2x 1 − x2 solution: sin 2u = 2 sin u cos u, so √ √ sin(2 cos−1 (x) = 2 sin(cos−1 (x)) cos(cos−1 (x)) = 2 1 − x2 x = 2x 1 − x2 Inverse Tangent The function tan is one-to-one when restricted to tan : − π2 , π2 → R. Thus there exists a function tan−1 or arctan h π πi −1 tan : R → − , 2 2 arctan x(= tan−1 x) tan x The arctan function satisfies arctan(x) = y ⇔ tan(y) = x π π ≤x≤ 2 2 tan(arctan x) = x if x ∈ R arctan(tan x) = x if − Examples: Find (a) tan−1 (1), (b) tan−1 √ 3 and (c)arctan(−20) solution: (a) tan( π4 ) = 1, thus tan−1 (1) = π4 . √ √ (b) tan( π3 ) = 3, thus tan−1 ( 3) = π 3 (c) Using a calculator, we find arctan(−20) ≈ 1.52084 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: 1 cot−1 6= tan−1 1 csc−1 6= sin−1 1 sec−1 6= cos−1 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) tan2 (x) − 3 = 0 (b) sin(x) = cos(x) (c) 1 + sin x = 2 cos2 (x) (d) sin 2x − cos x = 0 (e) cos(x) + 1 = sin x with t ∈ [0, 2π] solution: √ (a) tan2 (x) − 3 = 0 ⇒ tan(x) = ± 3. √ tan(x) = 3 when x = π3 . But tan has period π, so √ π tan(x) = 3 ⇒ x = + kπ k ∈ Z 3 Likewise, √ π tan(x) = − 3 ⇒ x = − + kπ k ∈ Z 3 Thus the full set of solutions is n π o π S = − + kπ, x = + kπ | k ∈ Z 3 3 (b) If sin(x) = cos(x), cos(x) 6= 0 because sin and cos are not 0 in the same places. Thus we can divide both sides by cos and get sin(x) = 1 ⇒ tan(x) = 1 cos(x) This happens when x = π4 . Once again, tan has a period of π so π x = + kπ k ∈ Z 4 (c) We can get the equation 1 + sin x = 2 cos2 (x) into an easier form to deal with by subbing in 1 − sin2 x for cos2 x: 1 + sin x = 2 cos2 (x) 1 + sin x = 2 − 2 sin2 x 2 sin2 x + sin x − 1 = 0 This is a quadratic in sin x that factors as (2 sin x − 1)(sin x + 1) = 0 Which means either 2 sin x − 1 = 0 or sin x + 1 = 0. The first gives us that sin x = 12 , and the second gives sin x = −1. • sin x = −1 ⇒ x = • sin x = 1 2 ⇒x= π 6 3π 2 + 2kπ, k ∈ Z + 2kπ, k ∈ Z or x = 5π 6 + 2kπ, k ∈ Z (d) We use a double angle identity on sin 2x − cos x = 0 to get 2 sin x cos x − cos x = 0 cos x(2 sin x − 1) = 0 Hence either • cos x = 0 ⇒ x = • sin x = 1 2 ⇒x= π 2 π 6 + 2kπ, k ∈ Z or x = + 2kπ, k ∈ Z or x = 3π 2 5π 6 + 2kπ, k ∈ Z. + 2kπ, k ∈ Z. (e) Here we have to be a little trickier and square both sides cos x + 1 = sin x (cos x + 1)2 = 1 − cos2 x cos2 x + 2 cos x + 1 = 1 − cos2 x 2 cos2 x + 2 cos x = 0 2 cos x(cos x − 1) = 0 Hence either cos x = 0 or cos x = −1. Between 0 and 2π, the first only happens at π2 and 3π and the second happens at −π. Hence the three solutions are 2 x= π 3π , ,π 2 2 2. Consider the equation 2 sin(3x) − 1 = 0. (a) Find all the solutions to the equation. (b) Find all the solutions to the equation in the interval [0, 2π). solution: (a) The equation rearranges to sin(3x) = 12 . As we’ve seen before, this means that π 5π + 2kπ, k ∈ Z or 3x = + 2kπ, k ∈ Z. 6 6 π 2kπ 5π 2kπ x= + , k ∈ Z or x = + , k ∈ Z. 18 3 18 3 3x = (b) We solve the inequalities π 2kπ + < 2π 18 3 1 2k 0≤ + <2 18 3 2k 1 35 1 <2− = − ≤ 18 3 18 18 1 35 − ≤ 2k < 6 6 35 1 ≈ 2.91 − ≤k< 12 12 The ks in this range are k = 0, k = 1 and k = 2. Hence 0≤ x= π 13π 25π , , 18 18 18 For the other solutions we have 0≤ 5π 2kπ + < 2π 18 3 5 2k + <2 18 3 5 2k 5 31 − ≤ <2− = 18 3 18 18 31 5 − ≤ 2k < 6 6 31 5 ≈ 2.58 − ≤k< 12 12 The ks in this range are k = 0, k = 1 and k = 2. Hence 0≤ x= 3. Consider the equation √ 5π 17π 29π , , 18 18 18 3 tan( x2 ) − 1 = 0. (a) Find all the solutions of the equation. (b) Find all the solutions in the interval [0, 4π). solution: (a) This rearranges to tan( x2 ) = √1 3 This gives us x π = + kπ, k ∈ Z 2 6 π x = + 2kπ, k ∈ Z 3 (b) As before, we find the ks we need: 0≤ π + 2kπ < 4π 3 1 + 2k < 4 3 1 1 11 − ≤ 2k < 4 − = 3 3 3 1 11 − ≤k< ≈ 1.83 6 6 The ks in this range are k = 0 and k = 1. Hence x = 0≤ π 3 or x = 7π . 3 4. Solve the equation tan2 x − tan x − 2 = 0. solution: This is a quadratic in tan x, so tan2 x − tan x − 2 = 0 (tan x − 2)(tan x + 1) = 0 So tan x = 2 or tan x = −1. There isn’t a convenient angle with tan x = 2, but we can use tan−1 to write x = tan−1 (2) + kπ, k ∈ Z The second one of course gives us x=− π + kπ, k ∈ Z 4 5. Solve the equation 3 sin θ − 2 = 0. solution: This rearranges to sin θ = 32 . Once again there is no easy angle that gives us sin θ = 23 . We know that sin is positive in the first and second quadrants, and thus the two solutions in [0, 2π) are 2 2 −1 −1 sin and π − sin 3 3 Hence the full solution is −1 θ = sin 2 2 −1 + 2kπ or θ = π − sin + 2kπ 3 3