Chapter 7

7.4 Inverse Trigonometric Functions

Trigonometric functions are not one-to-one functions. So we need to restrict the domain of each function to be a one-to-one function to get the inverse trigonometric functions. The following graphs shows that y = sin x, y = cos x, and y = tan x are one-to-one in the restricted domains.

In these restricted domains, we can define the inverse functions y = sin-1x, y = cos-1x and y = tan-1x.
fig 7.4-1

fig 7.4-2

fig 7.4-3

fig 7.4-4

fig 7.4-5

fig 7.4-6

We can define the other trigonometric functions in the same manner.


Example 8.
Evaluate (a) sin-1 1⁄2
(b) cos-1 - (1⁄2)
$$\small{\text{(c) tan}^{-1} (-\sqrt{3})}$$ (d) sin-1 (sin 5π⁄6 ).
Solution
(a) Since sin π⁄6 = 1⁄2 and - π⁄2 ≤ π⁄6 ≤ π⁄2,
we have sin-1 1⁄2 = π⁄6.

(b) Since cos 2π⁄3 = - 1⁄2 and
0 ≤ 2π⁄3 ≤ π, we have
cos-1 (- 1⁄2) = 2π⁄3.

(c) Since tan(- π⁄3) = - √ 3 and
π⁄2 ≤ - π⁄3 ≤ π⁄2, we have
tan-1 (- √ 3 ) = - π⁄3.

(d) sin-1(sin 5π⁄6) = sin-1 (1⁄2) = π⁄6.

From y = sin-1 x if sin y = x, we have sin(sin-1 x) = x for the domain -1 ≤ x ≤ 1 of y = sin-1 x, but from Example 8(d) one can see that
sin-1(sin x) = x is not true in general.
Since sin(-sin-1x) = -sin(sin-1x) = -1, we have
sin-1(-x) = -sin-1x

so y = sin-1x is an odd function.

From y = cos-1x if cos y = x, we have cos(cos-1x) = x for the domain -1 ≤ x ≤ 1 of y = cos-1x. Since cos(π - cos-1x) = - cos(cos-1x) = -x,
π - cos-1x = cos-1(-x)
or
cos-1x + cos-1(-x = π

Exercise 7.4
1. Evaluate (a) csc(sin-10.3)
Solution
Let y = sin-1 0.3       ( - π⁄2 ≤ y ≤ π⁄2 )
sin y = 0.3 = 3⁄10
csc(sin-1 0.3) = csc y = 1⁄sin y
= 10⁄3

(b) cot(tan-1 5)
Solution
Let y = tan-1 5       ( - π⁄2 < y < π⁄2 )
tan y = 5
cot(tan-1 5) = cot y = 1⁄tan y
= 1⁄5 = 0.2

(c) sec(cos-1(-0.75)).
Solution
Let y = cos-1 (-0.75)       (0 ≤ y ≤ π)
cos y = -0.75
cos y = - 75⁄100 = - 3⁄4
csc(cos-1 (-0.75)) = csc y
= 1⁄cos y = - 4⁄3

2. Evaluate (a) sin-1 √ 3 ⁄2
Solution
Let y = sin-1 √ 3 ⁄2       ( - π⁄2 ≤ y ≤ π⁄2 )
sin y = √ 3 ⁄2 = sin π⁄3
y = π⁄3

(b) cos-1 √ 2 ⁄2
Solution
Let y = cos-1 √ 2 ⁄2       (0 ≤ y ≤ π)
cos y = √ 2 ⁄2 = cos π⁄4
y = π⁄4

(c) tan-1 √ 3 ⁄3
Solution
Let y = tan-1 √ 3 ⁄3       ( - π⁄2 < y < π⁄2 )
tan y = √ 3 ⁄3 = tan π⁄6
y = π⁄6

3. Evaluate (a) sin(cos-1 √ 3 ⁄2 )
Solution
sin(cos-1 √ 3 ⁄2 ) = sin π⁄6 = 1⁄2

(b) tan(sin-1 √ 3 ⁄2)
Solution
tan(sin-1 √ 3 ⁄2 ) = tan π⁄3 = √ 3

(c) csc(tan-1(- √ 3 ⁄3 ))
Solution
csc(tan-1(- √ 3 ⁄3 ) = csc(- π⁄6) = - csc π⁄6 = -2

4. Evaluate (a) sin-1(sin 2π⁄3)
Solution
sin-1(sin 2π⁄3) = sin-1 √ 3 ⁄2 = π⁄3

(b) cos-1(cos(- 1⁄2))
Solution
cos-1(cos(- 1⁄2)) = cos-1(cos(1⁄2)) = 1⁄2

(c) tan-1(tan(- √ 3 ))
Solution
tan-1(tan(- √ 3 )) = tan-1(tan( π - √ 3 )) = π - √ 3

5. Prove that sin-1 x + cos-1 x = π⁄2
Solution
Let y = sin-1 x       ( - π⁄2 ≤ y ≤ π⁄2 )
sin y = x
cos (π⁄2 - y) = x
π⁄2 - y = cos-1 x
y + cos-1 x = π⁄2
sin-1 x + cos-1 x = π⁄2