Trigonometric functions are as you know periodic functions. This means the function values are repeated in regular periods and the graphs of trigonometric functions fail the horizontal line test to confirm the existence of their inverses. But often in trigonometric problems the angle or variable values are to be found when the function values are known. For this purpose, the inverse trigonometric functions are defined by restricting the domains of trigonometric functions suitably. You can see the graphs of the functions in the restricted domains and the domain and range of inverse trigonometric functions. Other than the common notation f-1 (x) used in algebra, inverse trigonometric functions are also called arcsin, arccos arctan functions and indicated using these words. For example, the inverse sin trigonometric function is either written as sin-1 x or arcsin x.The following tables show the graphs of trigonometric functions, the restricted domains and the domain and range for the inverse functions defined.

Graphs Functions with
restricted domains

Domain and range of
inverse functions

Sin Function
y = sin x

Restricted Domain
{x | -π/2 ≤ x ≤ π/2}

Range
{y | -1 ≤ y ≤ 1}
y = arcsin x
or
y = sin-1 x

Domain
{x | -1 ≤ x ≤ 1|

Range
{y | -π/2 ≤ y ≤ π/2}
Cos Function y = cos x

Restricted Domain
{x | 0 ≤ x ≤ π}

Range
{y | -1 ≤ y ≤ 1}


y = arccos x
or
y = cos-1 x

Domain
{x | -1 ≤ x ≤ 1}

Range
{y | 0 ≤ y ≤ π}
Tan Function
y = tan x

Restricted Domain
{x | -π/2 < x < π/2}

Range
{y | -∞ < y < ∞}



y = arctan x
or
y = tan-1 x

Domain
{x | -∞ < x < ∞}

Range
{y | -π/2 < y < π/2}

The restricted graphs for which the inverse is defined are shown in green color. The values in the restricted domains are called the principal values. Let us see how the inverse trigonometric graphs look like and solve few problems on them.

Graph of Inverse Sine function
The graph of sine inverse function is the reflection over the line y = x of the part of the graph shown in green in the graph of sine function.

Sin Inverse

You may note in the graph the domain of sin inverse function is the closed interval [-1, 1] and the range is the closed interval [-$\frac{\pi}{2}$, $\frac{\pi}{2}$]. The function is increasing in the entire domain as the graph is a raising up through out.

Graph of Inverse Cos function
The graph of cos inverse function is the reflection over the line y = x of the part of the graph shown in green in the graph of  cos function.

Cos Inverse

From the graph it can be observed, the domain of cos inverse function as the closed interval [-1, 1] and the range is the closed interval [0, π].  The graph lies entirely above the x axis and is decreasing in its entire domain as it is seen coming down through out.

Graph of Inverse Tan function
The graph of tan inverse function is the reflection over the line y =x of the part of the graph shown in green in the graph of tan function.

Tan Inverse

In contrast to sin inverse and cos inverse functions the domain of tan inverse is the set of all real numbers or (-∞, ∞).
And its range is the open interval (-$\frac{\pi}{2}$, $\frac{\pi}{2}$). The graph is squeezed between the two horizontal asymptotes y = -$\frac{\pi}{2}$ and y = $\frac{\pi}{2}$. (The asymptotes are shown in green), and is increasing in the entire domain.

The graphs of the inverses of the reciprocal functions csc, sec and cot are shown below along with their domain and ranges.

 Csc Inverse   Graph of csc-1 x
Domain (-∞,-1] U [1,∞)
Range [-$\frac{\pi}{2}$, 0) U (0, $\frac{\pi}{2}$]
Graph of sec-1 x
Domain (-∞, -1] U [1, ∞)
Range [0, $\frac{\pi}{2}$) U ($\frac{\pi}{2}$, π]
  Sec Inverse
 Cot Inverse Graph of cot-1 x
Domain (-∞, ∞)
Range (0, π)

Let us now solve few problems using inverse trigonometric functions.
Derivatives of inverse trigonometric functions are derived using the definitions of inverse trigonometric functions. Here is the list of derivatives of inverse trigonometric functions with their domains.

Derivative of inverse sin function
$\frac{d}{dx}$$sin^{-1}x$ = $\frac{1}{\sqrt{1-x^{2}}}$ for -1 < x < 1
Even though the inverse sin function is defined in [-1, 1], the derivative is undefined at the boundary values.

Derivative of inverse cos function
$\frac{d}{dx}$$cos^{-1}x$ = $\frac{-1}{\sqrt{1-x^{2}}}$ for -1 < x < 1.
Just as for the derivative of sin inverse, the derivative of cos inverse is undefined at the boundary points of its domain
[-1, 1].

Derivative of inverse tan function
$\frac{d}{dx}$$tan^{-1}x$ = $\frac{1}{1+x^{2}}$
Derivative of tan inverse is defined for all the values in the domain of inverse tan that is (-∞, ∞)

Derivative of inverse csc function
$\frac{d}{dx}$$csc^{-1}x$ = $\frac{-1}{|x|\sqrt{x^{2}-1}}$ for |x| > 1
Derivative of csc function is not defined for |x| = 1 even though these values are included in the domain of inverse csc.

Derivative of inverse sec function
$\frac{d}{dx}$$secx$ = $\frac{1}{|x|\sqrt{x^{2}-1}}$ for |x| > 1.
Derivative of sec function is not defined for |x| = 1 even though these values are included in the domain of inverse sec.

Derivative of inverse cot function
$\frac{d}{dx}$$cotx$ = $\frac{-1}{1+x^{2}}$
Derivative of cot inverse is defined for all the values in the domain of inverse cot that is (-∞, ∞)

Solved Example

Question: Evaluate sin-1 $\frac{\sqrt{3}}{2}$,  arccos $\frac{\sqrt{2}}{2}$   and tan-1 (-1)
Solution:
 
Let sin-1 $\frac{\sqrt{3}}{2}$ = x
   ⇒ sin x = $\frac{\sqrt{3}}{2}$
   We need to find the value of x whose sine value = $\frac{\sqrt{3}}{2}$ and which lies in the interval -$\frac{\pi}{2}$ ≤ x ≤ $\frac{\pi}{2}$.
   We know sin $\frac{\pi }{3}$ = $\frac{\sqrt{3}}{2}$
   Thus x = $\frac{\pi }{3}$ radians or 60º.

   Hence sin-1 $\frac{\sqrt{3}}{2}$ = $\frac{\pi }{3}$

   Let arccos $\frac{\sqrt{2}}{2}$ = x
   ⇒ cos x = $\frac{\sqrt{2}}{2}$
   The value of x, whose cos function = $\frac{\sqrt{2}}{2}$ and which lies in the interval 0 ≤ x ≤ π is  $\frac{\pi }{4}$ radians
   Thus arccos $\frac{\sqrt{2}}{2}$ = $\frac{\pi }{4}$ radians or 45º.

  Let tan-1 (-1) = x
  ⇒ tan x = -1
  The value of x whose tan function = -1 and which lies in the interval -$\frac{\pi}{2}$ < x < $\frac{\pi}{2}$ is -$\frac{\pi }{4}$ radians.
  Hence tan-1 (-1) = $\frac{\pi }{4}$ radians or -45º.