Since the trig functions are all periodic, none of them pass the horizontal line test. Thus, none of the graphs are one-to-one and do not have inverse functions on their entire domain. What we can do is restrict the domain of each of the trig functions to make each one one-to-one. As we said, the graphs are periodic, but we pick an appropriate domain that uses all values in the range.

Consider $y = sin x$ . If we restrict the domain to $[- \frac{π}{2}, \frac{π}{2}]$ , then $y = sin x$ is one-to-one.

Notice, every range value is defined if we use this section. The range is $- 1 \leq y \leq 1$ .

y = sin x

is one-to-one on

[- \frac{π}{2}, \frac{π}{2}]

[- \frac{π}{2}, \frac{π}{2}]

Since $y = sin x$ is one-to-one on $[- \frac{π}{2}, \frac{π}{2}]$ , it will have an inverse on this restricted domain. This inverse is called the inverse sine function and is denoted by $y = {sin}^{- 1} x$ . It is also referred to as $y = a r c sin x$ . As with any inverse function, we can obtain the graph of $y = {sin}^{- 1} x$ by reflecting the graph of $y = sin x$ about the line y=x. On the right is its graph.

Notice, that the domains and ranges switch.

Domain $({sin}^{- 1} x)$ = Range $(sin x)$ = [-1,1]

Range $({sin}^{- 1} x)$ = Restricted domain( $sin x$ ) = $[- \frac{π}{2}, \frac{π}{2}]$

Example: ${sin}^{- 1} (- \frac{1}{2}) = - \frac{π}{6}$ , rather than $\frac{7 π}{6}$ or $\frac{11 π}{6}$ , since $- \frac{π}{6}$ is the only angle in $[- \frac{π}{2}, \frac{π}{2}]$ whose sine is equal to $- \frac{1}{2}$ .

Graph of

y = {sin}^{- 1} x

Now consider $y = cos x$ . If we restrict the domain to $[0, π]$ , then $y = cos x$ is one-to-one.

Notice, every range value is defined if we use this section. The range is $- 1 \leq y \leq 1$ .

y = cos x

is one-to-one on

[0, π]

Since

y = cos x

is one-to-one on

[0, π]

, it will have an inverse on this restricted domain. This inverse is called the inverse cosine function and is denoted by

y = {cos}^{- 1} x

. It is also referred to as

y = a r c cos x

. As with any inverse function, we can obtain the graph of

y = {cos}^{.1} x

by reflecting the graph of

y = cos x

about the line y=x. To the right is its graph.

Notice, that the domains and ranges switch.

Domain $({cos}^{- 1} x)$ = Range $(cos x) = [0, π]$

Example: $\cos^{- 1} (- \frac{1}{2}) = \frac{2 π}{3}$ , rather than $\frac{4 π}{3}$ or $- \frac{4 π}{3}$ , since $\frac{2 π}{3}$ is the only angle in $[0, π]$ whose cosine is equal to $- \frac{1}{2}$ . $\frac{4 π}{3}$

Graph of

y = {cos}^{- 1} x

Consider $y = tan x$ . If we restrict the domain to $(- \frac{π}{2}, \frac{π}{2})$ , then $y = tan x$ is one-to-one.

Notice, every range value is defined if we use this section. The range is the set of all real numbers.

y = tan x

is one-to-one on

(- \frac{π}{2}, \frac{π}{2})

Since $y = tan x$ is one-to-one on $(- \frac{π}{2}, \frac{π}{2})$ , it will have an inverse on this restricted domain. This inverse is called the inverse tangent function and is denoted by $y = {tan}^{- 1} x$ . It is also referred to as $y = a r c tan x$ . As with any inverse function, we can obtain the graph of $y = {tan}^{- 1} x$ by reflecting the graph of $y = tan x$ about the line $y = x$ . Its graph is to the right.

Notice, that the domains and ranges switch.

Domain $({tan}^{- 1} x)$ = Range $(tan x)$ = all reals

Range $({tan}^{- 1} x)$ = Restricted domain $(tan x)$ = $(- \frac{π}{2}, \frac{π}{2})$

Example: ${tan}^{- 1} (- 1) = - \frac{π}{4}$ , rather than $\frac{3 π}{4}$ or $\frac{7 π}{4}$ , since $- \frac{π}{4}$ is the only angle in $(- \frac{π}{2}, \frac{π}{2})$ whose tangent is equal to -1 .

Graph of

y = {tan}^{- 1} x

Consider $y = cot x$ . If we restrict the domain to $(0, π)$ , then $y = cot x$ is one-to-one.

Notice, every range value is defined if we use this section. The range is the set of all real numbers.

y = cot x

is one-to-one on

(0, π)

Since

y = cot x

is one-to-one on

(0, π)

, it will have an inverse on this restricted domain. This inverse is called the inverse cotangent function and is denoted by

{cot}^{- 1} x

. It is also referred to as

y = a r c cot x

. As with any inverse function, we can obtain the graph of

y = {cot}^{- 1} x

by reflecting the graph of

y = cot x

about the line y=x. Its graph is to the right.

Notice, that the domains and ranges switch.

Domain $({cot}^{- 1} x)$ = Range $(cot x)$ = all reals

Range $({cot}^{- 1} x)$ = Restricted domain $(cot x)$ = $(0, π)$

Example: ${cot}^{- 1} (- \sqrt{3}) = \frac{5 π}{6}$ , rather than $\frac{7 π}{6}$ or $- \frac{7 π}{6}$ , since $\frac{5 π}{6}$ is the only angle in $(0, π)$ whose cotangent is equal to $- \sqrt{3}$ .

Graph of

y = {cot}^{- 1} x

Now, the other two inverse trigonometric functions, inverse secant $(y = {sec}^{- 1} x)$ and inverse cosecant $(y = {csc}^{- 1} x)$ , can be defined using a similar approach. Here are their domains and ranges.

Domain $({sec}^{- 1} x) = (- \infty, - 1] \cup [1, \infty)$ Range $({sec}^{- 1} x) = [0 . \frac{π}{2}) \cup (\frac{π}{2}, π]$

Domain $({csc}^{- 1} x) = (- \infty, - 1] \cup [1, \infty)$ Range $({csc}^{- 1} x) = [- \frac{π}{2}, 0) \cup (0, \frac{π}{2}]$

But rather than having to think of these restrictions, when computing ${sec}^{- 1} x$ or ${csc}^{- 1} x$ , values, we convert to ${cos}^{- 1} x$ or ${sin}^{- 1} x$ respectively.

For example, ${csc}^{- 1} (- 2) = {sin}^{- 1} (- \frac{1}{2}) = - \frac{π}{6}$