In order for inverses of functions to be functions, the original function must pass the **horizontal line test**. Though none of the trigonometric functions pass the horizontal line test, you can restrict their domains so that they can pass. Then the inverses are produced just like with normal functions. Once you have the basic inverse functions, the normal transformation rules apply.

Why is \begin{align*}\sin^{-1} (\sin 370^\circ) \neq 370^\circ\end{align*}? Don’t the arcsin and sin just cancel out?

### Graphs of Inverse Trigonometric Functions

Since none of the six trigonometric functions pass the horizontal line test, you must **restrict their domains** before finding inverses of these functions. This is just like the way \begin{align*}y=\sqrt{x}\end{align*} is the inverse of \begin{align*}y=x^2\end{align*} when you restrict the domain to \begin{align*}x \ge 0\end{align*}.

Consider the sine graph:

As a general rule, the restrictions to the domain are either the interval \begin{align*}\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\end{align*} or \begin{align*}[0, \pi]\end{align*} to keep things simple. In this case sine is restricted to \begin{align*}\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\end{align*}, as shown above. To find the inverse, reflect the bold portion across the line \begin{align*}y=x\end{align*}. The blue curve below shows \begin{align*}f(x)=\sin^{-1}x\end{align*}.

The result of this inversion is that arcsine will only ever produce angles between \begin{align*}-\frac{\pi}{2}\end{align*} and \begin{align*}\frac{\pi}{2}\end{align*}. You must use logic and common sense to interpret these numbers in context.

The blue curve below shows \begin{align*}f(x)=\cos^{-1}x\end{align*}?

The portion of cosine that fits the horizontal line test is the interval \begin{align*}[0, \pi]\end{align*}. To find the inverse, that portion is reflected across the line \begin{align*}y=x\end{align*}.

### Examples

#### Example 1

Earlier, you were asked why sine and arcsine don't always just cancel out. Since arcsine only produces angles between \begin{align*}-\frac{\pi}{2}\end{align*} and \begin{align*}\frac{\pi}{2}\end{align*} or \begin{align*}-90^\circ\end{align*} to \begin{align*}+90^\circ\end{align*} the result of \begin{align*}\sin^{-1} (\sin 370^\circ)\end{align*} is \begin{align*}10^\circ\end{align*} which is coterminal to \begin{align*}370^\circ\end{align*}.

#### Example 2

Graph the function \begin{align*}f(x)=2 \cos^{-1} (x-1)\end{align*}.

Since the graph of \begin{align*}f(x)=\cos^{-1}x\end{align*} was done in Example A, now you just need to shift it right one unit and stretch it vertically by a factor of 2. It intersected the \begin{align*}x\end{align*} axis at 1 before and now it will intersect at 2. It reached a height of \begin{align*}\pi\end{align*} before and now it will reach a height of \begin{align*}2 \pi\end{align*}.

#### Example 3

Evaluate the following expression with and without a calculator using right triangles and your knowledge of inverse trigonometric functions.

\begin{align*}\cot \left(\csc^{-1} \left(-\frac{13}{5}\right)\right)\end{align*}

When using a calculator it can be extremely confusing trying to tell the difference between \begin{align*}\sin^{-1}x\end{align*} and \begin{align*}(\sin x)^{-1}\end{align*}. In order to be able to effectively calculate this out it is best to write the expression explicitly only in terms of functions that your calculator does have.

The hardest part of this question is seeing the csc as a function (which produces an angle) on a ratio of a hypotenuse of 13 and an opposite side of -5. The sine of the inverse ratio must produce the same angle, so you can substitute it.

- \begin{align*}\csc^{-1} \left(-\frac{13}{5}\right)=\sin^{-1} \left(-\frac{5}{13}\right)\end{align*}
- \begin{align*}\cot (\theta)=\frac{1}{\tan \theta}\end{align*}

\begin{align*}\cot \left(\csc^{-1} \left (-\frac{13}{5}\right) \right)=\frac{1}{\tan \left(\sin^{-1} \left(-\frac{5}{13}\right)\right)}=-\frac{12}{5}\end{align*}

Not using a calculator is usually significantly easier. Start with your knowledge that \begin{align*}\csc^{-1} \left(-\frac{13}{5}\right)\end{align*} describes an angle in the fourth or the second quadrant because those are the two quadrants where cosecant is negative. Since \begin{align*}\csc^{-1} \theta\end{align*} has range \begin{align*}-\frac{\pi}{2}, \frac{\pi}{2}\end{align*}, it only produces angles in quadrant I or quadrant IV (see Guided Practice 2). This triangle must then be in the fourth quadrant. All you need to do is draw the triangle and identify the cotangent ratio.

Cotangent is adjacent over opposite.

\begin{align*}\cot \left(\csc^{-1} \left(-\frac{13}{5}\right)\right)=-\frac{12}{5}\end{align*}

#### Example 4

What is the graph of \begin{align*}y=\tan^{-1} x\end{align*}?

Graph the portion of tangent that fits the horizontal line test and reflect across the line \begin{align*}y=x\end{align*}. Note that the graph of arctan is in blue.

#### Example 5

What is the graph of \begin{align*}y=\csc^{-1}x\end{align*}

Graph the portion of cosecant that fits the horizontal line test and reflect across the line \begin{align*}y=x\end{align*}.

Note that \begin{align*}f(x)=\csc^{-1}x\end{align*} is in blue. Also note that the word “arc-co-secant” is too cumbersome to use because of the train of prefixes.

### Review

1. Graph \begin{align*}f(x)=\cot^{-1}x\end{align*}.

2. Graph \begin{align*}g(x)=\sec^{-1}x\end{align*}.

Name each of the following graphs.

3.

4.

5.

6.

7.

Graph each of the following functions using your knowledge of function transformations.

8. \begin{align*}h(x)=3 \sin^{-1} (x+1)\end{align*}

9. \begin{align*}k(x)=2 \sin^{-1}(x)+\frac{\pi}{2}\end{align*}

10. \begin{align*}m(x)=-\cos^{-1}(x-2)\end{align*}

11. \begin{align*}j(x)=\cot^{-1}(x)+\pi\end{align*}

12. \begin{align*}p(x)=-2 \tan^{-1}(x-1)\end{align*}

13. \begin{align*}q(x)=\csc^{-1}(x-2)\end{align*}

14. \begin{align*}r(x)=-\sec^{-1}(x)+4\end{align*}

15. \begin{align*}t(x)=\csc^{-1}(x+1) -\frac{3 \pi}{2}\end{align*}

16. \begin{align*}v(x)=2 \sec^{-1}(x+2)+\frac{\pi}{2}\end{align*}

17. \begin{align*}w(x)=-\cot^{-1}(x)-\frac{\pi}{2}\end{align*}

Evaluate each expression.

18. \begin{align*}\sec \left(\tan^{-1} \left[\frac{3}{4}\right]\right)\end{align*}

19. \begin{align*}\cot \left(\csc^{-1} \left[\frac{13}{12}\right]\right)\end{align*}

20. \begin{align*}\csc \left(\tan^{-1} \left[\frac{4}{3}\right]\right)\end{align*}

### Review (Answers)

To see the Review answers, open this PDF file and look for section 5.8.