New Zealand
Level 8 - NCEA Level 3

# Intro to Inverse Trig Functions

Lesson

### Finding an inverse for $\sin x$sinx

Consider the equation:

$\sin x=\frac{1}{\sqrt{2}}$sinx=12

We would like to find the solutions to this equation, shown as the points of intersection on the graph below. To begin with we are going to determine an inverse function for sine, which will help us find one of these solutions. We will then look at how to recover the other solutions in a later lesson.

Solutions to $\sin x=\frac{1}{\sqrt{2}}$sinx=12

The graph of an inverse function can be found by reflecting the original function about the line $y=x$y=x. The graph below shows that reflection applied to the graph of $y=\sin x$y=sinx.

The graph of $y=\sin x$y=sinx reflected about the line $y=x$y=x.

Notice, however, that this graph has multiple $y$y-values for a given $x$x-value, so it is not a function.

In order to find an inverse function for $\sin x$sinx we will need to restrict the domain of the function to a portion that meets the following criteria:

1. It only has one $y$y-value for each $x$x-value (it is one-to-one).
2. It covers the full range of values of $\sin x$sinx.

There are infinitely many ways we could restrict the domain that would satisfy these criteria, but the domain that is typically chosen by mathematicians to work with is $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$[π2,π2]. When just this part of the curve is reflected about $y=x$y=x we obtain the inverse function for $\sin x$sinx, which we represent as $\sin^{-1}\left(x\right)$sin1(x) or $\arcsin(x)$arcsin(x).

The graph of $y=\sin x$y=sinx restricted to the domain $-\frac{\pi}{2}\le x\le\frac{\pi}{2}$π2xπ2.

The graph of the inverse sine function $y=\sin^{-1}\left(x\right)$y=sin1(x).

We can now rewrite our original equation using the inverse sine function:

 $\sin x$sinx $=$= $\frac{1}{\sqrt{2}}$1√2​ $x$x $=$= $\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)$sin−1(1√2​) $x$x $=$= $\frac{\pi}{4}$π4​

Notice that because we needed to restrict the domain of $\sin x$sinx to obtain an inverse function, we have only found the one solution between $-\frac{\pi}{2}$π2 and $\frac{\pi}{2}$π2. We will look at how to use this single solution to find the other solutions to the equation in a later lesson.

### Finding an inverse for $\cos x$cosx and $\tan x$tanx

We can find inverse functions for $\cos x$cosx and $\tan x$tanx in a similar way; we first restrict their domains using the above criteria, then reflect the remaining graph about the line $y=x$y=x.

The domain of $\cos x$cosx is restricted to $\left[0,\pi\right]$[0,π] to produce the inverse function, which we represent as $\cos^{-1}\left(x\right)$cos1(x) or $\arccos(x)$arccos(x).

The graph of $y=\cos x$y=cosx restricted to the domain $0\le x\le\pi$0xπ.

The graph of the inverse cosine function $y=\cos^{-1}\left(x\right)$y=cos1(x).

The domain of $\tan x$tanx is restricted to $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$(π2,π2) to produce the inverse function, which we represent as $\tan^{-1}\left(x\right)$tan1(x) or $\arctan(x)$arctan(x).

The graph of $y=\tan x$y=tanx restricted to the domain $-\frac{\pi}{2}\le x\le\frac{\pi}{2}$π2xπ2.

The graph of the inverse tangent function $y=\tan^{-1}\left(x\right)$y=tan1(x).

#### Questions

##### QUESTION 1

Consider the equation $\cos x=0.4$cosx=0.4, where $0\le x\le\pi$0xπ.

Which of the following could be a correct step to solve for $x$x?

Select all that apply.

1. $\arccos\left(\cos x\right)=\arccos\left(0.4\right)$arccos(cosx)=arccos(0.4)

A

$\frac{\cos x}{\cos}=\frac{0.4}{\cos}$cosxcos=0.4cos

B

$\cos\left(\frac{x}{x}\right)=\frac{0.4}{x}$cos(xx)=0.4x

C

$\cos^{-1}\left(\cos x\right)=\cos^{-1}\left(0.4\right)$cos1(cosx)=cos1(0.4)

D

$\arccos\left(\cos x\right)=\arccos\left(0.4\right)$arccos(cosx)=arccos(0.4)

A

$\frac{\cos x}{\cos}=\frac{0.4}{\cos}$cosxcos=0.4cos

B

$\cos\left(\frac{x}{x}\right)=\frac{0.4}{x}$cos(xx)=0.4x

C

$\cos^{-1}\left(\cos x\right)=\cos^{-1}\left(0.4\right)$cos1(cosx)=cos1(0.4)

D

##### QUESTION 2

To plot the graph of $y=\sin^{-1}\left(x\right)$y=sin1(x), first the graph of $y=\sin x$y=sinx has been plotted on the interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$[π2,π2].

1. Why was $y=\sin x$y=sinx restricted to this domain?

Select all that apply.

This graph reflected about the $y$y-axis is a function.

A

It is a function on this domain.

B

This graph reflected about the line $y=x$y=x is a function.

C

This interval is one period of $y=\sin x$y=sinx.

D

This interval covers the range of $y=\sin x$y=sinx.

E

This graph reflected about the $y$y-axis is a function.

A

It is a function on this domain.

B

This graph reflected about the line $y=x$y=x is a function.

C

This interval is one period of $y=\sin x$y=sinx.

D

This interval covers the range of $y=\sin x$y=sinx.

E
2. Select the graph of $y=\sin^{-1}\left(x\right)$y=sin1(x).

A

B

C

D

A

B

C

D
3. Consider the graph from part (b). What interval do the $y$y-values of $y=\sin^{-1}\left(x\right)$y=sin1(x) always lie on?

##### QUESTION 3

Consider that $\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}$sin(π6)=12.

1. Find the value of $\sin^{-1}\left(\frac{1}{2}\right)$sin1(12).

### Outcomes

#### M8-2

Display and interpret the graphs of functions with the graphs of their inverse and/or reciprocal functions