Lesson

Consider the equation:

$\sin x=\frac{1}{\sqrt{2}}$`s``i``n``x`=1√2

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.

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`=`s``i``n``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 $\sin x$`s``i``n``x` we will need to restrict the domain of the function to a portion that meets the following criteria:

- It only has one $y$
`y`-value for each $x$`x`-value (it is one-to-one). - It covers the full range of values of $\sin x$
`s``i``n``x`.

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 $\sin x$`s``i``n``x`, which we represent as $\sin^{-1}\left(x\right)$`s``i``n`−1(`x`) or $\arcsin(x)$`a``r``c``s``i``n`(`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 $\sin x$`s``i``n``x` 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.

We can find inverse functions for $\cos x$`c``o``s``x` and $\tan x$`t``a``n``x` 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 $\cos x$`c``o``s``x` is restricted to $\left[0,\pi\right]$[0,π] to produce the inverse function, which we represent as $\cos^{-1}\left(x\right)$`c``o``s`−1(`x`) or $\arccos(x)$`a``r``c``c``o``s`(`x`).

The domain of $\tan x$`t``a``n``x` is restricted to $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$(−π2,π2) to produce the inverse function, which we represent as $\tan^{-1}\left(x\right)$`t``a``n`−1(`x`) or $\arctan(x)$`a``r``c``t``a``n`(`x`).

Consider the equation $\cos x=0.4$`c``o``s``x`=0.4, where $0\le x\le\pi$0≤`x`≤π.

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

Select all that apply.

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

`a``r``c``c``o``s`(`c``o``s``x`)=`a``r``c``c``o``s`(0.4)A$\frac{\cos x}{\cos}=\frac{0.4}{\cos}$

`c``o``s``x``c``o``s`=0.4`c``o``s`B$\cos\left(\frac{x}{x}\right)=\frac{0.4}{x}$

`c``o``s`(`x``x`)=0.4`x`C$\cos^{-1}\left(\cos x\right)=\cos^{-1}\left(0.4\right)$

`c``o``s`−1(`c``o``s``x`)=`c``o``s`−1(0.4)D$\arccos\left(\cos x\right)=\arccos\left(0.4\right)$

`a``r``c``c``o``s`(`c``o``s``x`)=`a``r``c``c``o``s`(0.4)A$\frac{\cos x}{\cos}=\frac{0.4}{\cos}$

`c``o``s``x``c``o``s`=0.4`c``o``s`B$\cos\left(\frac{x}{x}\right)=\frac{0.4}{x}$

`c``o``s`(`x``x`)=0.4`x`C$\cos^{-1}\left(\cos x\right)=\cos^{-1}\left(0.4\right)$

`c``o``s`−1(`c``o``s``x`)=`c``o``s`−1(0.4)D

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

Loading Graph...

Why was $y=\sin x$

`y`=`s``i``n``x`restricted to this domain?Select all that apply.

This graph reflected about the $y$

`y`-axis is a function.AIt is a function on this domain.

BThis graph reflected about the line $y=x$

`y`=`x`is a function.CThis interval is one period of $y=\sin x$

`y`=`s``i``n``x`.DThis interval covers the range of $y=\sin x$

`y`=`s``i``n``x`.EThis graph reflected about the $y$

`y`-axis is a function.AIt is a function on this domain.

BThis graph reflected about the line $y=x$

`y`=`x`is a function.CThis interval is one period of $y=\sin x$

`y`=`s``i``n``x`.DThis interval covers the range of $y=\sin x$

`y`=`s``i``n``x`.ESelect the graph of $y=\sin^{-1}\left(x\right)$

`y`=`s``i``n`−1(`x`).Loading Graph...ALoading Graph...BLoading Graph...CLoading Graph...DLoading Graph...ALoading Graph...BLoading Graph...CLoading Graph...DConsider the graph from part (b). What interval do the $y$

`y`-values of $y=\sin^{-1}\left(x\right)$`y`=`s``i``n`−1(`x`) always lie on?

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

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

`s``i``n`−1(12).

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