Inverse trigonometric functions

Trigonometry: Trigonometric functions

Inverse trigonometric functions

. Therefore, if we want to solve the equation $\sin(x)=\tfrac{1}{2}$ , we will find infinitely many solutions. Now we will limit the domain of the functions so that we can define an inverse function. This inverse function can help us solve equations.

Inverse function sine, cosine and tangent

We define the inverse functions of sine, cosine and tangent as follows:

$\begin{array}{rcl} \blue x=\arcsin(\green y) &\Leftrightarrow& \green y=\sin(\blue x) \;\text{ and }\; -\frac{\pi}{2} \leq \blue x \leq \frac{\pi}{2} \\ \\ \blue x=\arccos(\green y) &\Leftrightarrow& \green y=\cos(\blue x) \;\text{ and }\; 0 \leq \blue x \leq \pi \\ \\ \blue x=\arctan(\green y) &\Leftrightarrow& \green y=\tan(\blue x) \;\text{ and }\; -\frac{\pi}{2} \lt \blue x \lt \frac{\pi}{2}\end{array}$

Example

$\arcsin\left(\green{\frac{\sqrt{2}}{2}}\right) = \blue{\frac{\pi}{4}}$

because

$\sin\left(\blue{\frac{\pi}{4}}\right)=\green{\frac{\sqrt{2}}{2}}$ and $-\frac{\pi}{2} \leq \blue{\frac{\pi}{4}} \leq \frac{\pi}{2}$

Calculator

On your calculator, the buttons $\arcsin$ , $\arccos$ and $\arctan$ are often called $\sin^{-1}$ , $\cos^{-1}$ and $\tan^{-1}$ . $\frac{1}{\sin(x)}=\csc(x)$ (cosecant), $\frac{1}{\cos(x)}=\sec(x)$ (secant) and $\frac{1}{\tan(x)}$ .

Properties of the arcsin, arccos and the arctan

The function $f(x)=\arcsin(x)$ has

domain: $\ivcc{-1}{1}$
range: $\ivcc{-\frac{\pi}{2}}{\frac{\pi}{2}}$

The graph is the reflection of $f(x)=\sin(x)$ on the domain $\ivcc{-\frac{\pi}{2}}{\frac{\pi}{2}}$ across the line $y=x$ .

Plaatje

The function $f(x)=\arccos(x)$ has

domain: $\ivcc{-1}{1}$
range: $\ivcc{0}{\pi}$

The graph is the reflection of $f(x)=\cos(x)$ on the domain $\ivcc{0}{\pi}$ across the line $y=x$ .

plaatje

The function $f(x)=\arctan(x)$ has

domain: all values of $x$
range: $\ivoo{-\frac{\pi}{2}}{\frac{\pi}{2}}$

The graph's horizontal asymptotes are $y=-\frac{\pi}{2}$ and $y=\frac{\pi}{2}$ and the graph is the reflection of $f(x)=\tan(x)$ on the domain $\ivoo{-\frac{\pi}{2}}{\frac{\pi}{2}}$ across the line $y=x$ .

plaatje

Link arcsin and arccos

When we have a good look at the graphs of $f(x)=\arcsin(x)$ and $f(x)=\arccos(x)$ , we'll see:

$\arccos(x) = \dfrac{\pi}{2}-\arcsin(x) \text{ for } -1\le x\le 1$

This is because $f(x)=\sin(x)$ and $f(x)=\cos(x)$ share the same graph, only shifted by $\frac{\pi}{2}$ .

Determine $\arcsin\left(\dfrac{\sqrt{3}}{2}\right)$ in radians.

$\arcsin\left(\dfrac{\sqrt{3}}{2}\right)=$ $\dfrac{\pi}{3}$

Because, $\sin\left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}$ and $-\dfrac{\pi}{2}\leq\dfrac{\pi}{3}\leq\dfrac{\pi}{2}$ .

New example