inverse trigonometric functions

The inverse trigonometric functions are $\arcsin(x)$ , $\arccos(x)$ and $\arctan(x)$ . These functions perform the reverse operations to the original trigonometric functions $\sin(x)$ , $\cos(x)$ and $\tan(x)$ respectively. Recall that a function is invertible if it is one-to-one. Click here to revise inverse functions. Hence, before we can sketch the graphs of the inverse trigonometric functions, we must choose a domain for them for which they are one-to-one. Note that the original trigonometric functions work on angles and so each of the inverse trigonometric functions will return an angle. We use radians for all angles in the following – see more on radians. Also note that we use $\arcsin(x)$ instead of $\sin^{-1}(x)$ , for example, as this can be confused with $\text{cosec}(x)$ , the reciprocal trigonometric function.

Inverse Trigonometric Function: arcsin(x)

Since $\sin(x)$ is periodic, there are infinitely many regions for which it is one-to-one. We choose the default domain to be $\frac{-\pi}{2}\leq x\leq \frac{\pi}{2}$ . The range of $\sin(x)$ is $-1\leq\sin(x)\leq 1$ . It follows that the domain of $\text{arcsin}(x)$ is $-1\leq x\leq 1$ and the range is $\frac{-\pi}{2}\leq \text{arcsin}(x)\leq \frac{\pi}{2}$ . The graphs of $\sin(x)$ and $\text{arcsin}(x)$ are reflections of one another in the line $y=x$ .

Inverse Trigonometric Function: arccos(x)

Since $\cos(x)$ is periodic, there are infinitely many regions for which it is one-to-one. We choose the default domain to be $0\leq x\leq \pi$ . The range of $\cos(x)$ is $-1\leq\cos(x)\leq 1$ . It follows that the domain of $\text{arccos}(x)$ is $-1\leq x\leq 1$ and the range is $0\leq \text{arcsin}(x)\leq \pi$ . The graphs of $\cos(x)$ and $\text{arccos}(x)$ are reflections of one another in the line $y=x$ .

Inverse Trigonometric Function: arctan(x)

Since $\tan(x)$ is periodic, there are infinitely many regions for which it is one-to-one. We choose the default domain to be $\frac{-\pi}{2}< x<\frac{\pi}{2}$ . The range of $\tan(x)$ is all of ${\mathbb R}$ . It follows that the domain of $\text{arctan}(x)$ is $x\in{\mathbb R}$ and the range is $\frac{-\pi}{2}< \text{arctan}(x)< \frac{\pi}{2}$ . The graphs of $\tan(x)$ and $\text{arctan}(x)$ are reflections of one another in the line $y=x$ .

Examples

Find the values (in radians) of the following without using a calculator:

$\text{arcsin}(1)$
$\text{arccos}(\cos(\pi/7))$
$\sin(\text{arctan}(1))$

Solution:

We can see from the graph or use the fact that $\sin\left(\frac{\pi}{2}\right)=1$ on this interval to find $\text{arcsin}(1)=\frac{\pi}{2}$ .
$\text{arccos}$ and $\cos$ are inverses of one another and so the result is $\pi/7$ .
$\text{arctan}(1)=\frac{\pi}{4}$ and $\sin(\pi/4)=\frac{\sqrt{2}}{2}$ . Hence $\sin(\text{arctan}(1))=\frac{\sqrt{2}}{2}.$

Sketch the graph of $y=2\text{arccos}(-x)$ , stating its domain and range.

Solution:

Recall from the transformations page that replacing $x$ with $-x$ reflects the original graph across the $y$ -axis. Multiplying by 2 stretches the graph by a factor of 2 in the $y$ -direction. It follows that the graph of $y=2\text{arccos}(-x)$ is given by:

The domain of $y=2\text{arccos}(-x)$ is $-1\leq x\leq 1$ and the range is $0\leq y\leq 2\pi$ .

Given that $\theta=\text{arccos}(x)$ where $\theta$ is an acute angle, find an expression for $\sin(\theta)$ in terms of $x$ .

Solution:

If $\theta=\text{arccos}(x)$ then $\cos(\theta)=x$ . Using $\sin^2(\theta)+\cos^2(\theta)=1$ (see more trigonometric identities), $\sqrt{1-\sin^2(\theta)}=x$ and so $1-\sin^2(\theta)=x^2$ . It follows that $\sin(\theta)=\pm\sqrt{1-x^2}$ . Note that $x=\cos(\theta)$ is between $-1$ and $1$ (for acute angles it is actually between 0 and 1) and so $0\leq x^2\leq 1$ . It follows that in the expression for $\sin(\theta)$ we are square rooting a nonnegative number, as we would hope. However, for acute $\theta$ , $\sin(\theta)$ must be positive and so $\sin(\theta)=\sqrt{1-x^2}$ .