Inverse trigonometric functions

​​In a nutshell

Often you will use the inverse trigonometric functions. But it is also important to understand their graphs and their domains and ranges for solving problems.

Names

The inverse trigonometric functions can be referred to in different ways. They can either be called "inverse [trigonometric function]" or "arc[trigonometric function]". So for example the inverse of sine is referred to as "sine inverse" or "arcsine". On the calculator, this is $$\sin^{-1}$$.

Graphs

The graphs of $$y=\arcsin(x)$$, $$y=\arccos(x)$$ and $$y=\arctan(x)$$ can be sketched by respectively reflecting the curves $$y=\sin(x)$$, $$y=\cos(x)$$ and $$y=\tan(x)$$ in the line $$y=x$$. 

Arcsine

This is the graph of $$y=\arcsin(x)$$, in radians. Its domain is $$[-1,1]$$ and its range is $$\left[-\frac{\pi}2,\frac{\pi}2\right]$$ (or if in degrees, $$[-90,90]$$). Note: This is not a true reflection of the whole of $$y=\sin(x)$$ in the line $$y=x$$​. If it was, then what you would get would not be a function. Instead you would have a one-to-many mapping.

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Arccosine

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This is the graph of $$y=\arccos(x)$$ in radians. Its domain is also $$[-1,1]$$ and its range is $$[0,\pi]$$ (or if in degrees, $$[0,180]$$). As with the arcsine graph, in order to be a function, it is not a full reflection of $$y=\cos(x)$$ in the line $$y=x$$.

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Arctangent

This is the graph of $$y=\arctan(x)$$. Its domain is all real numbers and its range is $$\left(-\frac{\pi}2,\frac{\pi}2\right)$$ (or in degrees, $$(-90,90)$$). Notice that the interval boundaries are not included in the range. This is shown by the horizontal asymptotes on the graph.

Example 1

Work out in radians, the value of $$\arccos(0.5)$$.

This is the same as solving the equation $$\cos(x)=0.5$$​ in the interval $$[0,\pi]$$. Using the $$\cos^{-1}$$ button on the calculator, you have 

$$\begin{aligned}\arccos(0.5)&=\cos^{-1}(0.5)\\&=\underline{\frac{\pi}3}\end{aligned}$$​​

Example 2

Without using a calculator, find the value of $$\arcsin(-0.5)$$.

This is the same as solving $$\sin(x)=-0.5$$ in the interval $$\left[-\frac{\pi}2,\frac{\pi}2\right]$$. Use the graph of $$y=\sin(x)$$ in the interval $$\left[-\frac{\pi}2,\frac{\pi}2\right]$$​:

You can see that by finding the $$a$$ such that $$\sin(a)=0.5$$, you will have the negative of $$\arcsin(-0.5)$$. In other words, $$\arcsin(-0.5)=-a$$.

Using your known trigonometric evaluations, you have that $$\sin\left(\frac{\pi}{6}\right)=0.5$$, so $$a=\frac\pi6$$. Thus $$\arcsin(-0.5)=\underline{-\frac\pi6}$$.