Inverse Trig Functions

The idea

If $\sin \theta = 0.5$, what is $\theta$? Well, $\theta = \pi/6$ works — but so do $5\pi/6$, $13\pi/6$, and infinitely many more.

To make sine invertible, we restrict its domain. The inverse sine $\sin^{-1}(x)$ (also written $\arcsin x$) gives the one angle in $[-\pi/2,\ \pi/2]$ whose sine is $x$.

Notation and ranges

Function	Output range
$\sin^{-1} x$	$\left[-\dfrac{\pi}{2},\ \dfrac{\pi}{2}\right]$
$\cos^{-1} x$	$[0,\ \pi]$
$\tan^{-1} x$	$\left(-\dfrac{\pi}{2},\ \dfrac{\pi}{2}\right)$

Note. $\sin^{-1} x \neq \dfrac{1}{\sin x}$. The $-1$ here means "inverse function," not "reciprocal."

Evaluating

Example. $\sin^{-1}\!\left(\dfrac{\sqrt{2}}{2}\right)$. Ask: what angle in $[-\pi/2,\ \pi/2]$ has sine $\dfrac{\sqrt{2}}{2}$? Answer: $\pi/4$.

Example. $\cos^{-1}(-1)$. Want an angle in $[0,\ \pi]$ with cosine $-1$. Answer: $\pi$.

Example. $\tan^{-1}(1)$. Want an angle in $(-\pi/2,\ \pi/2)$ with tangent $1$. Answer: $\pi/4$.

Compositions

For values in the correct range, the inverse and original cancel:

$\sin(\sin^{-1} x) = x \quad \text{for } x \in [-1, 1]$ $\sin^{-1}(\sin \theta) = \theta \quad \text{for } \theta \in [-\pi/2,\ \pi/2]$

Mixed compositions like $\sin(\cos^{-1} x)$ usually simplify with a right-triangle picture.

Example. Simplify $\sin(\cos^{-1}(3/5))$.

Let $\theta = \cos^{-1}(3/5)$. So $\cos \theta = 3/5$ with $\theta \in [0,\ \pi]$ — meaning $\sin \theta \geq 0$.

Picture a right triangle with adjacent $3$ and hypotenuse $5$. Opposite is $\sqrt{25 - 9} = 4$. So $\sin \theta = 4/5$.

Solving with inverses

When a trig equation's reference solution isn't a nice angle, use the inverse function:

$\sin \theta = 0.4 \implies \theta = \sin^{-1}(0.4) \approx 0.4115$

Then use unit-circle symmetry to find the other solutions in the desired interval.

Key takeaways

• Inverse trig functions return one specific angle per input — that's why their output ranges are restricted.
• $\sin^{-1} x \neq 1/\sin x$ — different thing entirely.
• For mixed compositions like $\sin(\cos^{-1} x)$, sketch a right triangle.