# How do you find sin(tan^-1(a/3))?

Sep 27, 2016

sin(tan^(-1)(a/3)) = a/(sqrt(a^2+9)

#### Explanation:

Note that the range of ${\tan}^{- 1} \left(y\right)$ is $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$.

If $\theta \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$ then $\cos \theta > 0$.

If $\frac{a}{3} > 0$ then consider a right angled triangle with sides $\frac{a}{3}$, $1$ and $\sqrt{{a}^{2} / 9 + 1}$

We find:

sin(tan^(-1)(a/3)) = (a/3)/sqrt(a^2/9+1) = a/(sqrt(a^2+9)

This identity continues to hold for $\frac{a}{3} < 0$ since $\sin \theta$ and $\tan \theta$ are odd functions.

It also holds for $a = 9$ since $\sin \left({\tan}^{- 1} \left(0\right)\right) = \sin \left(0\right) = 0$