# How do you simplify the expression cos(arcsin(x/5)) ?

Jul 18, 2016

$\cos \left(\arcsin \left(\setminus \frac{x}{5}\right)\right) = \setminus \frac{\sqrt{25 - {x}^{2}}}{5}$.

#### Explanation:

Firstly, note that for the range of $\arcsin \left(\setminus \frac{x}{5}\right)$, the cosine function is purely positive.

Thus, I can say,

$\cos \left(\arcsin \left(\setminus \frac{x}{5}\right)\right) = \sqrt{1 - {\sin}^{2} \left(\arcsin \left(\setminus \frac{x}{5}\right)\right)}$, and also for the domain of x, $\sin \left(\arcsin \left(\setminus \frac{x}{5}\right)\right) = \setminus \frac{x}{5}$ since the arcsine function outputs values ranging from $- \setminus \frac{\pi}{2}$ to $\setminus \frac{\pi}{2}$.

This can be verified graphically, the red one being $\cos \left(\arcsin \left(\setminus \frac{x}{5}\right)\right)$, green $\cos \left(\setminus \frac{x}{5}\right)$, and blue $\arcsin \left(\setminus \frac{x}{5}\right)$:

Therefore,

$\cos \left(\arcsin \left(\setminus \frac{x}{5}\right)\right) = \sqrt{1 - \setminus \frac{{x}^{2}}{25}}$

$= \setminus \frac{\sqrt{25 - {x}^{2}}}{5}$.