# How do you evaluate cos (arcsin (3/5))?

May 26, 2018

$\cos \left(\arcsin \left(\frac{3}{5}\right)\right) = \frac{4}{5}$

#### Explanation:

The ratio $\frac{3}{5}$ should probably be a clue.

Note that $\arcsin \left(x\right) \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$ and if $0 < x < 1$ then $\arcsin \left(x\right) \in \left(0 , \frac{\pi}{2}\right)$.

So $\arcsin \left(\frac{3}{5}\right)$ is an angle in Q1, which we can consider in a right-angled triangle.

Consider a $3$, $4$, $5$ (right) triangle: We have:

$\sin A = \text{opposite"/"hypotenuse} = \frac{3}{5}$

$\cos A = \text{adjacent"/"hypotenuse} = \frac{4}{5}$

So:

$\cos \left(\arcsin \left(\frac{3}{5}\right)\right) = \cos \left(A\right) = \frac{4}{5}$

Alternatively, we could note more generally that:

${\cos}^{2} \theta + {\sin}^{2} \theta = 1$

Hence:

$\cos \theta = \pm \sqrt{1 - {\sin}^{2} \theta}$

If $\theta = \arcsin \left(x\right)$, then $\theta \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$ and hence $\cos \theta \ge 0$ and we find:

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

May 27, 2018

The inverse sine is multivalued, so

$\cos \arcsin \left(\frac{3}{5}\right) = \pm \sqrt{1 - {\left(\frac{3}{5}\right)}^{2}} = \pm \frac{4}{5}$