# How do you evaluate tan(arcsin(1/3)) ?

Apr 13, 2018

$\frac{1}{2 \sqrt{2}}$

#### Explanation:

Recalling that $\tan x = \sin \frac{x}{\cos} x ,$

$\tan \left(\arcsin \left(\frac{1}{3}\right)\right) = \sin \frac{\arcsin \left(\frac{1}{3}\right)}{\cos} \left(\arcsin \left(\frac{1}{3}\right)\right)$

$\sin \left(\arcsin \left(\frac{1}{3}\right)\right) = \frac{1}{3}$ from the fact that $\sin \left(\arcsin x\right) = x$, but for the cosine, some more work is needed.

Recall that

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

Then,

${\cos}^{2} x = 1 - {\sin}^{2} x$

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

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

Knowing $\sin \left(\arcsin \left(\frac{1}{3}\right)\right) = \frac{1}{3} , {\sin}^{2} \left(\arcsin \left(\frac{1}{3}\right)\right) = {\left(\frac{1}{3}\right)}^{2} = \frac{1}{9}$

Then,

$\cos \left(\arcsin \left(\frac{1}{3}\right)\right) = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2 \sqrt{2}}{3}$

Thus,

$\tan \left(\arcsin \left(\frac{1}{3}\right)\right) = \frac{\frac{1}{3}}{\frac{2 \sqrt{2}}{3}} = \frac{1}{\cancel{3}} \cdot \frac{\cancel{3}}{2 \sqrt{2}} = \frac{1}{2 \sqrt{2}}$