# How do you evaluate cos[arcsin ((5sqrt 29)/29)]?

May 15, 2015

If $\theta = \arcsin \left(\frac{5 \sqrt{29}}{29}\right)$, then that means

$\sin \theta = \frac{5 \sqrt{29}}{29} = \frac{5}{\sqrt{29}}$

From Pythagoras theorem we know ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

So

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

$= \sqrt{1 - {\left(\frac{5}{\sqrt{29}}\right)}^{2}}$

$= \sqrt{1 - \frac{{5}^{2}}{\sqrt{29}} ^ 2}$

$= \sqrt{1 - \frac{25}{29}}$

$= \sqrt{\frac{29 - 25}{29}}$

$= \sqrt{\frac{4}{29}}$

$= \frac{\sqrt{4}}{\sqrt{29}}$

$= \frac{2}{\sqrt{29}}$

$= \frac{2 \sqrt{29}}{29}$

Basically we are looking at a right angled triangle with sides $2$, $5$ and $\sqrt{29}$.