# How do you evaluate cot(arcsin(-5/8)?

Apr 22, 2015

Let $t = \arcsin \left(- \frac{5}{8}\right)$

Then $t$ is in $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$ and $\sin t = - \frac{5}{8}$

Find $\cos t$ (I'm sure you have done this kind of problem by now.)

Because $- \frac{\pi}{2} < t < \frac{\pi}{2}$, we know that $\cos t > 0$

$\cos t = \frac{\sqrt{39}}{8}$

And $\cot t = \cos \frac{t}{\sin} t = - \frac{\sqrt{39}}{5}$

Method 2

Use ${\cot}^{2} t = {\csc}^{2} t - 1$ together with $- \frac{\pi}{2} < t < \frac{\pi}{2}$, so we know that $\cos t > 0$ and so $\cot t < 0$

$\csc t = - \frac{8}{5}$, so ${\cot}^{2} t = {\left(- \frac{8}{5}\right)}^{2} - 1 = \frac{39}{25}$ and so on