# What is Cot [ arcsin( sqrt5 / 6) ]?

Jul 28, 2015

$\frac{\sqrt{155}}{5}$

#### Explanation:

Start by letting $\arcsin \left(\frac{\sqrt{5}}{6}\right)$ to be a certain angle $\alpha$

It follows that $\alpha = \arcsin \left(\frac{\sqrt{5}}{6}\right)$
and so
$\sin \left(\alpha\right) = \frac{\sqrt{5}}{6}$

This means that we are now looking for $\cot \left(\alpha\right)$

Recall that : $\cot \left(\alpha\right) = \frac{1}{\tan} \left(\alpha\right) = \frac{1}{\sin \frac{\alpha}{\cos} \left(\alpha\right)} = \cos \frac{\alpha}{\sin} \left(\alpha\right)$

Now, use the identity ${\cos}^{2} \left(\alpha\right) + {\sin}^{2} \left(\alpha\right) = 1$ to obtain $\cos \left(\alpha\right) = \sqrt{\left(1 - {\sin}^{2} \left(\alpha\right)\right)}$

$\implies \cot \left(\alpha\right) = \cos \frac{\alpha}{\sin} \left(\alpha\right) = \frac{\sqrt{\left(1 - {\sin}^{2} \left(\alpha\right)\right)}}{\sin} \left(\alpha\right) = \sqrt{\frac{1 - {\sin}^{2} \left(\alpha\right)}{\sin} ^ 2 \left(\alpha\right)} = \sqrt{\frac{1}{\sin} ^ 2 \left(\alpha\right) - 1}$

Next, substitute $\sin \left(\alpha\right) = \frac{\sqrt{5}}{6}$ inside $\cot \left(\alpha\right)$

$\implies \cot \left(\alpha\right) = \sqrt{\frac{1}{\frac{\sqrt{5}}{6}} ^ 2 - 1} = \sqrt{\frac{36}{5} - 1} = \sqrt{\frac{31}{5}} = \textcolor{b l u e}{\frac{\sqrt{155}}{5}}$