# What is tan(pi + arcsin (2/3))?

Jul 21, 2015

$\frac{2 \sqrt{5}}{5}$

#### Explanation:

First thing to note is that every $\textcolor{red}{\tan}$ function has a period of $\pi$

This means that $\tan \left(\pi + \textcolor{g r e e n}{\text{angle")-=tan(color(green)"angle}}\right)$

$\implies \tan \left(\pi + \arcsin \left(\frac{2}{3}\right)\right) = \tan \left(\arcsin \left(\frac{2}{3}\right)\right)$

Now, let $\theta = \arcsin \left(\frac{2}{3}\right)$

So, now we are looking for color(red)tan(theta)!

We also have it that : $\sin \left(\theta\right) = \frac{2}{3}$

Next, we use the identity : $\tan \left(\theta\right) = \sin \frac{\theta}{\cos} \left(\theta\right) = \sin \frac{\theta}{\sqrt{1 - {\sin}^{2} \left(\theta\right)}}$

And then we substitute the value for $\sin \left(\theta\right)$

$\implies \tan \left(\theta\right) = \frac{\frac{2}{3}}{\sqrt{1 - {\left(\frac{2}{3}\right)}^{2}}} = \frac{2}{3} \times \frac{1}{\sqrt{1 - \frac{4}{9}}} = \frac{2}{3} \times \frac{1}{\sqrt{\frac{9 - 4}{9}}} = \frac{2}{3} \times \sqrt{\frac{9}{9 - 4}} = \frac{2}{3} \times \frac{3}{\sqrt{5}} = \frac{2}{\sqrt{5}} = \frac{2 \sqrt{5}}{5}$