<s>[INST] Evaluate $\tan\left(2\sin^{-1}\frac{\sqrt{5}}{5}\right)$ without using a calculator Evaluate without using a calculator:
$\displaystyle{\tan\left(2\sin^{-1}\left(\sqrt{5} \over 5\right)\right).}$
So I built my triangle hyp=$5$, adj=$2\sqrt{5}$, opp=$\sqrt{5}$.
$$
\tan\left(2\theta\right) = 2\sin\left(\theta\right)\cos\left(\theta\right)\,,
\qquad\qquad
\tan\left(2\theta\right) = 2\,{\sqrt{5} \over 5}\,{2\sqrt{5} \over 5} = 4.
$$
I used my calculator and it says the answer is $4/3$. I'm not sure what I'm doing wrong
 [/INST] $\displaystyle \frac{\sqrt{5}}{5} = \frac{1}{\sqrt{5}}$.  So, draw a triangle with 
sides $1$, $2$, and $\sqrt{5}$ which should be a right triangle since $1^2 + 2^2 = \left(\sqrt{5}\right)^2$.  The sine of one of the angles of this triangle, call
it $\theta$, is $\sin(\theta) = \frac{1}{\sqrt{5}}$, so its tangent $\tan(\theta)$ equals
$\frac{1}{2}$. You want to find $\displaystyle \tan(2\theta) = \frac{2\tan(\theta)}{1-\tan^2(\theta)}$.
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