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
 A: $\displaystyle \frac{\sqrt{5}}{5} = \frac{1}{\sqrt{5}}$.  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 $\displaystyle \tan(2\theta) = \frac{2\tan(\theta)}{1-\tan^2(\theta)}$.
A: 
Using the identity $\tan{2A}=\dfrac{2\tan A}{1-\tan^2 A}$, $\tan \theta=\dfrac{\sqrt{5}}{\sqrt {20}} =\dfrac{1}{2}. \text{Hence} \tan\left(2\sin^{-1}\frac{\sqrt{5}}{5}\right)=\dfrac{2\times \dfrac{1}{2}}{1-\left(\dfrac{1}{2}\right)^2}=\dfrac{4}{3}$
A: Let $\displaystyle \sin^{-1}\frac{\sqrt5}5=\theta$
$\displaystyle\implies(i) \sin\theta=\frac{\sqrt5}5=\frac1{\sqrt5}$
and $\displaystyle (ii)-\frac\pi2\le \theta\le\frac\pi2$ based on the definition of principal value
In fact, $\displaystyle 0<\theta<\frac\pi2$(why?)
$\displaystyle\implies\cos\theta=+\sqrt{1-\sin^2\theta}=+\frac2{\sqrt5} $
So, $\displaystyle\tan\left(2\sin^{-1}\frac{\sqrt5}5\right)=\tan2\theta=\frac{\sin2\theta}{\cos2\theta}=\frac{2\sin\theta\cos\theta}{2\cos^2\theta-1}=\cdots$
