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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

Felix Marin
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Gᴇᴏᴍᴇᴛᴇʀ
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  • I'm not sure about the numerical evaluation involved here but certainly $\tan 2\theta \neq 2 \sin\theta \cos\theta$. Only $\sin 2\theta=2 \sin\theta \cos \theta$ – Sudarsan Nov 24 '13 at 22:50
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    Hint: You should use: $\sin^{-1} x=2 \tan^{-1}{\frac{x}{1+\sqrt{1-x^2}}}$. – Sudarsan Nov 24 '13 at 23:00

3 Answers3

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$\displaystyle \frac{\sqrt{5}}{5} = \frac{1}{\sqrt{5}}$. 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 $\displaystyle \tan(2\theta) = \frac{2\tan(\theta)}{1-\tan^2(\theta)}$.

Dilip Sarwate
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triangle

Using the identity $\tan{2A}=\dfrac{2\tan A}{1-\tan^2 A}$, $\tan \theta=\dfrac{\sqrt{5}}{\sqrt {20}} =\dfrac{1}{2}. \text{Hence} \tan\left(2\sin^{-1}\frac{\sqrt{5}}{5}\right)=\dfrac{2\times \dfrac{1}{2}}{1-\left(\dfrac{1}{2}\right)^2}=\dfrac{4}{3}$

K. Rmth
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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, $\displaystyle\tan\left(2\sin^{-1}\frac{\sqrt5}5\right)=\tan2\theta=\frac{\sin2\theta}{\cos2\theta}=\frac{2\sin\theta\cos\theta}{2\cos^2\theta-1}=\cdots$

lab bhattacharjee
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