# How to prove $\tan\Big[\frac{1}{2}\sin^{-1}\frac{3}{4}\Big]=\frac{4-\sqrt{7}}{3}$

Prove

$$\tan\Big[\frac{1}{2}\sin^{-1}\frac{3}{4}\Big]=\frac{4-\sqrt{7}}{3}$$ and justify why $\frac{4+\sqrt{7}}{3}$ is ignored.

My Attempt: $$\tan x=\frac{2\tan\frac{x}{2}}{1-\tan^2\frac{x}{2}}\implies\tan x-\tan x\tan^2\frac{x}{2}=2\tan\frac{x}{2}\\ \implies \tan^2\frac{x}{2}(\tan x)+2\tan\frac{x}{2}-\tan x=0\\ \implies \tan\frac{x}{2}=\frac{-2\pm2\sec x}{2\tan x}=\frac{-1\pm\sec x}{\tan x}$$ Using this, $$\tan\bigg[\frac{\sin^{-1}\frac{3}{4}}{2}\bigg]=\frac{-1\pm\frac{4}{\sqrt{7}}}{\frac{3}{\sqrt{7}}}=\frac{-\sqrt{7}\pm4}{3}$$ As $0\leq\frac{\sin^{-1}3/4}{2}\leq\frac{\pi}{4}\implies0\leq\tan(\frac{\sin^{-1}3/4}{2})\leq1$ $$\implies \tan\bigg[\frac{\sin^{-1}\frac{3}{4}}{2}\bigg]=\frac{-\sqrt{7}+4}{3}$$ Is my attempt correct and where is the value $\frac{4+\sqrt{7}}{3}$ to be excluded ?

• $$4+\sqrt7>4>3$$ Also, $$\tan y=\dfrac{1-\cos2y}{\sin2y}$$ Feb 2, 2018 at 11:19

Here's a nice (if I say so myself) geometric solution. Consider the right triangle $\triangle ABC$ with right $\angle ABC$, $|BC| = 3$ and $|AC| = 4$. $AD$ is the angle bisector of $\angle BAC$. Let $|BD| = x$ and hence $|CD| = 3-x$.

Now $\displaystyle |AB| = \sqrt{4^2 - 3^2} = \sqrt 7$ (Pythagoras). Note that $\displaystyle \theta = \frac 12 \arcsin \frac 34$ and $\tan \theta$ is the required ratio.

By the angle bisector theorem, $\displaystyle \frac x{3-x} = \frac{\sqrt 7}4$, giving $\displaystyle x = \frac{3\sqrt 7}{4 + \sqrt 7}$.

Hence the required ratio is $\displaystyle \tan \theta = \frac x{\sqrt 7} = \frac 3{4 + \sqrt 7} = \frac{4-\sqrt 7}3$ after rationalising the denominator.

$$\tan\left[\frac{1}{2}\sin^{-1} \left(\frac{3}{4}\right)\right]=\frac{4-\sqrt{7}}{3}$$ $\sin^{-1}(x) = \arcsin(x) = 2\arctan \left(\frac x{1+\sqrt{1-x^2}}\right)$
See Inverse trigonometric functions \begin{align} \tan \left[\frac{1}{2}\sin^{-1} \left(\frac{3}{4}\right)\right] &=\tan \left[\frac{1}{2}2\arctan \left(\frac {\frac{3}{4}}{1+\sqrt{1-\left(\frac{3}{4}\right)^2}}\right)\right]\\ &= \frac {3}{4\left(1+\sqrt{1-\left(\frac{3}{4}\right)^2}\right)} \end{align} $\sqrt{1-\left(\frac{3}{4}\right)^2}=\frac{\sqrt{7}}{4}$ $$=\frac{3}{4\left(\frac{\sqrt{7}}{4}+1\right)} =\frac{3}{4\cdot \frac{4+\sqrt{7}}{4}}=\frac{3}{4+\sqrt{7}} = \frac{4-\sqrt{7}}{3}$$ $\tag*{$\Box$}$

• $2\tan^{-1}x=\begin{cases}\sin^{-1}\frac{2x}{1+x^2}, \quad |x|\leq1\\ \pi-\sin^{-1}\frac{2x}{1+x^2},\quad|x|>1,x>1\\-\pi-\sin^{-1}\frac{2x}{1+x^2},\quad|x|>1,x<1 \end{cases}$ So i guess i need to consider both cases. Thats why i used the above approach. Could u pls comment on my attempt and why am i not getting $\frac{4+\sqrt{7}}{3}$ to be eliminated ? Feb 2, 2018 at 17:57

let $\sin y= \frac{3}{4}$ we need to prove
$\tan \left[\frac{1}{2}\sin^{-1} \left(\frac{3}{4}\right)\:\right]=\frac{4-\sqrt{7}}{3}$

otherwise seen as

$\tan \left[\frac{1}{2}y\:\right]=\frac{4-\sqrt{7}}{3}$

We also know the identity

$\tan \left[\frac{1}{2}y\:\right]=\frac{1-\cos\left[y\:\right]}{\sin \left[y\:\right]}$

Thus

$\tan \left[y\:\right]=\frac{1-\frac{\sqrt 7}{4}}{\frac{3}{4}}=\frac{4-\sqrt{7}}{3}$

The method you used to eliminate $\frac{-4-\sqrt7}{3}$ is correct. Here I want to explain why it comes out to be a root of the equation so that things become more clear.
Consider the following $$\tan(n\pi +2\theta)=\tan2\theta$$ $$\frac{\tan(\frac{n\pi}{2}+\theta)}{1-\tan^2(\frac{n\pi}{2}+\theta)}=\tan2\theta$$
Now it follows that not only $\tan\theta$ but also $\tan(\frac{n\pi}{2}+\theta)$ is a root of $$\tan2\theta=\frac{x}{1-x^2}$$ and to decide which of the two roots of the quadratic equation is of $\tan\theta$ is to be determined by the value of $2\theta$ and not by merely knowing the value of $\tan2\theta$

• thanx. but my doubt is "where is $\frac{4+\sqrt{7}}{3}$ ?", i only seem to have to eliminate $\frac{-\sqrt{7}-4}{3}$ Feb 2, 2018 at 17:48
• @ss1729 Sorry, that I misunderstood your question. Can you be more clear with the statement "where is the value $\frac{4+\sqrt7}{3}$ to be excluded". Why should there be $\frac{4+\sqrt7}{3}$ Feb 2, 2018 at 19:15