# What is the value of $\tan({\frac{1}{2}}\sin^{-1}{\frac{3}{4}})$?

$\begin{array}{1 1} \frac{-4+\sqrt 7}{3} \\ \frac{-4 - \pi}{3} \\ \frac{4+\sqrt 7}{3} \\ \frac{4- \sqrt 7}{3} \end{array}$

Toolbox:
• Take $$sin^{-1}\frac{3}{4}=x$$ and proceed
• $$cosx=2cos^2\frac{x}{2}-1$$
• $$cosx=\sqrt{1-sin^2x}$$
Let $$sin^{-1}\frac{3}{4}=x \Rightarrow sinx=\frac{3}{4}$$
$$cos x=\sqrt{1-sin^2x}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt 7}{4}$$
$$\Rightarrow\:cosx= 2cos^2\frac{x}{2}-1=\frac{\sqrt 7}{4}$$
$$\Rightarrow 2cos^2\frac{x}{2}=\frac{\sqrt{7}}{4}+1=\frac{4+\sqrt 7}{4}$$
$$\Rightarrow cos\frac{x}{2}=\sqrt{\frac{4+\sqrt 7}{8}}$$
$$sin\frac{x}{2}=\sqrt{1-cos^2\frac{x}{2}}=\sqrt{1-\big(\frac{4+\sqrt7}{8}\big)}=\sqrt{\frac{4-\sqrt 7}{8}}$$
$$\Rightarrow\: tan\frac{x}{2}=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=\frac{4-\sqrt 7}{4+\sqrt 7}$$
$$=\frac{4-\sqrt 7}{3}$$ (by rationalising the denominator)
$$tan\frac{x}{2}=\frac{4-\sqrt 7}{3}$$
edited Jun 12, 2014