Question

Find the value of $\sin \left[\frac{1}{2}{\cot }^{-1}\left(\frac{-3}{4}\right)\right]$.

Answer 1

$\sin \left[\frac{1}{2}{\cot }^{-1}(-\frac{3}{4})\right]$

Let $\theta ={\cot }^{-1}(-\frac{3}{4})$

$\Rightarrow \cot \theta =\frac{-3}{4}$

so, $\cos \theta =\frac{-3}{5}$

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{-3}{5}\right)$

$\therefore \sin \left[\frac{1}{2}{\cot }^{-1}(-\frac{3}{4})\right]=\sin \left[{\cos }^{-1}(-\frac{3}{5})\right].....\left(1\right)$

we know that :

$\Rightarrow \cos \theta =1-2{\sin }^2\left(\frac{\theta }{2}\right)$

$\Rightarrow \sin \frac{\theta }{2}=\sqrt{\frac{1-\cos \theta }{2}}$

using this we can simplify $\left(1\right)$ as

$\Rightarrow \sin \left[\frac{{\cos }^{-1}\left(\frac{-3}{5}\right)}{2}\right]=\sqrt{\frac{1-\cos \left({\cos }^{-1}\left(\frac{-3}{5}\right)\right)}{2}}$

$=\sqrt{\frac{1-\left(\frac{-3}{5}\right)}{2}}$

$=\sqrt{\frac{1+\frac{3}{5}}{2}}$

$=\sqrt{\frac{8}{10}}$

$=\sqrt{\frac{4}{5}}$

$\therefore \sin \left[\frac{1}{2}{\cot }^{-1}\left(\frac{-3}{4}\right)\right]=\sqrt{\frac{4}{5}}$

Hence, the answer is $\sqrt{\frac{4}{5}}.$