Question

$sin\left(\frac{1}{2}co{t}^{-1}\right(\frac{-3}{4}\left)\right)=$

• A. $\frac{1}{\sqrt{5}}$
• B. $\frac{2}{\sqrt{5}}$
• C. $\frac{-2}{\sqrt{5}}$
• D. $\frac{-1}{\sqrt{5}}$

Answer 1

The correct option is B $\frac{2}{\sqrt{5}}$

Given $sin\left(\frac{1}{2}co{t}^{-1}\right(-\frac{3}{4}\left)\right)$
Formula $co{t}^{-1}(-x)=\pi -co{t}^{-1}x$
$\therefore sin[\frac{\pi }{2}-\frac{co{t}^{-1}}{2}(\frac{3}{4}\left)\right]=cos\left(\frac{co{t}^{-1}}{2}\right(\frac{3}{4}\left)\right)\left[sin\right(90-\theta )=cos\theta ]$
$cos\theta =\frac{3}{5}$
$2co{s}^2\frac{\theta }{2}-1=\frac{3}{5}[\theta =co{t}^{-1}(\frac{3}{4}\left)\right]$
$2co{s}^2\frac{\theta }{2}=\frac{8}{5}$
$co{s}^2\frac{\theta }{2}=\frac{4}{5}$
$cos\frac{\theta }{2}=\frac{2}{\sqrt{5}}$
$\Rightarrow cos\left[\frac{co{t}^{-1}}{2}\right(\frac{3}{2}\left)\right]=\frac{2}{\sqrt{5}}$