# How do you solve and find the value of sin(cos^-1(3/4))?

Mar 2, 2017

$\frac{\sqrt{7}}{4.}$

#### Explanation:

Let, ${\cos}^{-} 1 \left(\frac{3}{4}\right) = \theta .$

Knowing that, ${\cos}^{-} 1 x = \theta \iff x = \cos \theta , \theta \in \left[0 , \pi\right] ,$ we get,

$\cos \theta = \frac{3}{4} , \mathmr{and} , 0 \le \theta \le \pi .$

But $\cos \theta > 0 \Rightarrow \theta \notin \left[\frac{\pi}{2} , \pi\right] \Rightarrow 0 \le \theta \le \frac{\pi}{2.}$

$\therefore \sin \theta = \pm \sqrt{1 - {\cos}^{2} \theta} = \pm \sqrt{1 - \frac{9}{16}} = \pm \frac{\sqrt{7}}{4.}$

$0 \le \theta \le \frac{\pi}{2} \Rightarrow \sin \theta > 0 \Rightarrow \sin \theta = + \frac{\sqrt{7}}{4.}$

$\therefore \sin \left({\cos}^{-} 1 \left(\frac{3}{4}\right)\right) = \sin \theta = \frac{\sqrt{7}}{4.}$

Enjoy Maths.!