# Question #8ab7c

Jan 5, 2018

$\cos \left(x - 5 \frac{\pi}{6}\right) = \frac{1}{2} \left\{\sqrt{1 - {\cos}^{2} x} - \sqrt{3} \cos x\right\}$.

#### Explanation:

Prerequisites $\left(1\right) : \cos \left(A - B\right) = \cos A \cos B + \sin A \sin B$,

$\left(2\right) \left(i\right) : \cos \left(\pi - \theta\right) = - \cos \theta , \left(2\right) \left(i i\right) : \sin \left(\pi - \theta\right) = \sin \theta$,

$\left(3\right) \left(i\right) : \cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} , \left(3\right) \left(i i\right) \sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$,

$\left(4\right) : \sin \theta = \sqrt{1 - {\cos}^{2} \theta}$.

Now, $\cos \left(x - 5 \frac{\pi}{6}\right) = \cos x \cos \left(5 \frac{\pi}{6}\right) + \sin x \sin \left(5 \frac{\pi}{6}\right)$,

$= \cos x \cos \left(\pi - \frac{\pi}{6}\right) + \sin x \sin \left(\pi - \frac{\pi}{6}\right)$,

$= \left(- \cos \left(\frac{\pi}{6}\right)\right) \cos x + \sin \left(\frac{\pi}{6}\right) \sin x$,

$= - \frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x$,

$= \frac{1}{2} \left(\sin x - \sqrt{3} \cos x\right)$.

$\Rightarrow \cos \left(x - 5 \frac{\pi}{6}\right) = \frac{1}{2} \left\{\sqrt{1 - {\cos}^{2} x} - \sqrt{3} \cos x\right\} ,$ is the

desired expression!