# Prove that (a) sin(x-(3pi)/2)=cosx and (b) sin(theta-pi/3)+cos(theta-pi/6)=sintheta?

Feb 7, 2018

Verified the above relations below.

#### Explanation:

(a) LHS
= $\sin \left(x - \frac{3 \pi}{2}\right)$

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

As $\cos \left(\frac{3 \pi}{2}\right) = 0$ and $\sin \left(\frac{3 \pi}{2}\right) = - 1$

Substituting we have

= $\sin x \times 0 - \cos x \times \left(- 1\right)$

= $0 + \cos x$

= $\cos x$ =RHS

(b) $L H S = \sin \left(\theta - \frac{\pi}{3}\right) + \cos \left(\theta - \frac{\pi}{6}\right)$

$\sin \left(A - B\right) = \sin A \cos B - \cos A \sin B$ and
$\cos \left(A - B\right) = \cos A \cos B + \sin A S \in B$

LHS
= $\sin \theta \cos \left(\frac{\pi}{3}\right) - \cos \theta \sin \left(\frac{\pi}{3}\right) + \cos \theta \cos \left(\frac{\pi}{6}\right) + \sin \theta \sin \left(\frac{\pi}{6}\right)$

As $\cos \left(\frac{\pi}{6}\right) = \sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$ and $\cos \left(\frac{\pi}{3}\right) = \sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$

Therefore LHS=$\sin \theta \times \frac{1}{2} - \cos \theta \times \frac{\sqrt{3}}{2} + \cos \theta \times \frac{\sqrt{3}}{2} + \sin \theta \times \frac{1}{2}$

= $\sin \theta \left(\frac{1}{2} + \frac{1}{2}\right)$

= $\sin \theta$

= RHS