Question:
Prove that
$\sin \left(150^{\circ}+x\right)+\sin \left(150^{\circ}-x\right)=\cos x$
Solution:
In this question the following formula will be used:
$\sin (A+B)=\sin A \cos B+\cos A \sin B$
$\sin (A-B)=\sin A \cos B-\cos A \sin B$
$=\sin 150^{\circ} \cos x+\cos 150^{\circ} \sin x+\sin 150^{\circ} \cos x-\cos 150^{\circ} \sin x$
$=2 \sin 150^{\circ} \cos x$
$=2 \sin \left(90^{\circ}+60^{\circ}\right) \cos x$
$=2 \cos 60^{\circ} \cos x$
$=2 \times \frac{1}{2} \cos x$
$=\cos x$