Prove that: $(\sin 3 x+\sin x) \sin x+(\cos 3 x-\cos x) \cos x=0$
$L.H.S.$ $=(\sin 3 x+\sin x) \sin x+(\cos 3 x-\cos x) \cos x$
$=\sin 3 x \sin x+\sin ^{2} x+\cos 3 x \cos x-\cos ^{2} x$
$=\cos 3 x \cos x+\sin 3 x \sin x-\left(\cos ^{2}-\sin ^{2} x\right)$
$=\cos (3 x-x)-\cos 2 x \quad[\cos (A-B)=\cos A \cos B+\sin A \sin B]$
$=\cos 2 x-\cos 2 x$
$=0$
$= R . H.S.$
If $\sin \theta = - \frac{1}{{\sqrt 2 }}$ and $\tan \theta = 1,$ then $\theta $ lies in which quadrant
If $a\cos \theta + b\sin \theta = m$ and $a\sin \theta - b\cos \theta = n,$ then ${a^2} + {b^2} = $
Prove that:
$2 \sin ^{2} \frac{\pi}{6}+\cos ec ^{2} \frac{7 \pi}{6} \cos ^{2} \frac{\pi}{3}=\frac{3}{2}$
Prove that: $\sin x+\sin 3 x+\sin 5 x+\sin 7 x=4 \cos x \cos 2 x \sin 4 x$
Find the degree measures corresponding to the following radian measures (Use $\pi=\frac{22}{7}$ ).
$\frac{11}{16}$