Prove that: $\sin 3 x+\sin 2 x-\sin x=4 \sin x \cos \frac{x}{2} \cos \frac{3 x}{2}$
$L.H.S.$ $=\sin 3 x+\sin 2 x-\sin x$
$=\sin 3 x+\left[2 \cos \left(\frac{2 x+x}{2}\right) \sin \left(\frac{2 x-x}{2}\right)\right]$
$\left[\sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\right]$
$=\sin 3 x+\left[2 \cos \left(\frac{3 x}{2}\right) \sin \left(\frac{x}{2}\right)\right]$
$=\sin 3 x+2 \cos \frac{3 x}{2} \sin \frac{x}{2}$
$=2 \sin \frac{3 x}{2} \cdot \cos \frac{3 x}{2}+2 \cos \frac{3 x}{2} \sin \frac{x}{2}$$\quad[\sin 2 A=2 \sin A \cdot \cos B]$
$=2 \cos \left(\frac{3 x}{2}\right)\left[\sin \left(\frac{3 x}{2}\right)+\sin \left(\frac{x}{2}\right)\right]$
$=2 \cos \left(\frac{3 x}{2}\right)\left[2 \sin \left\{\frac{\left(\frac{3 x}{2}\right)+\left(\frac{x}{2}\right)}{2}\right\} \cos \left\{\frac{\left(\frac{3 x}{2}\right)-\left(\frac{x}{2}\right)}{2}\right\}\right]$
$\left[\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$
$=2 \cos \left(\frac{3 x}{2}\right) \cdot 2 \sin x \cos \left(\frac{x}{2}\right)$
$=4 \sin x \cos \left(\frac{x}{2}\right) \cos \left(\frac{3 x}{2}\right)= R. H.S.$
Prove that $\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)$
Find the radian measures corresponding to the following degree measures:
$25^{\circ}$
The value of $2({\sin ^6}\theta + {\cos ^6}\theta ) - 3({\sin ^4}\theta + {\cos ^4}\theta ) + 1$ is
If $5\tan \theta = 4,$ then $\frac{{5\sin \theta - 3\cos \theta }}{{5\sin \theta + 2\cos \theta }} = $
If $x = \sec \,\phi - \tan \phi ,y = {\rm{cosec}}\phi + \cot \phi ,$ then