સાબિત કરો કે : $\sin 2 x+2 \sin 4 x+\sin 6 x=4 \cos ^{2} x \sin 4 x$
$L.H.S.$ $=\sin 2 x+2 \sin 4 x+\sin 6 x$
$=[\sin 2 x+\sin 6 x]+2 \sin 4 x$
$=\left[2 \sin \left(\frac{2 x+6 x}{2}\right) \cos \left(\frac{2 x-6 x}{2}\right)\right]+2 \sin 4 x$
$\left[\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$
$=2 \sin 4 x \cos (-2 x)+2 \sin 4 x$
$=2 \sin 4 x \cos 2 x+2 \sin 4 x$
$=2 \sin 4 x(\cos 2 x+1)$
$=2 \sin 4 x\left(2 \cos ^{2} x-1+1\right)$
$=2 \sin 4 x\left(2 \cos ^{2} x\right)$
$=4 \cos ^{2} x \sin 4 x$
$=R.H .S.$
જો $\tan \theta = \frac{{\sin \alpha - \cos \alpha }}{{\sin \alpha + \cos \alpha }},$ તો $\sin \alpha + \cos \alpha $ અને $\sin \alpha - \cos \alpha $ ની કિમત . . . . ને સમાન થવી જ જોઈએ.
If $k = \sin \frac{\pi }{{18}}\,.\,\sin \frac{{5\pi }}{{18}}\,.\,\sin \frac{{7\pi }}{{18}},$ then the numerical value of $k$ is
$\cos \frac{\pi }{7}\cos \frac{{2\pi }}{7}\cos \frac{{4\pi }}{7} = $
$3\,\left[ {{{\sin }^4}\,\left( {\frac{{3\pi }}{2} - \alpha } \right) + {{\sin }^4}\,(3\pi + \alpha )} \right]$ $ - 2\,\left[ {{{\sin }^6}\,\left( {\frac{\pi }{2} + \alpha } \right) + {{\sin }^6}(5\pi - \alpha )} \right] = $
જો $A + B + C = {180^o},$ તો $\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} = . . .$