સાબિત કરો કે : $\cot 4 x(\sin 5 x+\sin 3 x)=\cot x(\sin 5 x-\sin 3 x)$
$L.H.S$ $=\cot 4 x(\sin 5 x+\sin 3 x)$
$=\frac{\cot 4 x}{\sin 4 x}\left[2 \sin \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)\right]$
$\left[\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$
$=\left(\frac{\cos 4 x}{\sin 4 x}\right)[2 \sin 4 x \cos x]$
$=2 \cos 4 x \cos x$
$R.H.S.$ $=\cot x(\sin 5 x-\sin 3 x)$
$=\frac{\cos x}{\sin x}\left[2 \cos \left(\frac{5 x+3 x}{2}\right) \sin \left(\frac{5 x-3 x}{2}\right)\right]$
$\left[\because \sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\right]$
$=\frac{\cos x}{\sin x}[2 \cos 4 x \sin x]$
$=2 \cos 4 x \cdot \cos x$
$L.H.S.$ $=$ $R.H.S.$
જો $\cos x + \cos y + \cos \alpha = 0$ અને $\sin x + \sin y + \sin \alpha = 0,$ તો $\cot \,\left( {\frac{{x + y}}{2}} \right) = $
$2\,{\sin ^2}\beta + 4\,\,\cos \,(\alpha + \beta )\,\,\sin \,\alpha \,\sin \,\beta + \cos \,2\,(\alpha + \beta ) = $
${\rm{cosec }}A - 2\cot 2A\cos A = $
જો $\sin \theta + \sin 2\theta + \sin 3\theta = \sin \alpha $અને $\cos \theta + \cos 2\theta + \cos 3\theta = \cos \alpha $, તો $\theta$ મેળવો.
સાબિત કરો કે : $\sin 2 x+2 \sin 4 x+\sin 6 x=4 \cos ^{2} x \sin 4 x$