$2\cos x - \cos 3x - \cos 5x = $
$2\cos x - \cos 3x - \cos 5x = $
- A
$16{\cos ^3}x{\sin ^2}x$
- B
$16{\sin ^3}x{\cos ^2}x$
- C
$4{\cos ^3}x{\sin ^2}x$
- D
$4{\sin ^3}x{\cos ^2}x$
Similar Questions
$\left( {\frac{{\sin 2A}}{{1 + \cos 2A}}} \right)\,\left( {\frac{{\cos A}}{{1 + \cos A}}} \right)= $
$\left( {\frac{{\sin 2A}}{{1 + \cos 2A}}} \right)\,\left( {\frac{{\cos A}}{{1 + \cos A}}} \right)= $
$\frac{{3 + \cot {{76}^o}\cot {{16}^o}}}{{\cot {{76}^o} + \cot {{16}^o}}}$ =
$\frac{{3 + \cot {{76}^o}\cot {{16}^o}}}{{\cot {{76}^o} + \cot {{16}^o}}}$ =
$\cos \frac{\pi }{5}\cos \frac{{2\pi }}{5}\cos \frac{{4\pi }}{5}\cos \frac{{8\pi }}{5} = $
$\cos \frac{\pi }{5}\cos \frac{{2\pi }}{5}\cos \frac{{4\pi }}{5}\cos \frac{{8\pi }}{5} = $
સમીકરણ $\frac{{{{\tan }^2}20^\circ - {{\sin }^2}20^\circ }}{{{{\tan }^2}20^\circ \,\cdot\,{{\sin }^2}20^\circ }}$ =
સમીકરણ $\frac{{{{\tan }^2}20^\circ - {{\sin }^2}20^\circ }}{{{{\tan }^2}20^\circ \,\cdot\,{{\sin }^2}20^\circ }}$ =
સાબિત કરો કે, $=\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x$
સાબિત કરો કે, $=\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x$