$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
$\sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6 $ is equal to
$\sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6 $ is equal to
- [IIT 1966]
The value of $\frac{1}{4} \,\,tan \frac{\pi}{8} +\frac{1}{8} \,\,tan \frac{\pi}{16}+\frac{1}{16} \,\,tan \frac{\pi}{32}+.\,.\,.\,\infty $ terms is equal to-
The value of $\frac{1}{4} \,\,tan \frac{\pi}{8} +\frac{1}{8} \,\,tan \frac{\pi}{16}+\frac{1}{16} \,\,tan \frac{\pi}{32}+.\,.\,.\,\infty $ terms is equal to-
Prove that: $\cos 6 x=32 x \cos ^{6} x-48 \cos ^{4} x+18 \cos ^{2} x-1$
Prove that: $\cos 6 x=32 x \cos ^{6} x-48 \cos ^{4} x+18 \cos ^{2} x-1$
The value of $x$ that satisfies the relation $x = 1 - x + x^2 - x^3 + x^4 - x^5 + ......... \infty$
The value of $x$ that satisfies the relation $x = 1 - x + x^2 - x^3 + x^4 - x^5 + ......... \infty$
If $\sin A = n\sin B,$ then $\frac{{n - 1}}{{n + 1}}\tan \,\frac{{A + B}}{2} = $
If $\sin A = n\sin B,$ then $\frac{{n - 1}}{{n + 1}}\tan \,\frac{{A + B}}{2} = $