Question

If $f\left(x\right)=\cos x\cdot \cos 2x\cdot \cos 4x\cdot \cos 8x\cdot \cos 16x$, then $f^{\prime }\left(\pi /4\right)$ equals

• A. $-\sqrt{2}$
• B. 0
• C. ${}^1/\sqrt{2}$
• D. cosec $\pi {/}_4$

Answer 1

The correct option is D cosec $\pi {/}_4$

$f\left(x\right)=\frac{2sinxcosxcos2xcos4xcos2xcos16x}{2sinx}$
$=\frac{2sin2xcos2xcos4xcos8xcos16x}{2^{2}sinx}$
$=\frac{2sin4xcos4xcos8xcos16x}{2^{3}sinx}$
$=\frac{2sin8xcos8xcos16x}{2^{4}sinx}$
$=\frac{sin32x}{2^{5}sinx}$
$f^{\prime }\left(x\right)=\frac{1}{2^5}\left(\frac{32sin32xsinx-sin32xcosx}{\left(si{n}^{2}x\right)}\right)$
$f^{\prime }\left(\frac{\pi }{4}\right)=\frac{1}{2^5}\left(\frac{32\times \left(1\right)sin\left(\frac{\pi }{4}\right)-0}{si{n}^2\left(\frac{\pi }{4}\right)}\right)$
$=cosec\left(\pi /4\right)$