Question

If $f(x)=cosx\cdot cos2x\cdot cos4x\cdot cos8x\cdot cos16x$, then the value of $f^{\prime }\left(\frac{\pi }{4}\right)$ is

• A. $1$
• B. $\sqrt{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $0$

Answer 1

Answer: (B)

We know that,
$cosAcos2Acos{2}^{2}A...cos{2}^{n-1}A$$=\frac{sin\left(2^{n}A\right)}{2^{n}sinA}$
$\therefore cosxcos2xcos4xcos8xcos16x=\frac{sin32x}{32sinx}$
$\Rightarrow f\left(x\right)=\frac{1}{32}\cdot \frac{sin32}{sinx}$
$\therefore f^{\prime }\left(x\right)=\frac{1}{32}\times \frac{sinx\left(32cos32x\right)-sin32x\cdot cosx}{si{n}^{2}x}$
$\therefore f^{\prime }\left(\frac{\pi }{4}\right)=\sqrt{2}$