If f(x) = cos x cos 2x cos 4x cos 8x cos 16x, | vedclass

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(A) We know that $\cos 	heta \cos 2	heta \cos 4	heta \dots \cos(2^{n-1}	heta) = \frac{\sin(2^n 	heta)}{2^n \sin 	heta}$.Here, $n=5$ and $	heta

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$	ext{cosec} \left( \frac{\pi}{4} ight)$

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(A) We know that $\cos 	heta \cos 2	heta \cos 4	heta \dots \cos(2^{n-1}	heta) = \frac{\sin(2^n 	heta)}{2^n \sin 	heta}$.Here, $n=5$ and $	heta = x$, so $f(x) = \frac{\sin(32x)}{32 \sin x}$.Using the quotient rule, $f'(x) = \frac{1}{32} \left[ \frac{32 \cos(32x) \sin x - \cos x \sin(32x)}{\sin^2 x} ight]$.At $x = \frac{\pi}{4}$, $\sin(32 \cdot \frac{\pi}{4}) = \sin(8\pi) = 0$ and $\cos(32 \cdot \frac{\pi}{4}) = \cos(8\pi) = 1$.Substituting these values: $f' \left( \frac{\pi}{4} ight) = \frac{1}{32} \left[ \frac{32(1) \sin(\pi/4) - \cos(\pi/4)(0)}{\sin^2(\pi/4)} ight]$.$f' \left( \frac{\pi}{4} ight) = \frac{32 \sin(\pi/4)}{32 \sin^2(\pi/4)} = \frac{1}{\sin(\pi/4)} = 	ext{cosec} \left( \frac{\pi}{4} ight)$.

If $f(x) = \cos x \cos 2x \cos 4x \cos 8x \cos 16x$, then $f' \left( \frac{\pi}{4} \right) = $

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