The value of cos frac{pi}{2^2} cdot cos frac{ | vedclass

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(A) We use the identity $\cos A \cdot \cos 2A \cdot \cos 4A \cdot \dots \cdot \cos 2^{n-1}A = \frac{\sin(2^n A)}{2^n \sin A}$.Let $A = \frac{\pi}{2^{1

Vedclass · Accepted Answer

$\frac{1}{512}$

Vedclass · Answer

(A) We use the identity $\cos A \cdot \cos 2A \cdot \cos 4A \cdot \dots \cdot \cos 2^{n-1}A = \frac{\sin(2^n A)}{2^n \sin A}$.Let $A = \frac{\pi}{2^{10}}$. The given expression is $P = \cos \frac{\pi}{2^2} \cdot \cos \frac{\pi}{2^3} \cdot \dots \cdot \cos \frac{\pi}{2^{10}} \cdot \sin \frac{\pi}{2^{10}}$.Note that $\frac{\pi}{2^2} = 2^8 \cdot \frac{\pi}{2^{10}}$,$\frac{\pi}{2^3} = 2^7 \cdot \frac{\pi}{2^{10}}$,and so on.This is equivalent to $\cos(2^8 A) \cdot \cos(2^7 A) \cdot \dots \cdot \cos(A) \cdot \sin A$.Rearranging the product: $(\cos A \cdot \cos 2A \cdot \dots \cdot \cos 2^8 A) \cdot \sin A$.Using the formula with $n=9$: $\frac{\sin(2^9 A)}{2^9 \sin A} \cdot \sin A = \frac{\sin(2^9 \cdot \frac{\pi}{2^{10}})}{2^9} = \frac{\sin(\frac{\pi}{2})}{512} = \frac{1}{512}$.

The value of $\cos \frac{\pi}{2^2} \cdot \cos \frac{\pi}{2^3} \cdot \dots \cdot \cos \frac{\pi}{2^{10}} \cdot \sin \frac{\pi}{2^{10}}$ is

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