cos A cos 2 A cos 4 A ldots cos 2^{n-1} A equ | vedclass

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(A) To evaluate the product $P = \cos A \cos 2 A \cos 2^2 A \ldots \cos 2^{n-1} A$,multiply and divide by $2 \sin A$: $P = \frac{1}{2 \sin A} (2 \sin

Vedclass · Accepted Answer

$\frac{\sin 2^n A}{2^n \sin A}$

Vedclass · Answer

(A) To evaluate the product $P = \cos A \cos 2 A \cos 2^2 A \ldots \cos 2^{n-1} A$,multiply and divide by $2 \sin A$: $P = \frac{1}{2 \sin A} (2 \sin A \cos A) \cos 2 A \cos 4 A \ldots \cos 2^{n-1} A$ Using the identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$,we get: $P = \frac{1}{2 \sin A} (\sin 2 A \cos 2 A) \cos 4 A \ldots \cos 2^{n-1} A$ Multiply and divide by $2$: $P = \frac{1}{2^2 \sin A} (2 \sin 2 A \cos 2 A) \cos 4 A \ldots \cos 2^{n-1} A = \frac{\sin 4 A}{2^2 \sin A} \cos 4 A \ldots \cos 2^{n-1} A$ Continuing this process $n$ times,we obtain: $P = \frac{\sin 2^n A}{2^n \sin A}$

$\cos A \cos 2 A \cos 4 A \ldots \cos 2^{n-1} A$ equals

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