If cos x cos 2x cos 2²x … cos 2ⁿ⁻¹x = λ sin(2ⁿx)/sin x, Then Find λ
Question:
\[ \cos x \cos 2x \cos 2^2x \cdots \cos 2^{n-1}x = \lambda \frac{\sin(2^n x)}{\sin x} \]Find the value of \(\lambda\).
Solution
Use the standard identity:
\[ \sin 2A = 2\sin A\cos A \]Therefore,
\[ \sin 2x = 2\sin x\cos x \] \[ \sin 4x = 2\sin 2x\cos 2x = 2^2\sin x\cos x\cos 2x \] \[ \sin 8x = 2\sin 4x\cos 4x = 2^3\sin x\cos x\cos 2x\cos 4x \]Continuing this process up to \(2^n x\), we obtain
\[ \sin(2^n x) = 2^n \sin x \cos x \cos 2x \cos 2^2x \cdots \cos 2^{n-1}x \]Hence,
\[ \cos x \cos 2x \cos 2^2x \cdots \cos 2^{n-1}x = \frac{\sin(2^n x)}{2^n\sin x} \]Comparing with
\[ \cos x \cos 2x \cos 2^2x \cdots \cos 2^{n-1}x = \lambda \frac{\sin(2^n x)}{\sin x} \]we get
\[ \lambda=\frac{1}{2^n} \]