Question:

$96\cos\dfrac{\pi}{33}\cos\dfrac{2\pi}{33}\cos\dfrac{4\pi}{33}\cos\dfrac{8\pi}{33}\cos\dfrac{16\pi}{33}$ is equal to

Step-by-Step Solution

Key Concept: Use the identity $\prod_{k=0}^{n-1}\cos(2^k A)=\frac{\sin(2^nA)}{2^n\sin A}$ with $A=\frac{\pi}{33}$, $n=5$.

$96\cdot\frac{1}{32}=3$.

Correct Answer: 1