QJEEM23Madc99: JEE Main 2023 Mathematics - Class 11 JEEM23Madc99 Solution

Question

Verified Solution • 2025–26 Session moderate

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

Accepted Answer

Let

\begin{aligned}&amp; A=96 \cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33} \\\\&amp; \Rightarrow 2 A=96 \times 2\left(\cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33}\right) \\\\&amp; \Rightarrow 2 A \times \sin \frac{\pi}{33} =96 \times\left(2 \sin \frac{\pi}{33} \cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cdot \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33}\right) \\\\&amp; \Rightarrow 2 A \times \sin \frac{\pi}{33}=6 \times \sin \frac{32 \pi}{33}=6 \times \sin \frac{\pi}{33} \\\\&amp; \Rightarrow 2 A=6 \Rightarrow A=3\end{aligned}

Thus, the required answer is 3 .

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