JEE MAIN - Mathematics (2023 - 10th April Morning Shift - No. 10)
$$96\cos {\pi \over {33}}\cos {{2\pi } \over {33}}\cos {{4\pi } \over {33}}\cos {{8\pi } \over {33}}\cos {{16\pi } \over {33}}$$ is equal to :
4
2
1
3
Explanation
Let
$$ \begin{aligned} & A=96 \cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33} \\\\ & \Rightarrow 2 A=96 \times 2\left(\cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33}\right) \\\\ & \Rightarrow 2 A \times \sin \frac{\pi}{33} =96 \times\left(2 \sin \frac{\pi}{33} \cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cdot \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33}\right) \\\\ & \Rightarrow 2 A \times \sin \frac{\pi}{33}=6 \times \sin \frac{32 \pi}{33}=6 \times \sin \frac{\pi}{33} \\\\ & \Rightarrow 2 A=6 \Rightarrow A=3 \end{aligned} $$
Thus, the required answer is 3 .
$$ \begin{aligned} & A=96 \cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33} \\\\ & \Rightarrow 2 A=96 \times 2\left(\cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33}\right) \\\\ & \Rightarrow 2 A \times \sin \frac{\pi}{33} =96 \times\left(2 \sin \frac{\pi}{33} \cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cdot \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33}\right) \\\\ & \Rightarrow 2 A \times \sin \frac{\pi}{33}=6 \times \sin \frac{32 \pi}{33}=6 \times \sin \frac{\pi}{33} \\\\ & \Rightarrow 2 A=6 \Rightarrow A=3 \end{aligned} $$
Thus, the required answer is 3 .
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