Exams Joint Entrance Examination Mathematics Trigonometric Ratios and Identities Question 21 Question 21 of 298 Medium MCQ
96 cos π 33 cos 2 π 33 cos 4 π 33 cos 8 π 33 cos 16 π 33 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} 96 cos 33 π cos 33 2 π cos 33 4 π cos 33 8 π cos 33 16 π is equal to :
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Let
\[
A = 96 cos π 33 cos 2 π 33 cos 4 π 33 cos 8 π 33 cos 16 π 33 ⇒ 2 A = 96 × 2 ( cos π 33 cos 2 π 33 cos 4 π 33 cos 8 π 33 cos 16 π 33 ) ⇒ 2 A × sin π 33 = 96 × ( 2 sin π 33 cos π 33 cos 2 π 33 cos 4 π 33 ⋅ cos 8 π 33 cos 16 π 33 ) ⇒ 2 A × sin π 33 = 6 × sin 32 π 33 = 6 × sin π 33 ⇒ 2 A = 6 ⇒ A = 3 \begin{aligned}
& A=96 \cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33} \\\\
& \Rightarrow 2 A=96 \times 2\left(\cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33}\right) \\\\
& \Rightarrow 2 A \times \sin \frac{\pi}{33} =96 \times\left(2 \sin \frac{\pi}{33} \cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cdot \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33}\right) \\\\
& \Rightarrow 2 A \times \sin \frac{\pi}{33}=6 \times \sin \frac{32 \pi}{33}=6 \times \sin \frac{\pi}{33} \\\\
& \Rightarrow 2 A=6 \Rightarrow A=3
\end{aligned} A = 96 cos 33 π cos 33 2 π cos 33 4 π cos 33 8 π cos 33 16 π ⇒ 2 A = 96 × 2 ( cos 33 π cos 33 2 π cos 33 4 π cos 33 8 π cos 33 16 π ) ⇒ 2 A × sin 33 π = 96 × ( 2 sin 33 π cos 33 π cos 33 2 π cos 33 4 π ⋅ cos 33 8 π cos 33 16 π ) ⇒ 2 A × sin 33 π = 6 × sin 33 32 π = 6 × sin 33 π ⇒ 2 A = 6 ⇒ A = 3
\]
Thus, the required answer is 3 . Answer the question to unlock