\(96 \cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33}\) is equal to

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Trigonometric Ratios and Identities

96 cos \frac{π}{33} cos \frac{2 π}{33} cos \frac{4 π}{33} cos \frac{8 π}{33} cos \frac{16 π}{33}

is equal to

Answer: (A)

Let

P = 96 cos \frac{π}{33} cos \frac{2 π}{33} cos \frac{4 π}{33} cos \frac{8 π}{33} cos \frac{16 π}{33}

Multiply both sides by

2 sin \frac{π}{33}

2 P sin \frac{π}{33} = 96 \cdot 2 sin \frac{π}{33} cos \frac{π}{33} cos \frac{2 π}{33} cos \frac{4 π}{33} cos \frac{8 π}{33} cos \frac{16 π}{33}

Using the identity

2 sin θ cos θ = sin 2 θ

repeatedly:

2 P sin \frac{π}{33} = 96 sin \frac{2 π}{33} cos \frac{2 π}{33} cos \frac{4 π}{33} cos \frac{8 π}{33} cos \frac{16 π}{33}

Continuing this process:

2 P sin \frac{π}{33} = 6 sin \frac{32 π}{33} = 6 sin (π - \frac{π}{33}) = 6 sin \frac{π}{33}

Therefore

P = 3

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Single correct option