The value of $2\sin\left(\frac{\pi}{8}\right)\sin\left(\frac{2\pi}{8}\right)\sin\left(\frac{3\pi}{8}\right)\sin\left(\frac{5\pi}{8}\right)\sin\left(\frac{6\pi}{8}\right)\sin\left(\frac{7\pi}{8}\right)$ is :

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Trigonometric Ratios and Identities

The value of

2 sin (\frac{π}{8}) sin (\frac{2 π}{8}) sin (\frac{3 π}{8}) sin (\frac{5 π}{8}) sin (\frac{6 π}{8}) sin (\frac{7 π}{8})

is :

Answer: (D)

sin (π - θ) = sin θ

sin \frac{7 π}{8} = sin \frac{π}{8}

sin \frac{6 π}{8} = sin \frac{2 π}{8}

and

sin \frac{5 π}{8} = sin \frac{3 π}{8}

, so the product is

2 sin^{2} \frac{π}{8} sin^{2} \frac{π}{4} sin^{2} \frac{3 π}{8} = 2 \cdot \frac{1}{2} (sin \frac{π}{8} sin \frac{3 π}{8})^{2}

Since

sin \frac{3 π}{8} = cos \frac{π}{8}

sin \frac{π}{8} sin \frac{3 π}{8} = sin \frac{π}{8} cos \frac{π}{8} = \frac{1}{2} sin \frac{π}{4} = \frac{1}{2 2}

The value of the product is therefore

(\frac{1}{2 2})^{2} = \frac{1}{8}

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Single correct option