$2 \sin \left(\frac{\pi}{22}\right) \sin \left(\frac{3 \pi}{22}\right) \sin \left(\frac{5 \pi}{22}\right) \sin \left(\frac{7 \pi}{22}\right) \sin \left(\frac{9 \pi}{22}\right)$ is equal to

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Trigonometric Ratios and Identities

2 sin (\frac{π}{22}) sin (\frac{3 π}{22}) sin (\frac{5 π}{22}) sin (\frac{7 π}{22}) sin (\frac{9 π}{22})

is equal to

Answer: (B)

The five angles are the odd multiples

\frac{π}{22}, \frac{3 π}{22}, \frac{5 π}{22}, \frac{7 π}{22}, \frac{9 π}{22}

, that is

\frac{( 2 k - 1 ) π}{22}

for

k = 1

5

. For angles of this form there is the standard result

\prod_{k = 1}^{n} 2 sin \frac{( 2 k - 1 ) π}{4 n + 2} = 1

; here

n = 5

and

4 n + 2 = 22

, so

\prod_{k = 1}^{5} 2 sin \frac{( 2 k - 1 ) π}{22} = 1

. Dividing out the five factors of

2

\prod_{k = 1}^{5} sin \frac{( 2 k - 1 ) π}{22} = \frac{1}{2 ^{5}} = \frac{1}{32}

, and multiplying by the leading factor

2

in the expression gives

\frac{2}{32} = \frac{1}{16}

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Single correct option