JEE Main 2022 (Online) 26th June Evening ShiftQuestion 1
$16\sin(20^{\circ})\sin(40^{\circ})\sin(80^{\circ})$ is equal to :
Trigonometric Ratio & Identities
Solution
Concept. $\sin20^\circ\sin40^\circ\sin80^\circ$ is of the form $\sin x\sin(60^\circ-x)\sin(60^\circ+x)$ with $x=20^\circ$. Formulas used. (1) $\sin x\sin(60^\circ-x)\sin(60^\circ+x)=\frac14\sin3x$Step 1 - Apply (1). $\sin20^\circ\sin40^\circ\sin80^\circ=\frac14\sin60^\circ=\frac{\sqrt3}{8}$. Step 2 - Value. $16\cdot\frac{\sqrt3}{8}=2\sqrt3$. Answer: (B)