JEE Main 2026 (Online) 2nd April Morning ShiftQuestion 1
If $\sin\left(\frac{\pi}{18}\right)\sin\left(\frac{5\pi}{18}\right)\sin\left(\frac{7\pi}{18}\right)=K$, then the value of $\sin\left(\frac{10K\pi}{3}\right)$ is :
Trigonometric Ratio & Identities
Solution
Concept. $\sin10^\circ\sin50^\circ\sin70^\circ$ is of the form $\sin x\sin(60^\circ-x)\sin(60^\circ+x)$. Formulas used. (1) $\sin x\sin(60^\circ-x)\sin(60^\circ+x)=\frac14\sin3x$(2) $\sin75^\circ=\sin(45^\circ+30^\circ)=\dfrac{\sqrt3+1}{2\sqrt2}$Step 1 - Find $K$. $\frac{\pi}{18}=10^\circ$, $\frac{5\pi}{18}=50^\circ$, $\frac{7\pi}{18}=70^\circ$. By (1) with $x=10^\circ$: $$K=\frac14\sin30^\circ=\frac18.$$ Step 2 - The angle. $\dfrac{10K\pi}{3}=\dfrac{10\pi}{24}=\dfrac{5\pi}{12}=75^\circ$. Step 3 - Value. By (2), $\sin75^\circ=\dfrac{\sqrt3+1}{2\sqrt2}$. Answer: (A)