JEE Main 2022 (Online) 26th June Morning ShiftQuestion 1
If $\sin^{2}(10^{\circ})\sin(20^{\circ})\sin(40^{\circ})\sin(50^{\circ})\sin(70^{\circ})=\alpha-\frac{1}{16}\sin(10^{\circ})$, then $16+\alpha^{-1}$ is equal to $\_\_\_\_\_\_$ .
Trigonometric Ratio & Identities
Answer$80$
Solution
Concept. Group $\sin10^\circ\sin50^\circ\sin70^\circ$ (a standard product) and simplify the rest with product-to-sum. Formulas used. (1) $\sin x\sin(60^\circ-x)\sin(60^\circ+x)=\frac14\sin3x$(2) $2\sin A\sin B=\cos(A-B)-\cos(A+B)$, $2\sin A\cos B=\sin(A+B)+\sin(A-B)$Step 1 - Group. By (1), $\sin10^\circ\sin50^\circ\sin70^\circ=\frac14\sin30^\circ=\frac18$. The product is $\frac18\sin10^\circ\cdot\sin20^\circ\sin40^\circ$. Step 2 - $\sin20^\circ\sin40^\circ$. By (2), $\frac12(\cos20^\circ-\cos60^\circ)=\frac12\cos20^\circ-\frac14$. Step 3 - $\sin10^\circ\cos20^\circ$. By (2), $\frac12(\sin30^\circ-\sin10^\circ)=\frac14-\frac12\sin10^\circ$. Step 4 - Put together. $$\frac18\left[\frac12\left(\frac14-\frac12\sin10^\circ\right)-\frac14\sin10^\circ\right]=\frac{1}{64}-\frac{1}{16}\sin10^\circ.$$ So $\alpha=\frac1{64}$. Step 5 - Value. $16+64=80$. Answer: 80