JEE Main 2022 — Mathematics Trigonometry

We know that $\mathrm{sin}\theta \mathrm{sin}(60-\theta )\mathrm{sin}(60+\theta )=\frac{1}{4}\mathrm{sin}3\theta$sinθsin(60−θ)sin(60+θ)=41​sin3θ

Now given $16\mathrm{sin}20^{\circ}\mathrm{sin}40^{\circ}\mathrm{sin}80^{\circ}$16sin20∘sin40∘sin80∘

Comparing with above formula $\theta =20^{\circ}$θ=20∘

we get $16\mathrm{sin}20^{\circ}\mathrm{sin}40^{\circ}\mathrm{sin}80^{\circ}=16\times \frac{1}{4}\times \mathrm{sin}(3\times 20^{\circ})$16sin20∘sin40∘sin80∘=16×41​×sin(3×20∘)

$=16\times \frac{1}{4}\times \mathrm{sin}60^{\circ}$=16×41​×sin60∘ $=4\times \frac{\sqrt{3}}{2}=2\sqrt{3}$=4×23​​=23​