JEE Main 2019 — Mathematics Trigonometry

$$\mathrm{sin}10^{\circ}\mathrm{sin}30^{\circ}\mathrm{sin}50^{\circ}\mathrm{sin}70^{\circ}$$sin10∘sin30∘sin50∘sin70∘

$$=\mathrm{sin}10^{\circ}\times \frac{1}{2}(\frac{1}{2}\times 2\mathrm{sin}50^{\circ}\mathrm{sin}70^{\circ})$$=sin10∘×21​(21​×2sin50∘sin70∘)

$$=\frac{1}{4}\mathrm{sin}10^{\circ}(2\mathrm{sin}50^{\circ}\mathrm{sin}70^{\circ})$$=41​sin10∘(2sin50∘sin70∘)

Using $2\mathrm{sin}A\mathrm{sin}B=\mathrm{cos}(A-B)-\mathrm{cos}(A+B)$2sinAsinB=cos(A−B)−cos(A+B)

$$=\frac{1}{4}\mathrm{sin}10^{\circ}(\mathrm{cos}20^{\circ}-\mathrm{cos}120^{\circ})$$=41​sin10∘(cos20∘−cos120∘)

$$=\frac{1}{4}\mathrm{sin}10^{\circ}(\mathrm{cos}20^{\circ}-\mathrm{cos}(90^{\circ}+30^{\circ}))$$=41​sin10∘(cos20∘−cos(90∘+30∘))

$$=\frac{1}{4}\mathrm{sin}10^{\circ}(\mathrm{cos}20^{\circ}+\mathrm{sin}30^{\circ})$$=41​sin10∘(cos20∘+sin30∘)

$$=\frac{1}{4}\mathrm{sin}10^{\circ}(\mathrm{cos}20^{\circ}+\frac{1}{2})$$=41​sin10∘(cos20∘+21​)

$$=\frac{1}{8}(2\mathrm{sin}10^{\circ}\mathrm{cos}20^{\circ}+\mathrm{sin}10^{\circ})$$=81​(2sin10∘cos20∘+sin10∘)

Again, using $2\mathrm{sin}A\mathrm{cos}B=\mathrm{sin}(A+B)-\mathrm{sin}(A-B)$2sinAcosB=sin(A+B)−sin(A−B)

$$=\frac{1}{8}(\mathrm{sin}30^{\circ}-\mathrm{sin}10^{\circ}+\mathrm{sin}10^{\circ})$$=81​(sin30∘−sin10∘+sin10∘)

$$=\frac{1}{8}(\frac{1}{2})=\frac{1}{16}.$$=81​(21​)=161​.