JEE Main 2022 — Mathematics Trigonometry

We have, ${\mathrm{sin}}^{2}10^{\circ}\times \mathrm{sin}20^{\circ}\times \mathrm{sin}40^{\circ}\times \mathrm{sin}50^{\circ}\times \mathrm{sin}70^{\circ}$sin210∘×sin20∘×sin40∘×sin50∘×sin70∘

$$=(\mathrm{sin}10^{\circ}\times \mathrm{sin}50^{\circ}\times \mathrm{sin}70^{\circ})(\mathrm{sin}10^{\circ}\times \mathrm{sin}20^{\circ}\times \mathrm{sin}40^{\circ})$$=(sin10∘×sin50∘×sin70∘)(sin10∘×sin20∘×sin40∘)

Using, $\mathrm{sin}\theta =\mathrm{cos}(90^{\circ}-\theta )$sinθ=cos(90∘−θ)

$$=[\mathrm{cos}80^{\circ}\times \mathrm{cos}40^{\circ}\times \mathrm{cos}20^{\circ}][\mathrm{sin}10^{\circ}\times \mathrm{sin}(30^{\circ}-10^{\circ})\times \mathrm{sin}(30^{\circ}+10^{\circ})]$$=[cos80∘×cos40∘×cos20∘][sin10∘×sin(30∘−10∘)×sin(30∘+10∘)]

We know that $\mathrm{cos}\theta \mathrm{cos}2\theta \mathrm{cos}{2}^{2}\theta \ldots \mathrm{cos}{2}^{n-1}\theta =\frac{\mathrm{sin}{2}^{n}\theta }{{2}^{n}\mathrm{sin}\theta }$cosθcos2θcos22θ…cos2n−1θ=2nsinθsin2nθ​

and $\mathrm{sin}(A+B)\cdot \mathrm{sin}(A-B)={\mathrm{sin}}^{2}A-{\mathrm{sin}}^{2}B$sin(A+B)⋅sin(A−B)=sin2A−sin2B$\Rightarrow (\frac{\mathrm{sin}{2}^{3}(20^{\circ})}{{2}^{3}\mathrm{sin}20^{\circ}})[\mathrm{sin}10^{\circ}\times (\frac{1}{4}-{\mathrm{sin}}^{2}10^{\circ})]$⇒(23sin20∘sin23(20∘)​)[sin10∘×(41​−sin210∘)]

$$\Rightarrow (\frac{\mathrm{sin}160^{\circ}}{{2}^{3}\mathrm{sin}20^{\circ}})(\frac{\mathrm{sin}10^{\circ}-4{\mathrm{sin}}^{3}10^{\circ}}{4})$$⇒(23sin20∘sin160∘​)(4sin10∘−4sin310∘​)

$$\Rightarrow (\frac{\mathrm{sin}(180^{\circ}-20^{\circ})}{8\mathrm{sin}20^{\circ}})(\frac{3\mathrm{sin}10^{\circ}-4{\mathrm{sin}}^{3}10^{\circ}-2\mathrm{sin}10^{\circ}}{4})$$⇒(8sin20∘sin(180∘−20∘)​)(43sin10∘−4sin310∘−2sin10∘​)

$$\Rightarrow (\frac{\mathrm{sin}20^{\circ}}{8\mathrm{sin}20^{\circ}})(\frac{\mathrm{sin}30^{\circ}-2\mathrm{sin}10^{\circ}}{4})$$⇒(8sin20∘sin20∘​)(4sin30∘−2sin10∘​)

$$\Rightarrow \frac{1}{8}(\frac{\frac{1}{2}-2\mathrm{sin}10^{\circ}}{4})$$⇒81​(421​−2sin10∘​)

$\Rightarrow \frac{1}{8}(\frac{1-4\mathrm{sin}10^{\circ}}{8})=\alpha -\frac{\mathrm{sin}10^{\circ}}{16}$⇒81​(81−4sin10∘​)=α−16sin10∘​ (given)

$$\Rightarrow \frac{1-4\mathrm{sin}10^{\circ}}{64}=\frac{16\alpha -\mathrm{sin}10^{\circ}}{16}$$⇒641−4sin10∘​=1616α−sin10∘​

$$\Rightarrow 1-4\mathrm{sin}10^{\circ}=64\alpha -4\mathrm{sin}10^{\circ}$$⇒1−4sin10∘=64α−4sin10∘

$\Rightarrow \alpha =\frac{1}{64}$⇒α=641​.