JEE Main 2022 — Trigonometric Ratios & Identities Question with Solution

Given, $2\sin \frac{\pi }{22}\sin \frac{3\pi }{22}\sin \frac{5\pi }{22}\sin \frac{7\pi }{22}\sin \frac{9\pi }{22}$

Now by we know that $\sin \left(\theta \right)=\cos \left(\frac{\mathrm{\pi }}{2}-\theta \right)$, so by using this we get $\sin \frac{\mathrm{\pi }}{22}=\cos \left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{22}\right)=\cos \frac{5\mathrm{\pi }}{11}$ and similarly $\sin \frac{3\pi }{22}=\cos \frac{4\mathrm{\pi }}{11}, \sin \frac{5\pi }{22}=\cos \frac{3\mathrm{\pi }}{11}, \sin \frac{7\pi }{22}=\frac{2\mathrm{\pi }}{11}, \sin \frac{9\pi }{22}=\cos \frac{\mathrm{\pi }}{11}$

So, $2\sin \frac{\pi }{22}\sin \frac{3\pi }{22}\sin \frac{5\pi }{22}\sin \frac{7\pi }{22}\sin \frac{9\pi }{22}$

$=2\cos \frac{5\pi }{11}\cos \frac{4\pi }{11}\cos \frac{3\pi }{11}\cos \frac{2\pi }{11}\cos \frac{1\pi }{11}$

$=2\cos \frac{1\pi }{11}\cos \frac{2\pi }{11}\cos \frac{4\pi }{11}\cos \frac{8\pi }{11}\cos \frac{16\pi }{11}$

{As $\cos \frac{3\pi }{22}=-\cos \left(\mathrm{\pi }-\frac{3\pi }{11}\right)=-\cos \frac{8\pi }{11}$ and $\cos \frac{5\pi }{11}=-\cos \frac{16\pi }{11}$}

$=\frac{2\sin {\displaystyle \frac{\mathrm{\pi }}{11}}}{2\sin {\displaystyle \frac{\mathrm{\pi }}{11}}}\cos \frac{\pi }{11}\cos \frac{2\pi }{11}\cos \frac{4\pi }{11}\cos \frac{8\pi }{11}\cos \frac{16\pi }{11}$

$=\frac{2\sin \frac{4\pi }{11}\cos \frac{4\pi }{11}\cos \frac{8\pi }{11}\cos \frac{16\pi }{11}}{2^2\sin {\displaystyle \frac{\mathrm{\pi }}{11}}}$

$=\frac{2\sin \frac{8\pi }{11}\cos \frac{8\pi }{11}\cos \frac{16\pi }{11}}{2^3\sin {\displaystyle \frac{\mathrm{\pi }}{11}}}$

$=\frac{2\sin \frac{16\pi }{11}\cos \frac{16\pi }{11}}{2^4\sin {\displaystyle \frac{\mathrm{\pi }}{11}}}$

$=\frac{\sin \frac{32\pi }{11}}{2^4\sin \frac{\pi }{11}}$

$=\frac{1}{16}\frac{\sin \left(3\pi -\frac{\pi }{11}\right)}{\sin \frac{\pi }{11}}=\frac{1}{16}$