JEE Main 2022 — Mathematics Trigonometry

Given, $2\mathrm{sin}\frac{\pi }{22}\mathrm{sin}\frac{3\pi }{22}\mathrm{sin}\frac{5\pi }{22}\mathrm{sin}\frac{7\pi }{22}\mathrm{sin}\frac{9\pi }{22}$2sin22π​sin223π​sin225π​sin227π​sin229π​

Now by we know that $\mathrm{sin}(\theta )=\mathrm{cos}(\frac{\pi }{2}-\theta )$sin(θ)=cos(2π​−θ), so by using this we get $\mathrm{sin}\frac{\pi }{22}=\mathrm{cos}(\frac{\pi }{2}-\frac{\pi }{22})=\mathrm{cos}\frac{5\pi }{11}$sin22π​=cos(2π​−22π​)=cos115π​ and similarly $\mathrm{sin}\frac{3\pi }{22}=\mathrm{cos}\frac{4\pi }{11},\mathrm{sin}\frac{5\pi }{22}=\mathrm{cos}\frac{3\pi }{11},\mathrm{sin}\frac{7\pi }{22}=\frac{2\pi }{11},\mathrm{sin}\frac{9\pi }{22}=\mathrm{cos}\frac{\pi }{11}$sin223π​=cos114π​,sin225π​=cos113π​,sin227π​=112π​,sin229π​=cos11π​

So, $2\mathrm{sin}\frac{\pi }{22}\mathrm{sin}\frac{3\pi }{22}\mathrm{sin}\frac{5\pi }{22}\mathrm{sin}\frac{7\pi }{22}\mathrm{sin}\frac{9\pi }{22}$2sin22π​sin223π​sin225π​sin227π​sin229π​

$$=2\mathrm{cos}\frac{5\pi }{11}\mathrm{cos}\frac{4\pi }{11}\mathrm{cos}\frac{3\pi }{11}\mathrm{cos}\frac{2\pi }{11}\mathrm{cos}\frac{1\pi }{11}$$=2cos115π​cos114π​cos113π​cos112π​cos111π​

$$=2\mathrm{cos}\frac{1\pi }{11}\mathrm{cos}\frac{2\pi }{11}\mathrm{cos}\frac{4\pi }{11}\mathrm{cos}\frac{8\pi }{11}\mathrm{cos}\frac{16\pi }{11}$$=2cos111π​cos112π​cos114π​cos118π​cos1116π​

{As $\mathrm{cos}\frac{3\pi }{22}=-\mathrm{cos}(\pi -\frac{3\pi }{11})=-\mathrm{cos}\frac{8\pi }{11}$cos223π​=−cos(π−113π​)=−cos118π​ and $\mathrm{cos}\frac{5\pi }{11}=-\mathrm{cos}\frac{16\pi }{11}$cos115π​=−cos1116π​}

$$=\frac{2\mathrm{sin}\frac{\pi }{11}}{2\mathrm{sin}\frac{\pi }{11}}\mathrm{cos}\frac{\pi }{11}\mathrm{cos}\frac{2\pi }{11}\mathrm{cos}\frac{4\pi }{11}\mathrm{cos}\frac{8\pi }{11}\mathrm{cos}\frac{16\pi }{11}$$=2sin11π​2sin11π​​cos11π​cos112π​cos114π​cos118π​cos1116π​

$$=\frac{2\mathrm{sin}\frac{4\pi }{11}\mathrm{cos}\frac{4\pi }{11}\mathrm{cos}\frac{8\pi }{11}\mathrm{cos}\frac{16\pi }{11}}{{2}^{2}\mathrm{sin}\frac{\pi }{11}}$$=22sin11π​2sin114π​cos114π​cos118π​cos1116π​​

$$=\frac{2\mathrm{sin}\frac{8\pi }{11}\mathrm{cos}\frac{8\pi }{11}\mathrm{cos}\frac{16\pi }{11}}{{2}^{3}\mathrm{sin}\frac{\pi }{11}}$$=23sin11π​2sin118π​cos118π​cos1116π​​

$$=\frac{2\mathrm{sin}\frac{16\pi }{11}\mathrm{cos}\frac{16\pi }{11}}{{2}^{4}\mathrm{sin}\frac{\pi }{11}}$$=24sin11π​2sin1116π​cos1116π​​

$$=\frac{\mathrm{sin}\frac{32\pi }{11}}{{2}^{4}\mathrm{sin}\frac{\pi }{11}}$$=24sin11π​sin1132π​​

$$=\frac{1}{16}\frac{\mathrm{sin}(3\pi -\frac{\pi }{11})}{\mathrm{sin}\frac{\pi }{11}}=\frac{1}{16}$$=161​sin11π​sin(3π−11π​)​=161​