JEE Main 2023 — Mathematics Trigonometry

Given,

Expression $96\cdot \mathrm{cos}\frac{\pi }{33}\cdot \mathrm{cos}\frac{2\pi }{33}\cdot \mathrm{cos}\frac{4\pi }{33}.........\mathrm{cos}\frac{16\pi }{33}$96⋅cos33π​⋅cos332π​⋅cos334π​.........cos3316π​

Now we know that,

$$\mathrm{cos}A\cdot \mathrm{cos}2A\cdot \mathrm{cos}{2}^{2}A\cdot \mathrm{cos}{2}^{3}A.....\cdot \mathrm{cos}{2}^{n-1}A=\frac{\mathrm{sin}{2}^{n}A}{{2}^{n}\mathrm{sin}A}$$cosA⋅cos2A⋅cos22A⋅cos23A.....⋅cos2n−1A=2nsinAsin2nA​

Now using the above formula in given expression we get,

$$96\cdot \mathrm{cos}\frac{\pi }{33}\cdot \mathrm{cos}\frac{2\pi }{33}\cdot \mathrm{cos}\frac{4\pi }{33}.........\mathrm{cos}\frac{16\pi }{33}$$96⋅cos33π​⋅cos332π​⋅cos334π​.........cos3316π​

$$=96\times \frac{\mathrm{sin}\frac{32\pi }{33}}{{2}^{5}\mathrm{sin}\frac{\pi }{33}}$$=96×25sin33π​sin3332π​​

$$=96\times \frac{\mathrm{sin}(\pi -\frac{\pi }{33})}{{2}^{5}\mathrm{sin}\frac{\pi }{33}}$$=96×25sin33π​sin(π−33π​)​

$$=96\times \frac{\mathrm{sin}(\frac{\pi }{33})}{{2}^{5}\mathrm{sin}\frac{\pi }{33}}{\text{as}\mathrm{sin}(\pi -\alpha )=\mathrm{sin}\alpha }$$=96×25sin33π​sin(33π​)​assin(π−α)=sinα

$$=96\times \frac{1}{32}=3$$=96×321​=3