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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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$\cos\large\frac{\pi}{11}$$\cos\large\frac{2\pi}{11}$$\cos\large\frac{3\pi}{11}..........$$\cos\large\frac{11\pi}{11}$ is equal to

$(a)\;-\large\frac{1}{32}$$\qquad(b)\;\large\frac{1}{512}$$\qquad(c)\;\large\frac{1}{1024}$$\qquad(d)\;-\large\frac{1}{2048}$

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1 Answer

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$\cos\large\frac{\pi}{11}$$\cos\large\frac{2\pi}{11}$$\cos\large\frac{3\pi}{11}..........$$\cos\large\frac{11\pi}{11}$
$\Rightarrow\big [\cos\large\frac{\pi}{11}$$\cos\large\frac{2\pi}{11}$$\cos\large\frac{3\pi}{11}$$\cos\large\frac{4\pi}{11}$$\cos\large\frac{5\pi}{11}\big]^2$
$\Rightarrow \big[\cos\large\frac{\pi}{11}$$\cos\large\frac{2\pi}{11}$$\cos\large\frac{4\pi}{11}$$\cos\large\frac{8\pi}{11}$$\cos\large\frac{5\pi}{11}\big]^2$
$\Rightarrow \bigg[\large\frac{\sin 16\Large\frac{\pi}{11}}{16\sin\Large\frac{\pi}{11}}$$\cos\large\frac{5\pi}{11}\bigg]^2$
$\Rightarrow \bigg[\large\frac{2\sin\Large\frac{5\pi}{11}.\cos\Large\frac{5\pi}{11}}{32\sin\Large\frac{\pi}{11}}\bigg]^2$
$\Rightarrow \large\frac{1}{1024}$
Hence (c) is the correct answer.
answered Oct 15, 2013 by sreemathi.v
 

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