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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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The value of $\sin\large\frac{\pi}{14}$$\sin\large\frac{3\pi}{4}$$4\sin\large\frac{5\pi}{14}$$\sin\large\frac{7\pi}{14}$$\sin\large\frac{9\pi}{14}$$\sin\large\frac{11\pi}{14}$$\sin\large\frac{13\pi}{14}$ is equal to

$(a)\;\large\frac{1}{64}$$\qquad(b)\;\large\frac{1}{32}$$\qquad(c)\;8\qquad(d)\;\large\frac{1}{5}$

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1 Answer

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$\sin\large\frac{\pi}{14}$$\sin\large\frac{3\pi}{4}$$4\sin\large\frac{5\pi}{14}$$\sin\large\frac{7\pi}{14}$$\sin\large\frac{9\pi}{14}$$\sin\large\frac{11\pi}{14}$$\sin\large\frac{13\pi}{14}$
$\Rightarrow \big(\sin\large\frac{\pi}{14}$$\sin\large\frac{3\pi}{14}$$\sin\large\frac{5\pi}{14}\big)^2$
$\Rightarrow \big[\cos\big(\large\frac{\pi}{2}-\frac{\pi}{14}\big)$$\cos\big(\large\frac{\pi}{2}-\frac{3\pi}{14}\big)$$\cos\big(\large\frac{\pi}{2}-\frac{5\pi}{14}\big)\big]$
$\Rightarrow \big(\cos\large\frac{3\pi}{7}$$\cos\large\frac{2\pi}{7}$$\cos\large\frac{\pi}{7}\big)^2$
$\Rightarrow \bigg[\large\frac{1}{8\sin\Large\frac{\pi}{7}}$$\sin\large\frac{8\pi}{7}\bigg]^2$
$\Rightarrow \bigg[\large\frac{\sin\big(\pi+\pi/7\big)}{8\sin\pi/7}\bigg]^2$
$\Rightarrow \bigg[\large\frac{-\sin\Large\frac{\pi}{7}}{8\sin\Large\frac{\pi}{7}}\bigg]^2$
$\Rightarrow \bigg[\large\frac{1}{8}\bigg]^2$
$\Rightarrow \large\frac{1}{64}$
Hence (a) is the correct answer.
answered Oct 11, 2013 by sreemathi.v
edited Oct 11, 2013 by sreemathi.v
 

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