Question

The value of $sin\frac{\pi }{14}.sin\frac{3\pi }{14}.sin\frac{5\pi }{14}.sin\frac{7\pi }{14}.sin\frac{9\pi }{14}.sin\frac{11\pi }{14}.sin\frac{13\pi }{14}$ is equal to ............

Answer 1

$sin\frac{\pi }{14}.sin\frac{3\pi }{14}.sin\frac{5\pi }{14}.sin\frac{7\pi }{14}.sin\frac{9\pi }{14}.sin\frac{11\pi }{14}.sin\frac{13\pi }{14}$
$=sin\frac{\pi }{14}.sin\frac{3\pi }{14}.sin\frac{5\pi }{14}.sin(\pi -\frac{5\pi }{14}).sin(\pi -\frac{3\pi }{14}).sin(\pi -\frac{\pi }{14})$
$=si{n}^2\frac{\pi }{14}.si{n}^2\frac{3\pi }{14}.si{n}^2\frac{5\pi }{14}=(sin\frac{5\pi }{14}.sin\frac{3\pi }{14}.sin\frac{\pi }{14})$
$=\left(cos\right(\frac{\pi }{2}-\frac{\pi }{14}).cos(\frac{\pi }{2}-\frac{3\pi }{14}).cos(\frac{\pi }{2}-\frac{5\pi }{14}){)}^2$
$=(cos\frac{3\pi }{7}.cos\frac{2\pi }{7}.cos\frac{\pi }{7}{)}^2$
$=(-cos\frac{\pi }{7}.cos\frac{2\pi }{7}.cos\frac{\pi }{7}{)}^2$
$=(-cos\frac{\pi }{7}.cos\frac{2\pi }{7}.cos\frac{4\pi }{7}{)}^2=(\frac{sin{2}^3\pi /7}{2^3.sin\pi /7}{)}^2$
$=(-\frac{1}{8}.\frac{sin8\pi /7}{sin\pi /7}{)}^2[\because sin\frac{8\pi }{7}=sin\left(\right(\pi +\frac{\pi }{7})=-sin\frac{\pi }{7}]$
$1/64.$