The value of $2 \sin(\frac{\pi}{8}) \sin (\frac{2 \pi}{8}) \sin (\frac{3 \pi}{8}) \sin (\frac{5 \pi}{8}) \sin (\frac{6 \pi}{8}) \sin (\frac{7 \pi}{8})$ is:

$\frac{{\tan \,\,\left( {x\,\, - \,\,{\textstyle{\pi  \over 2}}} \right)\,\,.\,\,\cos \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)\,\, - \,\,{{\sin }^3}\,\left( {{\textstyle{{7\pi } \over 2}}\,\, - \,\,x} \right)}}{{\cos \,\,\left( {x\,\, - \,\,{\textstyle{\pi  \over 2}}} \right)\,\,.\,\,\tan \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)}}$ when simplified reduces to :

Let $A, B, C$ are three angles such that $sinA + sinB + sinC = 0,$ then

$ \frac {sinAsin BsinC}{(sin 3A+ sin 3B+ sin 3C)}$ (wherever definied) is -

For $A = 133^\circ ,\;2\cos \frac{A}{2}$ is equal to

Prove that $\sin ^{2} 6 x-\sin ^{2} 4 x=\sin 2 x \sin 10 x$

$A, B, C$ are the angles of a triangle, then ${\sin ^2}A + {\sin ^2}B + {\sin ^2}C - 2\cos A\,\cos B\,\cos C = $