$\frac{{\tan \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 2}}} \right)\,\,.\,\,\cos \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)\,\, - \,\,{{\sin }^3}\,\left( {{\textstyle{{7\pi } \over 2}}\,\, - \,\,x} \right)}}{{\cos \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 2}}} \right)\,\,.\,\,\tan \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)}}$ when simplified reduces to :
$sin \,x\, cos\, x$
$- sin^2\, x$
$- sin\, x\, cos\, x$
$sin^2x$
If $\sin \theta = \frac{1}{2}\left( {\sqrt {\frac{x}{y}\,} + \,\sqrt {\frac{y}{x}} } \right)\,,\,\left( {x,y \in R\, - \{ 0\} } \right)$. Then
$2{\cos ^2}\theta - 2{\sin ^2}\theta = 1$, then $\theta =$ .......$^o$
If $\sin 2\theta + \sin 2\phi = 1/2$ and $\cos 2\theta + \cos 2\phi = 3/2$, then ${\cos ^2}(\theta - \phi ) = $
$\frac{{\sin 3\theta + \sin 5\theta + \sin 7\theta + \sin 9\theta }}{{\cos 3\theta + \cos 5\theta + \cos 7\theta + \cos 9\theta }} = $
Prove that $\cos ^{2} 2 x-\cos ^{2} 6 x=\sin 4 x \sin 8 x$