If $\cos \theta = \frac{1}{2}\left( {x + \frac{1}{x}} \right)$, then $\frac{1}{2}\left( {{x^2} + \frac{1}{{{x^2}}}} \right) = $
$\sin 2\theta $
$\cos \,2\theta $
$\tan \,2\theta $
$\sec \,2\theta $
${\sin ^6}\theta + {\cos ^6}\theta + 3{\sin ^2}\theta {\cos ^2}\theta = $
The equation ${\sin ^2}\theta = \frac{{{x^2} + {y^2}}}{{2xy}},x,y, \ne 0$ is possible if
If $\sin (\alpha - \beta ) = \frac{1}{2}$ and $\cos (\alpha + \beta ) = \frac{1}{2},$ where $\alpha $ and $\beta $ are positive acute angles, then
Prove that: $(\sin 3 x+\sin x) \sin x+(\cos 3 x-\cos x) \cos x=0$
Convert $6$ radians into degree measure.