${(\cos \alpha + \cos \beta )^2} + {(\sin \alpha + \sin \beta )^2} = $
$4{\cos ^2}\frac{{\alpha - \beta }}{2}$
$4{\sin ^2}\frac{{\alpha - \beta }}{2}$
$4{\cos ^2}\frac{{\alpha + \beta }}{2}$
$4{\sin ^2}\frac{{\alpha + \beta }}{2}$
${\sin ^4}\frac{\pi }{4} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8} = $
$\cos A + \cos (240^\circ + A) + \cos (240^\circ - A) = $
If $\cos \theta = \frac{1}{2}\left( {a + \frac{1}{a}} \right),$then the value of $\cos 3\theta $is
If $\tan \theta = \frac{{\sin \alpha - \cos \alpha }}{{\sin \alpha + \cos \alpha }},$ then $\sin \alpha + \cos \alpha $ and $\sin \alpha - \cos \alpha $ must be equal to
Prove that $\sin 2 x+2 \sin 4 x+\sin 6 x=4 \cos ^{2} x \sin 4 x$