The expression $[1 - sin (3\pi - \alpha ) + cos (3\pi + \alpha )]$ $\left[ {1\,\, - \,\,\sin \,\left( {\frac{{3\,\pi }}{2}\,\, - \,\,\alpha } \right)\,\, + \,\,\cos \,\left( {\frac{{5\,\pi }}{2}\,\, - \,\,\alpha } \right)} \right]$ when simplified reduces to :
$sin \,2\alpha$
$- sin\, 2\alpha$
$1 - sin\, 2\alpha$
$1 + sin\, 2\alpha$
Prove that: $\cos 4 x=1-8 \sin ^{2} x \cos ^{2} x$
If $\cos \theta = \frac{1}{2}\left( {a + \frac{1}{a}} \right),$then the value of $\cos 3\theta $is
$\frac{{\sqrt 2 - \sin \alpha - \cos \alpha }}{{\sin \alpha - \cos \alpha }} = $
$\frac{1}{{\tan 3A - \tan A}} - \frac{1}{{\cot 3A - \cot A}} = $
$\cos 2(\theta + \phi ) - 4\cos (\theta + \phi )\sin \theta \sin \phi + 2{\sin ^2}\phi = $