The value of $\sum_{r-1}^{18} cos^2(5r)^o,$  where $x^o $ denotes the $x$ degree, is equals to

Prove that: $\cos 6 x=32 x \cos ^{6} x-48 \cos ^{4} x+18 \cos ^{2} x-1$

The exact value of $cos^273^o  + cos^247^o  + (cos73^o  . cos47^o )$ is

$\cos 2(\theta + \phi ) - 4\cos (\theta + \phi )\sin \theta \sin \phi + 2{\sin ^2}\phi = $

If ${\cos ^6}\alpha + {\sin ^6}\alpha + K\,{\sin ^2}2\alpha = 1,$ then $K =$

The sum of all values of $\theta \, \in \,\left( {0,\frac{\pi }{2}} \right)$ satisfying ${\sin ^2}\,2\theta  + {\cos ^4}\,2\theta  = \frac{3}{4}$ is