$1 + \cos 2x + \cos 4x + \cos 6x = $

Let $\frac{\pi}{2} < x < \pi$ be such that $\cot x=\frac{-5}{\sqrt{11}}$. Then $\left(\sin \frac{11 x}{2}\right)(\sin 6 x-\cos 6 x)+\left(\cos \frac{11 x}{2}\right)(\sin 6 x+\cos 6 x)$ is equal to

If $\sin 2\theta + \sin 2\phi = 1/2$ and $\cos 2\theta + \cos 2\phi = 3/2$, then ${\cos ^2}(\theta - \phi ) = $

If $x\cos \theta = y\cos \,\left( {\theta + \frac{{2\pi }}{3}} \right) = z\cos \,\left( {\theta + \frac{{4\pi }}{3}} \right),$ then the value of $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ is equal to

The value of $ \cos ^{3}\left(\frac{\pi}{8}\right) \cdot \cos \left(\frac{3 \pi}{8}\right)+\sin ^{3}\left(\frac{\pi}{8}\right) \cdot \sin \left(\frac{3 \pi}{8}\right)$ is 

$\frac{{\sin 3\theta - \cos 3\theta }}{{\sin \theta + \cos \theta }} + 1 = $