If $\cos(\theta - \alpha) = a$ and $\sin(\theta - \beta) = b$, then the value of $\cos^2(\alpha - \beta) + 2ab \sin(\alpha - \beta) + \cos^2(\theta - \alpha)$ is...

If $\sin \theta + \cos \theta = \frac{1}{2}$,then $16(\sin(2\theta) + \cos(4\theta) + \sin(6\theta))$ is equal to:

$\cos \frac{7 \pi}{8}+\cos \frac{\pi}{4}+\cos \left(\frac{-\pi}{8}\right)-1=$

If $\tan B = \frac{n \sin A \cos A}{1 - n \cos^2 A}$,then $\tan(A + B)$ equals

If $\sum_{r=1}^{13} \left\{ \frac{1}{\sin \left(\frac{\pi}{4} + (r-1) \frac{\pi}{6}\right) \sin \left(\frac{\pi}{4} + \frac{r\pi}{6}\right)} \right\} = a\sqrt{3} + b$,where $a, b \in \mathbb{Z}$,then $a^2 + b^2$ is equal to:

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