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
$\frac{1}{4}$
$\frac{1}{\sqrt{2}}$
$\frac{1}{2\sqrt{2}}$
$\frac{1}{2}$
If $2\tan A = 3\tan B,$ then $\frac{{\sin 2B}}{{5 - \cos 2B}}$ is equal to
Prove that $\sin 2 x+2 \sin 4 x+\sin 6 x=4 \cos ^{2} x \sin 4 x$
If $A + B + C = \pi ,$ then $\cos \,\,2A + \cos \,\,2B + \cos \,\,2C = $
If ${\tan ^2}\theta = 2{\tan ^2}\phi + 1,$ then $\cos 2\theta + {\sin ^2}\phi $ equals
If $\sin \theta + \sin 2\theta + \sin 3\theta = \sin \alpha $and $\cos \theta + \cos 2\theta + \cos 3\theta = \cos \alpha $, then $\theta$ is equal to