निम्नलिखित को सिद्ध कीजिए
$\sin 2 x+2 \sin 4 x+\sin 6 x=4 \cos ^{2} x \sin 4 x$
$L.H.S.$ $=\sin 2 x+2 \sin 4 x+\sin 6 x$
$=[\sin 2 x+\sin 6 x]+2 \sin 4 x$
$=\left[2 \sin \left(\frac{2 x+6 x}{2}\right) \cos \left(\frac{2 x-6 x}{2}\right)\right]+2 \sin 4 x$
$\left[\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$
$=2 \sin 4 x \cos (-2 x)+2 \sin 4 x$
$=2 \sin 4 x \cos 2 x+2 \sin 4 x$
$=2 \sin 4 x(\cos 2 x+1)$
$=2 \sin 4 x\left(2 \cos ^{2} x-1+1\right)$
$=2 \sin 4 x\left(2 \cos ^{2} x\right)$
$=4 \cos ^{2} x \sin 4 x$
$=R.H .S.$
$\frac{{\sin \theta + \sin 2\theta }}{{1 + \cos \theta + \cos 2\theta }} = $
$\frac{{\sqrt 2 - \sin \alpha - \cos \alpha }}{{\sin \alpha - \cos \alpha }} = $
$2 \sin \left(\frac{\pi}{8}\right) \sin \left(\frac{2 \pi}{8}\right) \sin \left(\frac{3 \pi}{8}\right) \sin \left(\frac{5 \pi}{8}\right) \sin \left(\frac{6 \pi}{8}\right) \sin \left(\frac{7 \pi}{8}\right)$ का मान है -
$\frac{{\sin 3\theta + \sin 5\theta + \sin 7\theta + \sin 9\theta }}{{\cos 3\theta + \cos 5\theta + \cos 7\theta + \cos 9\theta }} = $
यदि $a\tan \theta = b$, तो $a\cos 2\theta + b\sin 2\theta = $