Question

(i) $\frac{\cos 3\theta +2\cos 5\theta +\cos 7\theta }{\cos \theta +2\cos 3\theta +\cos 5\theta }=\cos 2\theta -\sin 2\theta \tan 3\theta$
(ii) $\frac{\sin A+\sin 3A+\sin 5A+\sin 7A}{\cos A+\cos 3A+\cos 5A+\cos 7A}=\tan 4A$

Answer 1

$\left(i\right)\frac{\cos 3\theta +2\cos 5\theta +\cos 7\theta }{\cos \theta +2\cos 3\theta +\cos 5\theta }$

$=\frac{(\cos 3\theta +\cos 7\theta )+2\cos 5\theta }{(\cos \theta +\cos 5\theta )+2\cos 3\theta }$

$=\frac{2\cos \left(\frac{3\theta +7\theta }{2}\right)\cos \left(\frac{3\theta -7\theta }{2}\right)+2\cos 5\theta }{2\cos \left(\frac{\theta +5\theta }{2}\right)\cos \left(\frac{\theta -5\theta }{2}\right)+2\cos 3\theta }$

$=\frac{2\cos 5\theta \cos 2\theta +2\cos 5\theta }{2\cos 3\theta \cos 2\theta +2\cos 3\theta }$

$=\frac{2\cos 5\theta (\cos 2\theta +1)}{2\cos 3\theta (\cos 2\theta +1)}$

$=\frac{\cos 5\theta }{\cos 3\theta }$

$=\frac{\cos (3\theta +2\theta )}{\cos 3\theta }$

$=\frac{\cos 3\theta \cos 2\theta -\sin 3\theta \sin 2\theta }{\cos 3\theta }$

$=\cos 2\theta -\tan 3\theta \sin 2\theta$

$\left(ii\right)\frac{\sin A+\sin 3A+\sin 5A+\sin 7A}{\cos A+\cos 3A+\cos 5A+\cos 7A}$

$=\frac{(\sin A+\sin 7A)+(\sin 3A+\sin 5A)}{(\cos A+\cos 7A)+(\cos 3A+\cos 5A)}$

$=\frac{2\sin \left(\frac{A+7A}{2}\right)\cos \left(\frac{A-7A}{2}\right)+2\sin \left(\frac{3A+5A}{2}\right)\cos \left(\frac{3A-5A}{2}\right)}{2\cos \left(\frac{A+7A}{2}\right)\cos \left(\frac{A-7A}{2}\right)+2\cos \left(\frac{3A+5A}{2}\right)\cos \left(\frac{3A-5A}{2}\right)}$

$=\frac{2\sin 4A\cos 3A+2\sin 4A\cos A}{2\cos 4A\cos 3A+2\cos 4A\cos A}$

$=\frac{2\sin 4A(\cos 3A+\cos A)}{2\cos 4A(\cos 3A+\cos A)}$

$=\frac{\sin 4A}{\cos 4A}=\tan 4A$

Hence proved.