Question

Show that $\frac{cos3\theta -cos11\theta }{sin11\theta -sin3\theta }=tan7\theta$

Answer 1

L.H.S$=\frac{\cos 3\theta -\cos 11\theta }{\sin 11\theta -\sin 3\theta }$

Using transformation angle formula, $\cos C-\cos D=-2\sin \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right)$ and $\sin C-\sin D=2\cos \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right)$

$=\frac{-2\sin \left(\frac{3\theta +11\theta }{2}\right)\sin \left(\frac{3\theta -11\theta }{2}\right)}{2\cos \left(\frac{11\theta +3\theta }{2}\right)\sin \left(\frac{11\theta -3\theta }{2}\right)}$

$=\frac{-2\sin \left(\frac{14\theta }{2}\right)\sin \left(\frac{-8\theta }{2}\right)}{2\cos \left(\frac{14\theta }{2}\right)\sin \left(\frac{8\theta }{2}\right)}$

$=\frac{-2\sin \left(7\theta \right)\sin (-4\theta )}{2\cos \left(7\theta \right)\sin \left(4\theta \right)}$

$=\frac{\sin 7\theta \sin 4\theta }{\cos 7\theta \sin 4\theta }$

$=\frac{\sin 7\theta }{\cos 7\theta }$

$=\tan 7\theta$

$=$R.H.S

Hence proved.