Question

Show that $\frac{\cos 2A+2\cos 4A+\cos 6A}{\cos A+2\cos 3A+\cos 5A}=\cos A-\sin A\cdot \tan 3A$

Answer 1

We have,

L.H.S.

$\frac{\cos 2A+2\cos 4A+\cos 6A}{\cos A+2\cos 3A+\cos 5A}$

$=\frac{\cos 2A+\cos 6A+2\cos 4A}{\cos A+\cos 5A+2\cos 3A}$

$=\frac{2\cos \left(\frac{2A+6A}{2}\right)\cos \left(\frac{2A-6A}{2}\right)+2\cos 4A}{2\cos \left(\frac{A+5A}{2}\right)\cos \left(\frac{A-5A}{2}\right)+2\cos 3A}$

$=\frac{2\cos 4A(\cos 2A+1)}{2\cos 3A(\cos 2A+1)}$

$=\frac{\cos 4A}{\cos 3A}$

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

$=\frac{\cos A\cos 3A-\sin A\sin 3A}{\cos 3A}$

$=\frac{\cos A\cos 3A}{\cos 3A}-\frac{\sin A\sin 3A}{\cos 3A}$

$=\cos A-\sin A\tan 3A$

R.H.S.

Hence proved.