Question

Prove that $\frac{\cos 3A+2\cos 5A+\cos 7A}{\cos A+2\cos 3A+\cos 5A}=\frac{\cos 5A}{\cos 3A}=\cos 2A-\sin 2A\tan 3A$

Answer 1

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

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

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

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

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

$=\cos 2A-\tan 3A\sin 2A=$Rhs

LHS=RHS

Hence proof.