Question

Prove that $\frac{\sin A+\sin 3A+\sin 5A+\sin 7A}{\cos A+\cos 3A+\cos 5A+\cos 7A}=\tan 4A$

Answer 1

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

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

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

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

$=\tan 4A$

LHS=RHS

Hence proof