Question

Show that:
$\frac{\sin A+\sin 5A+\sin 9A}{\cos A+\cos 5A+\cos 9A}=\tan 5A$

Answer 1

L.H.S

$\frac{\sin A+\sin 5A+\sin 9A}{\cos A+\cos 5A+\cos 9A}$

We know that

$\sin C+\sin D=2\sin \left(\frac{C+D}{2}\right)\cdot \cos \left(\frac{C-D}{2}\right)$

$\cos C+\cos D=2\cos \left(\frac{C+D}{2}\right)\cdot \cos \left(\frac{C-D}{2}\right)$

Therefore,

$=\frac{2\sin \left(\frac{A+9A}{2}\right)\cdot \cos \left(\frac{A-9A}{2}\right)+\sin 5A}{2\cos \left(\frac{A+9A}{2}\right)\cdot \cos \left(\frac{A-9A}{2}\right)+\cos 5A}$

$=\frac{2\sin \left(5A\right)\cdot \cos \left(4A\right)+\sin 5A}{2\cos \left(5A\right)\cdot \cos \left(4A\right)+\cos 5A}$

$=\frac{2\cos \left(4A\right)+1}{2\cos \left(4A\right)+1}\left(\frac{\sin 5A}{\cos 5A}\right)$

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

$=\tan 5A$

Hence, proved.