Question

$\frac{\sin A+\sin 2A+\sin 4A+\sin 5A}{\cos A+\cos 2A+\cos 4A+\cos 5A}=\tan xA$. Find $x$

Answer 1

$\frac{\sin A+\sin 2A+\sin 4A+\sin 5A}{\cos A+\cos 2A+\cos 4A+\cos 5A}$

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

we know that $\sin C+\sin D=2\sin \frac{C+D}{2}\cos \frac{C-D}{2}$

$\sin C-\sin D=2cos\frac{C+D}{2}\sin \frac{C-D}{2}$

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

$\Rightarrow x=3$
Ans: 3