Prove that: \[ \frac{\sin A + \sin 3A + \sin 5A} {\cos A + \cos 3A + \cos 5A} = \tan 3A \]
Solution
L.H.S.
\[ = \frac{\sin A + \sin 3A + \sin 5A} {\cos A + \cos 3A + \cos 5A} \]Group first and third terms and use identity:
\[ \sin C + \sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2} \] \[ \cos C + \cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2} \]
\[
=
\frac{
2\sin3A\cos2A + \sin3A
}{
2\cos3A\cos2A + \cos3A
}
\]
Take common factor:
\[ = \frac{ \sin3A(2\cos2A+1) }{ \cos3A(2\cos2A+1) } \]Cancel common factor:
\[ = \frac{\sin3A}{\cos3A} \]Use identity:
\[ \tan\theta=\frac{\sin\theta}{\cos\theta} \] \[ = \tan3A \]Hence Proved.