Prove that: \[ \frac{ \sin A + 2\sin3A + \sin5A }{ \sin3A + 2\sin5A + \sin7A } = \frac{\sin3A}{\sin5A} \]
Solution
L.H.S.
\[ = \frac{ \sin A + 2\sin3A + \sin5A }{ \sin3A + 2\sin5A + \sin7A } \]Group first and third terms and use identity:
\[ \sin C+\sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2} \]
\[
=
\frac{
2\sin3A\cos2A+2\sin3A
}{
2\sin5A+2\sin5A\cos2A
}
\]
Take common factor:
\[ = \frac{ 2\sin3A(\cos2A+1) }{ 2\sin5A(\cos2A+1) } \]Cancel common factors:
\[ = \frac{\sin3A}{\sin5A} \]Hence Proved.