Prove that: \[ \sin A + \sin 2A + \sin 4A + \sin 5A = 4\cos\frac{A}{2}\cos\frac{3A}{2}\sin 3A \]
Solution
Group the terms:
\[
(\sin A + \sin 5A)
+
(\sin 2A + \sin 4A)
\]
Using the identity:
\[
\sin A + \sin B
=
2\sin\frac{A+B}{2}\cos\frac{A-B}{2}
\]
For
\(\sin A + \sin 5A\):
\[
\sin A + \sin 5A
=
2\sin\frac{A+5A}{2}
\cos\frac{A-5A}{2}
\]
\[
=
2\sin 3A \cos(-2A)
\]
Since,
\[
\cos(-\theta)=\cos\theta
\]
\[
=
2\sin 3A \cos 2A
\]
For
\(\sin 2A + \sin 4A\):
\[
\sin 2A + \sin 4A
=
2\sin\frac{2A+4A}{2}
\cos\frac{2A-4A}{2}
\]
\[
=
2\sin 3A \cos(-A)
\]
\[
=
2\sin 3A \cos A
\]
Adding both results:
\[
=
2\sin 3A \cos 2A
+
2\sin 3A \cos A
\]
\[
=
2\sin 3A(\cos 2A+\cos A)
\]
Again using:
\[
\cos A + \cos B
=
2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
\]
\[
\cos 2A+\cos A
=
2\cos\frac{2A+A}{2}
\cos\frac{2A-A}{2}
\]
\[
=
2\cos\frac{3A}{2}\cos\frac{A}{2}
\]
Therefore,
\[
=
2\sin 3A
\times
2\cos\frac{3A}{2}\cos\frac{A}{2}
\]
\[
=
4\cos\frac{A}{2}\cos\frac{3A}{2}\sin 3A
\]
Hence,
\[
\boxed{
\sin A + \sin 2A + \sin 4A + \sin 5A
=
4\cos\frac{A}{2}\cos\frac{3A}{2}\sin 3A
}
\]