Prove that: \[ \sin 3A + \sin 2A – \sin A = 4\sin A \cos\frac{A}{2}\cos\frac{3A}{2} \]
Solution
Group the first and third terms:
\[
(\sin 3A – \sin A) + \sin 2A
\]
Using the identity:
\[
\sin A – \sin B
=
2\cos\frac{A+B}{2}\sin\frac{A-B}{2}
\]
For
\(\sin 3A – \sin A\):
\[
\sin 3A – \sin A
=
2\cos\frac{3A+A}{2}
\sin\frac{3A-A}{2}
\]
\[
=
2\cos 2A \sin A
\]
Therefore,
\[
\sin 3A + \sin 2A – \sin A
=
2\sin A \cos 2A + \sin 2A
\]
Using:
\[
\sin 2A = 2\sin A \cos A
\]
\[
=
2\sin A \cos 2A + 2\sin A \cos A
\]
\[
=
2\sin A(\cos 2A + \cos A)
\]
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,
\[
=
2\sin A
\times
2\cos\frac{3A}{2}\cos\frac{A}{2}
\]
\[
=
4\sin A \cos\frac{A}{2}\cos\frac{3A}{2}
\]
Hence,
\[
\boxed{
\sin 3A + \sin 2A – \sin A
=
4\sin A \cos\frac{A}{2}\cos\frac{3A}{2}
}
\]