Prove that \[ \sin3x+\sin2x-\sin x = 4\sin x\cos\frac{x}{2}\cos\frac{3x}{2} \]
Proof:
Start with the left-hand side:
\[
LHS=\sin3x+\sin2x-\sin x
\]
Group the last two terms:
\[
=\sin3x+(\sin2x-\sin x)
\]
Using the identity
\[
\sin A-\sin B
=
2\cos\frac{A+B}{2}\sin\frac{A-B}{2}
\]
with
\[
A=2x,\quad B=x
\]
we get
\[
\sin2x-\sin x
=
2\cos\frac{3x}{2}\sin\frac{x}{2}
\]
Therefore,
\[
LHS
=
\sin3x
+
2\cos\frac{3x}{2}\sin\frac{x}{2}
\]
Now use
\[
\sin3x
=
2\sin\frac{3x}{2}\cos\frac{3x}{2}
\]
Thus,
\[
LHS
=
2\cos\frac{3x}{2}
\left(
\sin\frac{3x}{2}
+
\sin\frac{x}{2}
\right)
\]
Using the identity
\[
\sin A+\sin B
=
2\sin\frac{A+B}{2}\cos\frac{A-B}{2}
\]
with
\[
A=\frac{3x}{2},\quad B=\frac{x}{2}
\]
we get
\[
\sin\frac{3x}{2}
+
\sin\frac{x}{2}
=
2\sin x\cos\frac{x}{2}
\]
Substituting:
\[
LHS
=
2\cos\frac{3x}{2}
\cdot
2\sin x\cos\frac{x}{2}
\]
\[
=
4\sin x\cos\frac{x}{2}\cos\frac{3x}{2}
\]
Hence proved,
\[
\boxed{
\sin3x+\sin2x-\sin x
=
4\sin x\cos\frac{x}{2}\cos\frac{3x}{2}
}
\]