QPaperGen

To Prove:
\[
\sin 3x + \sin 2x - \sin x = 4\sin x \cos\left(\frac{x}{2}\right) \cos\left(\frac{3x}{2}\right)
\]

Step 1: Use sum and difference identities

\[
\sin A + \sin B = 2\sin\left(\frac{A + B}{2}\right)\cos\left(\frac{A - B}{2}\right)
\]
\[
\sin A - \sin B = 2\sin\left(\frac{A - B}{2}\right)\cos\left(\frac{A + B}{2}\right)
\]

Step 2: Group and simplify
\[
\text{LHS} = \sin 3x + \sin 2x - \sin x
\]

Group \(\sin 2x - \sin x\):
\[
= \sin 3x + 2\sin\left(\frac{2x - x}{2}\right)\cos\left(\frac{2x + x}{2}\right)
= \sin 3x + 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{3x}{2}\right)
\]

Now express \(\sin 3x\) using the identity:
\[
\sin 3x = 2\sin\left(\frac{3x}{2}\right)\cos\left(\frac{3x}{2}\right)
\]

So:
\[
\text{LHS} = 2\sin\left(\frac{3x}{2}\right)\cos\left(\frac{3x}{2}\right) + 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{3x}{2}\right)
\]

Factor out \(2\cos\left(\frac{3x}{2}\right)\):
\[
= 2\cos\left(\frac{3x}{2}\right)\left[\sin\left(\frac{3x}{2}\right) + \sin\left(\frac{x}{2}\right)\right]
\]

Use the identity again:
\[
\sin A + \sin B = 2\sin\left(\frac{A + B}{2}\right)\cos\left(\frac{A - B}{2}\right)
\]

Apply to the bracket:
\[
= 2\cos\left(\frac{3x}{2}\right) \cdot 2\sin x \cos\left(\frac{x}{2}\right)
= 4\sin x \cos\left(\frac{x}{2}\right) \cos\left(\frac{3x}{2}\right)
\]
Therefore,
\[
\sin 3x + \sin 2x - \sin x = 4\sin x \cos\left(\frac{x}{2}\right) \cos\left(\frac{3x}{2}\right) \qquad \textbf{Hence Proved.}
\]