QPaperGen

Step 1: Expand the expression
\begin{align*}
&(\sin 3x + \sin x)\sin x + (\cos 3x - \cos x)\cos x \\
&= \sin 3x \sin x + \sin^2 x + \cos 3x \cos x - \cos^2 x
\end{align*}

Step 2: Group terms
\[
= (\sin 3x \sin x + \cos 3x \cos x) + (\sin^2 x - \cos^2 x)
\]

Step 3: Use identity
\[
\cos(A - B) = \cos A \cos B + \sin A \sin B
\]
\[
\Rightarrow \sin 3x \sin x + \cos 3x \cos x = \cos(3x - x) = \cos 2x
\]

Also,
\[
\sin^2 x - \cos^2 x = -\cos 2x
\]

Step 4: Combine
\[
\cos 2x + (-\cos 2x) = 0
\]

Hence Proved:
\[
(\sin 3x + \sin x)\sin x + (\cos 3x - \cos x)\cos x = 0
\]