Prove that \[ (\sin3x+\sin x)\sin x+(\cos3x-\cos x)\cos x=0 \]
Proof:
\[
LHS=(\sin3x+\sin x)\sin x+(\cos3x-\cos x)\cos x
\]
Expanding:
\[
=\sin3x\sin x+\sin^2x+\cos3x\cos x-\cos^2x
\]
Grouping terms:
\[
=(\sin3x\sin x+\cos3x\cos x)
+(\sin^2x-\cos^2x)
\]
Using the identity
\[
\cos(A-B)=\cos A\cos B+\sin A\sin B
\]
we get
\[
\sin3x\sin x+\cos3x\cos x
=
\cos(3x-x)
\]
\[
=\cos2x
\]
Also,
\[
\sin^2x-\cos^2x
=
-(\cos^2x-\sin^2x)
\]
Using
\[
\cos2x=\cos^2x-\sin^2x
\]
therefore,
\[
\sin^2x-\cos^2x=-\cos2x
\]
Substituting:
\[
LHS=\cos2x-\cos2x
\]
\[
=0
\]
Hence proved,
\[
\boxed{
(\sin3x+\sin x)\sin x+(\cos3x-\cos x)\cos x=0
}
\]