Prove that:
\[\left( {sin3x + sinx} \right)sinx + \left( {cos3x - cosx} \right)cosx = 0\]

Hint: We will start the problem by multiplying the terms and trying to simplify them. Then we arrange them to get our conventional form of \[cos\left( {A{\text{ }} - {\text{ }}B} \right){\text{ }} = {\text{ }}cosA{\text{ }}cosB{\text{ }} + {\text{ }}sinA{\text{ }}sinB\] and \[\cos 2A = {\cos ^2}A - {\sin ^2}A\], as \[A = 3x\] and \[B = x\], we will apply the formula and simplify to get our desired result.

Complete step by step Answer:

To prove: \[\left( {sin3x + sinx} \right)sinx + \left( {cos3x - cosx} \right)cosx = 0\]
Now, our left hand side is,
\[\left( {sin3x + sinx} \right)sinx + \left( {cos3x - cosx} \right)cosx\]
By multiplying we get,
\[ = \sin 3x\sin x + {\sin ^2}x + \cos 3x\cos x - {\cos ^2}x\]
On Arranging we get,
\[ = (\cos 3x\cos x + \sin 3x\sin x) - ({\cos ^2}x - {\sin ^2}x)\]
Now using, \[cos\left( {A{\text{ }} - {\text{ }}B} \right){\text{ }} = {\text{ }}cosA{\text{ }}cosB{\text{ }} + {\text{ }}sinA{\text{ }}sinB\]and \[\cos 2A = {\cos ^2}A - {\sin ^2}A\], we get,
\[ = \cos (3x - x) - \cos 2x\]
On simplification we get,
\[ = \cos 2x - \cos 2x\]
\[ = 0\]
\[ = \]R.H.S
Hence, \[\left( {sin3x + sinx} \right)sinx + \left( {cos3x - cosx} \right)cosx = 0\]

Note: In this given problem we are dealing with trigonometric quantities. The formulas we are using here are, \[cos\left( {A{\text{ }} - {\text{ }}B} \right){\text{ }} = {\text{ }}cosA{\text{ }}cosB{\text{ }} + {\text{ }}sinA{\text{ }}sinB\] and \[\cos 2A = {\cos ^2}A - {\sin ^2}A\]. First always open the brackets to simplify the given expression and then look for what pattern is it following and accordingly apply trigonometric formulas, to get the desired result.
Some other necessary trigonometric formulas are:
\[
  \cos (A + B) = \cos A\cos B - \sin A\sin B \\
  \cos (A - B) = \cos A\cos B + \sin A\sin B \\
  \sin (A + B) = \sin A\cos B + \cos A\sin B \\
  \sin (A - B) = \sin A\cos B - \cos A\sin B \\
 \]