Prove that: (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0
Prove that: (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0

Consider

 

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

By additional computation

 

\[=\text{ }sin\text{ }3x\text{ }sin\text{ }x\text{ }+\text{ }sin2\text{ }x\text{ }+\text{ }cos\text{ }3x\text{ }cos\text{ }x\text{ }\text{ }cos2\text{ }x\]

Taking out the normal terms

 

\[=\text{ }cos\text{ }3x\text{ }cos\text{ }x\text{ }+\text{ }sin\text{ }3x\text{ }sin\text{ }x\text{ }\text{ }\left( cos2\text{ }x\text{ }\text{ }sin2\text{ }x \right)\]

By formula,

 

\[cos\text{ }\left( A\text{ }\text{ }B \right)\text{ }=\text{ }cos\text{ }A\text{ }cos\text{ }B\text{ }+\text{ }sin\text{ }A\text{ }sin\text{ }B\]

\[=\text{ }cos\text{ }\left( 3x\text{ }\text{ }x \right)\text{ }\text{ }cos\text{ }2x\]

So we get

 

\[=\text{ }cos\text{ }2x\text{ }\text{ }cos\text{ }2x\]

\[=\text{ }0\]

\[=\text{ }RHS\]