Question
Mathematics Question on Trigonometric Functions of Sum and Difference of Two Angles
Prove that (sin 3x+sin x) sinx+(cos 3x–cos x) cos x=0
Answer
L.H.S.=(sin 3x+sin x) sin x + (cos 3x-cos x) cos x
=sin3x sinx+sin2x+cos3x cosx-cos2x.
=cos3x cosx+sin3x sinx-(cos2x-sin2x).
= cos(3x-x)-cos2x [cos(A-B)=cos A cos B+sin A sin B]
=cos 2x-cos 2x
=0
=R.H.S.