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Question

Mathematics Question on Trigonometric Functions of Sum and Difference of Two Angles

Prove that. sin(n+1) xsin(n+2)x+cos(n+1) xcos(n+2)x=cosx.

Answer

L.H.S. = sin (n+1)x sin(n+2)x + cos (n+1)x cos(n+2)x

=12[2 sin(n+1)x sin(n+2)x+cos(n+1)x cos(n+2)x]=\frac{1}{2}[2\,sin (n+1)x\,sin(n+2)x + cos (n+1)x\,cos(n+2)x]

=\frac{1}{2}\begin{bmatrix} cos\left \\{ (n+1)x-(n+2)x \right \\}-cos\left \\{ (n+1)x+(n+2)x \right \\} \\\ +cos\left \\{ (n+1)x+(n+2)x \right \\}+cos\left \\{ (n+1)x-(n+2)x \right \\} \end{bmatrix}

[−2sinAsinB−cos(A−B)][-2sinA sinB-cos(A-B)]

[2 cos A cosB=cos(A+B)+cos(A−B)][2\,cos\,A\,cosB=cos(A+B)+cos(A-B)]

=12×2 cos(n+1)x−(n+2)x=\frac{1}{2}×\\{2\,cos{(n+1)x-(n+2)x}\\}

=cos(−x)=cos x=R.H.S.=cos(-x)=cos\,x=R.H.S.