cos(2x)-cos(10x) = tan(4x)(sin(2x)+sin(10x))

Question

Prove that the left-hand side equals the right-hand side.

Answer 1

cos(A+B)=cosAcosB-sinAsinB

cos(A-B)=cosAcosB+sinAsinB

cos(A+B)+cos(A-B)=2cosAcosB

cos(A+B)-cos(A-B)=-2sinAsinB

If A+B=X and A-B=Y, then A=(X+Y)/2 and B=(X-Y)/2.

If X=2x and Y=10x, then A=6x and B=-4x

cos(2x)-cos(10x)=-2sin(6x)sin(-4x)=2sin(6x)sin(4x).

Similarly, sin(A+B)+sin(A-B)=2sin((A+B)/2)cos((A-B)/2) and sin(2x)+sin(10x)=2sin(6x)cos(-4x)=2sin(6x)cos(4x).

Therefore, (cos(2x)-cos(10x))/(sin(2x)+sin(10x))=2sin(6x)sin(4x)/2sin(6x)cos(4x)=tan(4x)

and cos(2x)-cos(10x)=tan(4x)(sin(2x)+sin(10x)) QED.