To express the sum of sines as the product of sines and cosines, we start with deriving a trig identity.
If A and B are angles,
sin(A+B)≡sinAcosB+cosAsinB and
sin(A-B)≡sinAcosB-cosAsinB.
Adding these two equations: sin(A+B)+sin(A-B)=2sinAcosB.
If we write X=A+B and Y=A-B, A=(X+Y)/2 and B=(X-Y)/2. We can apply this identity to the given sum of sines.
A=(6x+2x)/2=4x, B=(6x-2x)/2=2x, so sin(6x)+sin(2x)=2sin(4x)cos(2x).
A=(14x+10x)/2=12x, B=(14x-10x)/2=2x, so sin(14x)+sin(10x)=2sin(12x)cos(2x).
The sum of the sines becomes 2cos(2x)(sin(4x)+sin(12x)).
A=(12x+4x)/2=8x, B=(12x-4x)/2=4x, so the sum of sines becomes (2cos(2x))(2sin(8x)cos(4x)).
This simplifies to 4cos(2x)cos(4x)sin(8x).
We can take this further by expanding sin(8x)=2sin(4x)cos(4x), therefore the sum becomes:
8cos(2x)cos2(4x)sin(4x)=16cos2(2x)cos2(4x)sin(2x)=32sin(x)cos(x)cos2(2x)cos2(4x).