cos(3x)sin(11x)
That's a product of a cosine and a sine. Usually sines
come before cosines, so let's reverse the two factors:
Let's rewrite it as
sin(11x)cos(3x)
We think through the formulas for AħB
sin(AħB) = sin(A)cos(B) ħ cos(A)sin(B)
cos(AħB) = cos(A)cos(B) ∓ sin(A)sin(B)
and see that the first formula contains such a product,
sin(A)cos(B),
so we let A=11x and B=3x, then
sin(AħB) = sin(A)cos(B) ħ cos(A)sin(B)
becomes the two equations:
sin(11x+3x) = sin(11x)cos(3x) + cos(11x)sin(3x)
sin(11x-3x) = sin(11x)cos(3x) - cos(11x)sin(3x)
or
sin(14x) = sin(11x)cos(3x) + cos(11x)sin(3x)
sin(8x) = sin(11x)cos(3x) - cos(11x)sin(3x)
Addinng those equations term by term:
sin(14x) + sin(8x) = 2sin(11x)cos(3x)
Solving for sin(11x)cos(3x) by multiplying both sides by [sin(14x) + sin(8x)] = sin(11x)cos(3x)
sin(14x) + sin(8x)] = sin(11x)cos(3x)
Edwin