Question 770556
<pre>
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) &#8723; 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 {{{1/2}}}

{{{1/2}}}[sin(14x) + sin(8x)] = sin(11x)cos(3x)

{{{1/2}}}sin(14x) + {{{1/2}}}sin(8x)] = sin(11x)cos(3x)

Edwin</pre>