Question 1067660
<pre><b><font size=5>
4cos(2&#952;)sin(4&#952;)

We notice that this has a product of a sine and a cosine
and we remember that such a product appears in the formulas: 

sin(A+B) = sin(A)cos(B) + cos(A)sin(B)
sin(A-B) = sin(A)cos(B) - cos(A)sin(B)

So we substitute A=4&#952; and B=2&#952; in those

sin(4&#952;+2&#952;) = sin(4&#952;)cos(2&#952;) + cos(4&#952;)sin(2&#952;)
sin(4&#952;-2&#952;) = sin(4&#952;)cos(2&#952;) - cos(4&#952;)sin(2&#952;)

or

sin(6&#952;) = sin(4&#952;)cos(2&#952;) + cos(4&#952;)sin(2&#952;)
sin(2&#952;) = sin(4&#952;)cos(2&#952;) - cos(4&#952;)sin(2&#952;)

If we add those equations together, equal to equals,
the last terms cancel and we get

sin(6&#952;) + sin(2&#952;) = 2sin(4&#952;)cos(2&#952;)

So our problem

4cos(2&#952;)sin(4&#952;) = 2[2sin(4&#952;)cos(2&#952;)] = 2[sin(6&#952;) + sin(2&#952;)]

or

2sin(6&#952;) + 2sin(2&#952;)

Edwin</pre></font><b>