Hi, thanks CPhill and guest.
I am just going to look at this bit
sin^2x = [1 - cos(2x)] / 2 = 1/2 - cos(2x)/2
\(sin^2x =\frac{ 1 - cos(2x)}{2} \)
I am hopeless (also stuborn) aboul learning formulas. I just remember what I feel i have to remember and work the others out each time.
So I do not remember this one :/
\(Cos(A+B)=cosAcosB-sinAsinB\qquad \mbox{I have committed this to memory}\\ cos(2x)=cos^2x-sin^2x\\ cos(2x)=1-sin^2x\;\;\;\;\;-sin^2x\\ cos(2x)=1-2sin^2x\\ cos(2x)-1=-2sin^2x\\ \frac{cos(2x)-1}{-2}=sin^2x\\ sin^2x=\frac{1-cos(2x)}{2}\\\)
If I need sin^2(2x) or cos^2(x) etc I got through a similar procedure.
It doesn't take me long because I have had a LOT of practice :)