I am stuck on the 3rd line of example 4 in the following question?

http://www.math24.net/fourier-series-of-functions-with-arbitrary-period.html

Specifically the part where: [cos2xcos2nx] to 1/2[cos(2n - 2)x + cos(2n + 2)x]

I realise this is something simple but I just cant get it :(

Question

I am stuck on the 3rd line of example 4 in the following question?

http://www.math24.net/fourier-series-of-functions-with-arbitrary-period.html

Specifically the part where: [cos2xcos2nx] to 1/2[cos(2n - 2)x + cos(2n + 2)x] 

I realise this is something simple but I just cant get it :(

anonymous · Answer

LHS= cos(2x)*cos(2nx)=(1/2)(2*cos(2x)*cos(2nx))

now cos(A+B)+cos(A-B)=2*cos(A)*cos(B)

Hope that helps

anonymous · Answer

Now that was stupid of me.....thanks for that :)