We know the following identities for cosine of the sum and difference of two variables:
#cos(theta+phi)=cos(theta)cos(phi)-sin(theta)sin(phi)#
#cos(theta-phi)=cos(theta)cos(phi)+sin(theta)sin(phi)#
We also know this pythagorean identity:
#sin^2(theta)+cos^2(theta)=1#
We start by using the sum and difference identities:
#cos(x+y)cos(x-y)=#
#=(cos(x)cos(y)-sin(x)sin(y))(cos(x)cos(y)+sin(x)sin(y))#
By the difference of squares rule, this is:
#=cos^2(x)cos^2(y)-sin^2(x)sin^2(y)#
Now we can use the pythagorean identity to convert #cos^2# into #sin^2#:
#=(1-sin^2(x))(1-sin^2(y))-sin^2(x)sin^2(y)#
If we expand the parenthtesis, we get:
#=1-sin^2(y)+cancel(sin^2(x)sin^2(y))-sin^2(x)-cancel(sin^2(x)sin^2(y))#
Now we can once again use the pythagorean identity:
#=1-sin^2(y)-(1-cos^2(x))=cancel(1)-sin^2(y)-cancel(1)+cos^2(x)#
#=cos^2(x)-sin^2(y)#, which is what we wanted to prove.