Proving Trig Identity: 1-(cos(x)+sin(x))(cos(x)-sin(x))=2sin^2(x)

[KevinMWHM]

prove
1-(cos(x)+sin(x))(cos(x)-sin(x))=2sin^2(x)

foil out the center
I get
1-cos^2(x)-cos(x)sin(x)+cos(x)sin(x)+sin^2(x)

the -cos(x)sin(x)+cos(x)sin(x) cancels to 0 leaving

1-cos^2(x)-sin^2(x)

then I'm lost...

I know I can switch 1-cos^2(x) to sin^2(x) but that doesn't help because I get
sin^2(x)-sin^2(x)=0

where am i going wrong?

 

[QuarkCharmer]

"1-cos^2(x)-cos(x)sin(x)+cos(x)sin(x)+sin^2(x)

the -cos(x)sin(x)+cos(x)sin(x) cancels to 0 leaving

1-cos^2(x)-sin^2(x)"

Are you sure? Looks to me like the +cos(x)sin(x) and the -cos(x)sin(x) cancel out to leave:
1-cos^2(x)+sin^2(x)

 

[KevinMWHM]

I wrote my foil wrong
foiling leaves me
1-cos^2(x)-cos(x)sin(x)+cos(x)sin(x) - sin^2(x)

 

[Vonalak]

Don't remove the parentheses until after you've foiled it out or you're going to lose a negative sign.

Foiling 1-[(cos(x)+sin(x))(cos(x)-sin(x))], we get
1-(cos^2(x)-cos(x)sin(x)+cos(x)sin(x)-sin^2(x)). Two of the terms cancel, yielding:
1-(cos^2(x)-sin^2(x))
=1-cos^2(x)+sin^2(x)

Now try using those Pythagorean identities.