Prove the identity: cos(x+x) = cos^2(x) - sin^2(x)?

I tried to do it with minimal drawings and it is hard to describe. This website has it in detail (with drawings): http://mathforum.org/library/drmath/view/54051.htm...

Nah, it doesn't need to be as hard as Ramzi made it.

It's just an application of the formula for the cosine of the sum of two angles.

cos(a + b) = cosa cosb - sina sinb

cos(x + x) = cosx cosx - sinx sinx = cos^2x - sin^2x

It's that easy.

----

Edit: I've never tried to prove the formula itself, but I found a derivation here:

http://www.phy6.org/stargaze/Strig5.htm

http://whyslopes.com/etc/CalculusAndBeyond/ch23a.h...

Are you proving from cos x = (e^{ix}+e^{-ix})/2, or power series? Not sure what level proof you need here.

cos(x+x)=cosx*cosx-sinx*sinx

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

it comes from cos(a+b)=cosa*cosb-sina*sinb

solved!!!

cos^2(x) = .5 + cos(2x)/2

sin^2(x) = .5 - cos(2x)/2

cos(x+x) = cos(2x)

cos^2(x) - sin^2(x) = 5 + cos(2x)/2 - (.5 - cos(2x)/2) = 2cos(2x)/2 = cos(2x)

cos(x+x)=cos(x)cos(x) - sin(x)sin(x) from the sum formula of the cosine:

cos(u+v)=cos(u)cos(v)-sin(u)sin(v)