Derive trigonometric identity?

Consider cos(x+(-x)) = cos (x-x) = cos 0 = 1.

Using your formula, cos (x+(-x)) = cos(x)cos(-x) - sin(x)sin(-x).

Since cos(x) is an even function, cos(-x) = cos(x).
Since sin(x) is an odd function, sin(-x) = - sin(x).

Thus you have cos (x-x) = cos(x)cos(x) + sin(x)sin(x)
cos (0) = 1 = cos^2(x) + sin^2(x).

cos(x+y)= cosx cosy-sinx siny.
Put y=-x;
cos 0=cosx cosx - sinx [-sinx]
  1= cos^2x+sin^2x. QED.

First, before I begin working on the approach suggested,

Pick any point on a circle inscribed about the origin of the X-Y plane.

By Pythagoras we know that the square of the radius of the circle is equal to the sum of the square of the x coordinate of the point and the square of the y coordinate. or
(x/r)² + (y/r)²  = 1r² = x² + y²
if you multiply both sides by 1/r²  you get
r² /r² = x²'/r/²  + y²/r²
or
(x/r)² + (y/r)²  = 1

Measuring the angle (?) from the positive x axis,
The trigonometric functions sine and cosine are defined as
y/r and x/r respectively, so
sin(?)=y/r, and
cos(?)=x/r
Plugging these back into
(x/r)² + (y/r)²  = 1, we get...
cos²(?) + sin²(?)  = 1

That's the proof with which I'm familiar.

And the derivation of the sin(a+b) and cos(a+b) functions are geometric and impossible to do on this site.

That said, I'll spend a little time working on the problem you have presented.

Ok, this is as far as I can get it.
cox(x+y) = cosxcosy - sinxsiny
If x=y then you have
cox(x+y)=cos(2x) = cosxcos x- sinxsinx=cos²(x) - sin²(x)
But, cos(2x) is not necessarily = 1.

Give the points to Derek...
He's right.;.. and first...
I hadn't thought of that.