http://globalwarming-facts.info/50-tips.html

I want to share this with you. The derivation of one of my favourite identites in mathematics so far:
Let f(x) = cos x = a + bx + cx^2 + dx^3 + ex^4 + fx^5 + ...
(don't mix up e in ex^4 with e = 2.7182821828...; in this case e is just any constant)
so
f(0) = cos 0 = 1 = a, thus a = 1
f'(x) = -sinx = b + 2cx + 3dx^2 + 4ex^3 + 5fx^4 + ...
f'(0) = -sin0 = 0 = b; thus b = 0
f''(x) = -cosx = (2*1)c + (3*2)d + (4*3)ex^2 + (5*4)fx^3 + ...
f''(0) = -cos0 = -1 = (2*1)c; thus c = -1/(2*1) = -1/2!
f'''(x) = sinx = (3*2*1)d + (4*3*2)ex + (5*4*3)fx^2 + ...
f'''(0) = sin0 = 0 = (3*2*1)d; thus d = 0
f''''(x) = cosx = (4*3*2*1)e + (5*4*3*2)fx + ...
f''''(0) = cos0 = 1 = (4*3*2*1)e; thus e = 1/4!
...
and so on.
Therefore substuting the constants into the original polynomial equation,
(i) cos x = 1 - x^2/2! + x^4/4! - x^6/6! + x^8/8! + ...
You can check this by substitution a value in x, such as 1 (radians).
Similarly, you can work out an expansion for
(ii) sin x = x - x^3/3! + x^5/5! - x^7/7! + x^9/9! + ...
(iii) e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + x^5/5! + ...
Now using expansion (iii), you can expand e^(ix), where i = √(-1)
(iv) e^(ix) = 1 + ix + (ix)^2/2! + (ix)^3/3! + (ix)^4/4! + (ix)^5/5! + ...
= 1 + ix - x^2/2! - i(x)^3/3! + x^4/4! + i(x)^5/5! + ...
Separating the imaginary parts and real parts,
e^(ix) = (1 - x^2/2! + x^4/4! - x^6/6! + ...) + i(x - x^3/3! + x^5/5! - x^7/7! + ...)
= cos x + isin x
Wow!
so you get Euler's identity