How to prove that the CIS equasion is right -
e^(i*x) =cos x - isin x

Alexander (p195)

I've never heard it called the CIS equation, but I suppose that makes sense; I always thought of it as a theorem -- probably Euler's. And I'm afraid that you've got a sign error... it's

e^(ix) = cos x + i sin x

The easiest way to see this is from the Maclaurin expansions of exponential, cos and sin;
e(ix) = 1 + ix + (1/2)x² + (i/6)x³ + (1/24)x⁴ +...
cos x = 1 + (1/2)x² + (1/24)x⁴ +...
i sin x = ix + (i/6)x³ + +...

Writing (summing over n from 0 to infinity):

ei x=

å

[(i x)n/n!]

the next step is to break this sum into real and imaginary parts; let's make this easy on ourselves by splitting into even and odd powers by renumbering n=2m and now summing from m=0 to infinity:

=

å

[(i x)2m/(2m)!+(i x)2m+1/(2m+1)!]

=

å

[(-1)m x2m/(2m)!]+i

å

[(-1)m x2m+1/(2m+1)!]

=cosx+isinx
I don't know if this seems rigorous enough to you or seems perhaps a little vague -- I'm not really sure quite what else can be said. This is the only proof I've been presented with, and I can't find anything else in the books at hand.

Hope this helps,
Rup.