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eix = cos x + i sin x
By Alexander (p195) on April 14, 1998:

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

Alexander (p195)

By Rup on April 14, 1998:

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):

e(ix) = S [ (ix)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:
= S [ (ix)2m / (2m)! + (ix)2m+1) / (2m+1)! ]
= S [ (-1)m x2m / (2m)! ] + i S [ (-1)m x(2m+1) / (2m+1)! ]
= cos x + i sin x

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.

[Editor: The following alternative suggestion was made by Brice Yokem. Define f(x)=eix and g(x)=cos(x)+isin(x) then f'(x)=if(x) and g'(x)=ig(x). That is, the complex differentiable functions f and g satisfy the same differential equation. It is easily verified that f(0)=g(0) (initial conditions) therefore the two functions f and g must be equal.]