Back at the university we have proven (lot of work) that if $$S(X)C(Y)+C(X)S(Y) = S(X+Y)$$ and $$C(X)C(Y)-S(X)S(Y) = C(X+Y)$$ then $S(X)$ is $\sin(x)$ and $C(X)$ is $\cos(x)$ (or constant $0$, meh). What is this theorem called..?

Later note by someone other than the original poster:

An amazingly large number of people, in posted answers and comments (some now deleted) have MISSED THE POINT. These are not the angle addition formulas for the sine and the cosine. In those formulas, one assumes the function are the sine and cosine and shows that these equations hold. In this problem, it's the other way around: One assumes these equations hold and then proves, rather than assuming from the outset, that the functions are the sine and cosine. I even rejected an edit to the original posting that would have written $\sin$ and $\cos$ in place of $S$ and $C$. That would have made the question incomprehensible!

Please: stop doing this. --- Michael Hardy

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To potential answerers: if I have understood the question correctly, this is a partial converse to the trig addition formulas, not the trig addition formulas themselves. –  Zev Chonoles Mar 27 '12 at 1:22
@ZevChonoles, thanks for pointing this out! It's is a much more interesting question that way. –  lhf Mar 27 '12 at 1:33
Someone tried to edit this to write $\sin$ instead of $S$ and $\cos$ instead of $C$. That misses the point. The point is to prove that they are the sine and the cosine, not to assume at the outset that they are the sine and the cosine and go on from there to something else. If you're tempted to do such an edit, please don't! –  Michael Hardy Mar 27 '12 at 1:38
BTW, $C(x)$ would be $\cos(cx)$ for some constant $c$, and $S(x)$ would be $\sin(cs)$ with that same constant. –  Michael Hardy Mar 28 '12 at 1:24

Write $E(x) = C(x)+i S(x)$. Then $E(x+y)=E(x)E(y)$. This is a multiplicative variant of Cauchy's functional equation. Without further hypotheses on $S$ and $C$ it is likely that there are many solutions.

If $S$ and $C$ are assumed differentiable, then $E$ satisfies $E'=E$. If moreover $E$ is not identically zero, then $E(0)=1$ and so $E=\exp$. By Euler's formula, $S=\sin$ and $C=\cos$.

So, one answer to your question is uniqueness of the exponential function from its differential equation.

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I haven't worked through the details all the way, but it seems to me that another way to prove it might be to differentiate the equations with respect to $x$ and show thereby that $S'(x) = C(x)$ and $C'(x) = -S(x)$. This isn't quite enough to prove the theorem, because you could still have something like $S(x) = a \sin x + b\cos x$, but perhaps it could be pushed through from there. –  MJD Mar 27 '12 at 1:52
Without further hypotheses on S and C it is likely that there are many solutions. - are you sure? Isn't this where things actually get interesting? You might need to presume they are continuous or something but we spent months on this proof so it's definitely not this easy. –  chx Mar 27 '12 at 5:17
$E'$ would be some constant multiple of $E$. The constant need not be $1$. –  Michael Hardy Mar 28 '12 at 1:25

Hint $\:$ Said addition laws are true iff $\rm\: E(X) = C(X) + {\it i}\: S(X)\:$ satisfies $\rm\:E(X+Y)\: =\: E(X)\:E(Y)\:.\:$

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The Op also asked for a name of this result. Do you know one? –  Dirk Mar 27 '12 at 7:06
@Dirk I don't believe there is any widely-used name for such characterizations. –  Bill Dubuque Mar 27 '12 at 13:15
That's what I expected - I was just curious... –  Dirk Mar 27 '12 at 21:24

I found an article that seems directly relevant to the solution of the equations at: Dr. W Harold Wilson 'On Certain Related Functional Equations' AMS (read Dec 27,1917) Hope this is useful.

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