Uniqueness of $\sin x$ and $\cos x$? (Putnam Exam problem) I'm struggling with this former Putnam Exam problem:
Suppose $f$ and $g$ are nonconstant, differentiable, real-valued functions on $R$.  Furthermore, suppose that for each pair of real numbers $x$ and $y$, $f(x + y) = f(x)f(y) - g(x)g(y)$ and $g(x + y) = f(x)g(y) + g(x)f(y)$.  If $f'(0) = 0$, prove that $(f(x))^2 + (g(x))^2 = 1$ for all $x$.
Right.  So obviously, $f(x) = \cos x$ and $g(x) = \sin x$ satisfy the conditions and also the conclusion of the problem.  But are these the unique such functions, and if so, how to prove it?  And if not, then how to prove the conclusion otherwise?
 A: Since KennyTM posted a very elegant proof I will post an uncreative one that requires no thought to produce.
First write out all the obvious identities that come from the information told:


*

*$f(x + y) = f(x)f(y) - g(x)g(y)$

*$f(x) = f(x)f(0) - g(x)g(0)$

*$f(0) = f(0)f(0) - g(0)g(0)$

*$g(x + y) = f(x)g(y) + g(x)f(y)$

*$g(x) = f(x)g(0) + g(x)f(0)$

*$g(0) = f(0)g(0) + g(0)f(0)$

*$Df(x + c) = Df(x)f(c) - Dg(x)g(c)$

*$Dg(x + c) = Df(x)g(c) + Dg(x)f(c)$

*$Df(0) = 0 = Df(0)f(0) - Dg(0)g(0) = -Dg(0)g(0)$

*$Dg(0) = Df(0)g(0) + Dg(0)f(0) = Dg(0)f(0)$
The $Df(0)$ equation tells us that $Dg(0) = 0 \lor g(0) = 0$ Since we don't want $Dg(0) = 0$ lets discharge that automatically now by proving it contradictory: It would imply that $Dg(0 + c) = Df(0)g(c) + Dg(0)f(c) = 0$ so $g(x)$ would be a constant but that is not allowed.
We have seen that $g(0) = 0$ and now we update the identifies with this information, and that tells us is that $f(0) = 1$. We cannot determine the value of $Dg(0)$ so just call it some constant $k$. Now it is easy to recognize that we have a simple ODE problem that could be solved symbolically by a calculator or by taking eigenvalues/vectors of the matrix that defines the system. Giving the solutions $\sin(kx)$ & $\cos(kx)$.
A: Here is a different way.
We have
$f(x+y) = f(x) f(y) - g(x) g(y)$
Differentiate wrt $y$
$f'(x+y)  = f(x)f'(y) - g(x) g'(y)$ , put $y = 0$.
$f'(x) = -g(x) g'(0)$
Similarly we get
$g'(x) = f(x) g'(0)$
Thus $f(x)f'(x) + g(x)g'(x) = 0$
Thus the function $f^{2}(x) + g^{2}(x)$ is constant, as it's derivative is zero.
Now
$f(0) = f^2(0) - g^{2}(0)$
and
$g(0) = 2f(0)g(0)$
Squaring and adding both we get
$f^{2}(0) + g^{2}(0) = (f^{2}(0) + g^{2}(0))^2$
Now if $f^{2}(0) + g^{2}(0) = 0$ then because $f^{2}(x) + g^{2}(x)$ is a constant, we get $f(x) = 0$ which implies $f$ is constant.
Thus $f^{2}(0) + g^{2}(0) = 1$ and hence $f^{2}(x) + g^{2}(x) = 1$
A: Let $u(x) = f(x)^2 + g(x)^2$. It should be easy to show that $u(x+y) = u(x)u(y)$, thus $u$ must be an exponential function $u(x) = e^{Bx}$.
Now, as $f(2x) = f(x)^2 - g(x)^2$, we should get $0 = g(0) g'(0)$. Comparing this with $u'(0)$ we should get $B = 0$, thus $f(x)^2 + g(x)^2 = 1$.
