TITLE: Uniqueness of $\sin x$ and $\cos x$? (Putnam Exam problem)
QUESTION [7 upvotes]: 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?

REPLY [11 votes]: 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 $f^{2}(x) + g^{2}(x) = 1$