We know the following are true about sine and cosine (and that they can be proven geometrically):
- $\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$
- $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$
- $\lim\limits_{x\to0}\dfrac{\sin x}x=1$
- $\lim\limits_{x\to0}\dfrac{\cos x-1}x=0$
- They are continuous
Let's say we have two real functions: $s(x)$ and $c(x)$. If we know that the above are true for $s$ and $c$ (i.e. $s(a+b)=s(a)c(b)+s(b)c(a)$, etc.), can we conclude that $s$ and $c$ are equal to $\sin$ and $\cos$ respectively? In other words, are sine and cosine the only two functions that satisfy the above? Do the five points above uniquely define the sine and cosine?
I was thinking of the unit circle definition of sine and cosine, and I knew that there are many non-geometric definitions of them. I was wondering if the four facts shown above were enough to count as a non-geometric definition.
(Without the third point, stuff like $\sin(x \text{ degrees})$ and $\cos(x \text{ degrees})$ would also work; in other words, the third point specifies that we're using radians.)
EDIT: Added fourth point, since $s(x)=e^x\sin(x)$, $c(x)=e^x\cos(x)$ would work if it was omitted.
Best Answer
Yes, this uniquely defines the sine and cosine functions. In particular, let's write $$f(x)=\cos(x)+i\sin(x)$$ where $i$ is the imaginary unit. Then, the sum identities will yield, after a bit of computation that $f(x)f(y)=f(x+y)$. This is somewhat tedious, but easy to verify. But guess what! The only continuous functions which can satisfy $f(x)f(y) = f(x+y)$ are exponential functions; to prove this, notice that, for integer $n$, it is clear that $f(nx)=f(x)^n$. You can use this to show that, over the rationals, $f$ is an exponential function (i.e. $f(x)=e^{ax}$).*
Since we have $\sin(0)=0$, $\sin'(0)=1$, $\cos(0)=1$ and $\cos'(0)=0$, this implies that $f'(0)=i$. The only exponential function satisfying this is $f(x)=e^{ix}$. Extracting real and imaginary parts yields, uniquely, cosine and sine.
*If you wish to be formal about this, it would be wise to prove that $|f(x)|=e^{\alpha x}$ first, then prove that $\arg(f(x))\equiv \beta x$ - you can avoid the issue of $n^{th}$ roots being non-unique in the complex plane by separating your argument into these two sections.