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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.

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(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.

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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.

If you add the conditions $\lim_{x \to 0}\frac{\cos(x) – 1}{x}=0$ and $c(x),s(x)$ are continuously differentiable (i.e. $c,s \in C^1(\mathbb{R})$, we can then force $c(x) = \cos(x)$ and $s(x) = \sin(x)$.

Writing $f(x) = c(x)+i s(x)$, the first two relations become simply: $f(x+y) = f(x)f(y)$ (comparing real and imaginary parts on both sides). In particular, note that $f(0)=f(0)^2$, so $f(0)=0$ or $f(0)=1$. If $f(0)=0$, then $f(x)=f(x+0)=f(x)f(0)=0$, so $f(x)=0$ identically, contradicting $\lim_{x \to 0} \frac{\sin(x)}{x} = 0$. So $f(0)=1$.

Taking the derivative with respect to $x$ gives $f'(x+y)=f'(x)f(y)$ and so:

$$

f'(y)=f'(0)f(y)

$$

Which is an ODE whose solution is uniquely determined by the values $f(0), f'(0)$.

We showed above that $f(0)=1$.

And

$$

f'(0) = \lim_{x \to 0} \frac{f(x)-f(0)}{x} = \lim_{x \to 0} \frac{c(x)-1 + is(x)}{x} = i

$$

We know (by construction) that $f(0)$ and $f'(0)$ match what they would be if $c(x)=\cos(x)$ and $s(x) = \sin(x)$, and since $f(0$ and $f'(0)$ uniquely determine $f(x)$ (from the ODE), this completes the proof.

Note that we do need to know $\lim_{x \to 0}\frac{\cos(x)-1}{x}$ to determine $f'(0)$, which is why without that condition you get the counterexamples mentioned in the comments.

I’m not sure if there are any weird counterexamples if you omit the differentiability condition, but that assumption seems like a natural one to make.

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