# Defining sine and cosine

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.

• If not, what if we threw $\lim_{x\to0}\frac{\cos x-1}x=1$ into the mix? – Akiva Weinberger Dec 23 '14 at 3:37
• I think you mean $\lim_{x \rightarrow 0} \frac{\cos x - 1}{x} = 0$ :-) – Bungo Dec 23 '14 at 3:42
• @Bungo close enough – Akiva Weinberger Dec 23 '14 at 3:43
• @Bungo It just makes the question all the more interesting if we throw a different definition in. – Milo Brandt Dec 23 '14 at 3:43
• What about $s(x) = \sinh x$? – MathMajor Dec 23 '14 at 3:58

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.

• Nice! Thank you! (Interesting: my counter example—the one that made me edit my post—corresponds to $f(x)=e^{(1+i)x}$.) – Akiva Weinberger Dec 23 '14 at 4:18
• @columbus Yeah, I noticed that; when you posted the comment with that solution on your question, I'd written the first two paragraphs basically the same as above and had been unsuccessfully trying to prove that $\cos'(0)=0$. I probably would have been at that forever without every actually checking for extraneous solutions if you hadn't commented. – Milo Brandt Dec 23 '14 at 4:20
• This may be a stupid question, but what if I replaced the fourth, "new" bullet-point with $\sin(-x)=-\sin(x),\cos(-x)=\cos(x)$? EDIT: Given the other bullet-points, is that equivalent to $\sin^2(x)+\cos^2(x)=1$? I'm too tired to check. – Akiva Weinberger Dec 23 '14 at 4:20
• Well, you'd still have that $f$ is exponential, and thus that $$f(-x)=\frac{1}{f(x)}$$ However, your condition forces $|f(-x)|=|f(x)|$ which, for them to be reciprocals, implies that $|f(x)|=1$. Therefore, the solutions for $f$ are of the form $f(x)=e^{\alpha i x}$ for $\alpha$ with real part $0$, since $f$ is restricted to the unit circle. (Of note is the solution $f(x)=1$). If you keep that $\sin'(0)=1$, then yes, that would also uniquely define $f$. – Milo Brandt Dec 23 '14 at 4:24

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.