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What are a minimal set of trig identities that can uniquely define the trig functions? I know that you can define, for example, $\sin(x)$ as the unique solution to the differential equation $f''(x) = -f(x), f(0)=0, f'(0)=1$, but I'd like to avoid analytical definitions to as much of a degree as possible (though this definition is interesting in that it nowhere explicitly mentions $\pi$).

Obviously, if we define one trig function, then we can define the rest in terms of it (like $\cos(x) = \sin(\frac{\pi}{2} - x)$). I am just curious if we can give some clever set of trig identities (like half angle formulas or the Pythagorean identity), which completely define the trig functions. It would then follow that all other trig identities can be derived from these.

And is it possible to do this without using any analysis? My intuition says no, at least for a finite set of identities, because any finite set of identities would only define a countable set of points, so you'd at least have to assert that the functions are continuous. So what are either a good infinite set of identities to define one of the trig functions, or a good finite set with the addition of the condition that the function is continuous (which to me is the least possible analysis we can get away with in this case)? Or if you have another clever way that doesn't fit what I said above, I'd like to hear about it too.

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$f(x) = \cos x, g(x) = \sin x$ is the unique continuous (but even measurable would suffice) pair of functions satisfying

$$f(x)^2 + g(x)^2 = 1$$ $$f(x + y) = f(x) f(y) - g(x) g(y)$$ $$g(x + y) = f(x) g(y) + g(x) f(y)$$

with initial conditions $f(0) = 1, g(0) = 0$ and period $2 \pi$. If you don't want to explicitly mention $\pi$ then I am reasonably sure you will need to mention a derivative. You can't get $\pi$ with absolutely no analytic input whatsoever.

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Do you have a reference for this? –  Neal Oct 21 '12 at 5:05
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@Neal: nope. I would prefer to leave it as an exercise. Hint: consider the function $f(x) + i g(x)$. –  Qiaochu Yuan Oct 21 '12 at 5:06
    
True, you can't get period $\pi$. But if one is willing to live with the angles in units other than radians, it seems one could say "with period P" for any $P$ and get the trig functions in these alternate units. –  coffeemath Oct 21 '12 at 6:32
    
Oh sure, I knew you would have to mention $\pi$. When I said not analytic, I meant more about requirements on the functions rather than analytically defined constants. –  asmeurer Oct 21 '12 at 7:09
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@asmeurer: it's known that pathological solutions to the Cauchy functional equation ( en.wikipedia.org/wiki/Cauchy's_functional_equation) are not measurable (imi.kyushu-u.ac.jp/~ssaito/eng/maths/Cauchy.pdf). The proof should apply to this case as well. –  Qiaochu Yuan Oct 21 '12 at 7:25

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