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It's a pretty standard result, but how does one prove, using Stone-Weierstrass, that the exponential polynomials are uniformly dense in $C([0, 2\pi])$ ($2\pi$ periodic continuous functions)? Show that the exponential polynomials contain the constants seems reasonable, but showing the separation of points, I'm unsure of.

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    As you've stated the question, the result is false, because the trigonometric polynomials don't separate the points $0$ and $2\pi$. So if $f$ is a continuous function whose values at $0$ and $2\pi$ differ, then you have no hope of approximating it by trigonometric polynomials. – 2017-02-19
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    @user49640 I believe what the OP means by $C[0,2\pi]$ is a continuous $2\pi$-periodic function. Then all functions $f\in C[0,2\pi]$ are such that $f(0)=f(2\pi)$. – 2017-02-19
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    What is "Stone-Weierstrass"? That the trigonometric polynomials are uniformly dense in $C[0,2\pi]$ is a corollary to [Fejér's theorem](https://en.wikipedia.org/wiki/Fej%C3%A9r%27s_theorem) (which, given $f \in C[0,2\pi]$, gives us a particular sequence of trigonometric polynomials converging uniformly to $f$). I don't know if that's what you're looking for though. – 2017-02-19
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    Can you confirm that you're considering periodic continuous functions? And do you mean real- or complex-valued functions? – 2017-02-19
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    @user49640 Real valued functions. Note that appropriate edits have been made. – 2017-02-19
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    Really $C([0,2\pi])$ is not good notation here. Best to call it $\tilde C ([0,1])$ or something. – 2017-02-19

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Proving that the algebra generated by $\sin t$ and $\cos t$ separates points means proving that for any distinct values $a,b \in [0,2\pi)$, we have $\sin a \ne \sin b$ or $\cos a \ne \cos b$. This is immediate from the definition of the trigonometric functions.

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    I don't think it's that obvious. You are taking sums of powers of trigonometric/exponential functions, not simply one trigonometric/exponential function. – 2017-02-21
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    @Elliot Proving separation of points means proving that whenever $a \ne b$ there exists *at least* one function $f$ in the algebra for which $f(a)\ne f(b)$. – 2017-02-21