When looking up how the extremely famous series $$\sin(x)=\sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{(2k+1)!}$$ is derived, I found this great explanation by Proof Wiki.
My question is this: the explanation shows clearly how to derive the Maclaurin series for $\sin(x)$ and how it converges for all real arguments, however - as someone new to the intricacies of Maclaurin series - it does not prove that whatever the series converges to at the real number $a$ is $\sin(a)$. Why is this true?
For acute $x$ a geometric proof that $0<\sin x<x$ is easy, and we can alternately use $\cos x=1-\int_0^x\sin t dt,\,\sin x=\int_0^x\cos t dt$ to bound both trigonometric functions with increasingly high-degree polynomials. In each case the even- and odd-$n$ sums of the first $n$ terms provide subsequences that bound the functions on either side, so a ratio-test proof of convergence and the squeeze theorem proves each series converges to exactly what you think it does
Extending the above to non-acute $x$ is an exercise left for the reader. (Hint: prove compound-angle formulae for the Taylor series, then verify the effects of adding multiples of $\pi/2$ are the same as for the trigonometric functions.)