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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?

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    $\begingroup$ To this and similar questions I always respond with: What is your definition of $\sin\>$? $\endgroup$ Jul 22, 2018 at 18:57
  • $\begingroup$ Thanks Jose! Also, @ChristianBlatter, I would have used the definition from circles. $\endgroup$ Jul 22, 2018 at 19:04
  • $\begingroup$ One approach is to look at to show that for any complex $z$, $\left(1+\frac zn\right)^n$ converges to $\sum_{k=0}^{\infty}\frac{z^k}{k!}.$ The imaginary part of this function, when evaluated at $z=ix$ is your series. But then you can show the inequality that $\cos x + i\sin x = 1+x+ o(x^2),$ which is enough to show that $\left(1+\frac {ix} n\right)^n$ converges to the same value as $\left(\cos x/n + i\sin x/n\right)^n=\cos x + i\sin x.$ $\endgroup$ Jul 22, 2018 at 19:19

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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.)

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