<s>[INST] Elementary proof that the MacLaurin series of $\sin x$ converges to $\sin x$ for all $x$ In my book it is given: 
$\sin x = x- \dfrac {x^3}{3!}+\dfrac{x^5}{5!}- \dfrac{x^7}{7!}...$
I googled around for a proof but couldn't understand any of them. I would like to know if there's any elementary high school level proof the series
 [/INST] $$f(x)=e^x$$
$$f'(x)=e^x  \tag 1$$  (I assumed you know this property of $e^x$. ) 
$$f''(x)=f'(x)=e^x$$
$$f^{(n)}(x)=e^x$$
$$f^{(n)}(0)=1$$
If we find the Taylor series for a function $f(x)$ is:
$$f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)x^n}{n!}=\sum_{n=0}^{\infty} \frac{x^n}{n!}$$
$$f(ix)=e^{ix}=\sum_{n=0}^{\infty} \frac{i^nx^n}{n!}$$
$$i=\sqrt{-1}$$
$$i^2=-1$$
$$i^3=-i$$
$$i^4=1$$
$$i^5=i$$
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$$e^{ix}=1+\frac{ix}{1!}+\frac{i^2x^2}{2!}+\frac{i^3x^3}{3!}+\frac{i^4x^4}{4!}+\frac{i^5x^5}{5!}+\frac{i^6x^6}{6!}+\frac{i^7x^7}{7!}+\frac{i^8x^8}{8!}+......$$
$$e^{ix}=1+\frac{ix}{1!}+\frac{-x^2}{2!}+\frac{-ix^3}{3!}+\frac{x^4}{4!}+\frac{ix^5}{5!}+\frac{-x^6}{6!}+\frac{-ix^7}{7!}+\frac{x^8}{8!}+......$$
$$e^{ix}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+..... +i(\frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+......) \tag2$$
This is Euler Formula: link
$$e^{ix}=\cos x +i \sin x  \tag 3$$
Take both side derivative Equation $2$
$$(e^{ix})'=(-\frac{x}{1!}+\frac{x^3}{3!}-\frac{x^5}{5!}+\frac{x^7}{7!}+......) +i(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+.....) $$
$$(e^{ix})'=-\sin x +i\cos x  \tag 4 $$
$$(e^{ix})'=\cos 'x +i \sin' x  \tag 5 $$   ///If we want equal (4) and (5)
$$ \cos 'x +i \sin' x =-\sin x +i\cos x$$
we equal imaginary and real parts separately, we will get
$$\cos 'x= - \sin x$$
$$\sin 'x=  \cos x$$
They are trigonometric function properties. link
$$\cos x=\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!}$$
$$\sin x=\sum_{n=0}^{\infty} \frac{(-1)^{n}x^{2n+1}}{(2n+1)!}$$
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