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
 A: A proper proof would require ideas from analysis/calculus and you should get familiar with these (if not already). The typical proof is via Taylor's theorem. A non-rigorous version uses the formula $$\sin nx=n\sin x-\frac{n(n^2-1^2)}{3!}\sin^3x+\frac{n(n^2-1^2)(n^2-3^2)}{5!}\sin^5x-\cdots$$ (convince yourself that the above formula is true by trying out odd values of $n$ like $3,5$) and then puts $nx=t$ where $t$ is constant, $n\to\infty, x\to 0$ to get $$\sin t=t-\frac{t^3}{3!}+\dots$$
Newton used another procedure to obtain infinite series for $\sin x$. Using the geometric definition of $\sin x$ (as mentioned in your comment) one obtains $$x=\int_{0}^{\sin x}\frac{dt} {\sqrt{1-t^2}}$$ for $x\in[-\pi/2,\pi/2]$ and expanding the integrand via binomial theorem and integrating term by term we get $$x=\sin x+\frac{1}{2}\frac{\sin^3x}{3}+\frac{1\cdot 3}{2\cdot 4}\frac{\sin^5x}{5}+\dots$$ and then assuming $$\sin x=ax+bx^3+cx^5+\dots$$ and comparing coefficients we can easily find $a, b, c, d$. 
A: $$f(x)=e^x$$
$$f'(x)=e^x  \tag 1$$  (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:
$$f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)x^n}{n!}=\sum_{n=0}^{\infty} \frac{x^n}{n!}$$
$$f(ix)=e^{ix}=\sum_{n=0}^{\infty} \frac{i^nx^n}{n!}$$
$$i=\sqrt{-1}$$
$$i^2=-1$$
$$i^3=-i$$
$$i^4=1$$
$$i^5=i$$
.
.
$$e^{ix}=1+\frac{ix}{1!}+\frac{i^2x^2}{2!}+\frac{i^3x^3}{3!}+\frac{i^4x^4}{4!}+\frac{i^5x^5}{5!}+\frac{i^6x^6}{6!}+\frac{i^7x^7}{7!}+\frac{i^8x^8}{8!}+......$$
$$e^{ix}=1+\frac{ix}{1!}+\frac{-x^2}{2!}+\frac{-ix^3}{3!}+\frac{x^4}{4!}+\frac{ix^5}{5!}+\frac{-x^6}{6!}+\frac{-ix^7}{7!}+\frac{x^8}{8!}+......$$
$$e^{ix}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+..... +i(\frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+......) \tag2$$
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
$$\cos x=\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!}$$
$$\sin x=\sum_{n=0}^{\infty} \frac{(-1)^{n}x^{2n+1}}{(2n+1)!}$$
A: Well the expansion above is merely applying Taylor's Theorem to the function $\sin(x)$.
If you want to rigorously understand this expansion, you will most likely just have to look at a proof of Taylor's theorem, which involves creating a better approximation for any function $f(x)$, by continually adding terms which mitigate the error of the polynomial approximation.
A: Initially, I was looking for a simple proof that doesn't use the Taylor theorem but according to the answers I have received so far, I don't think its possible (except the answer by @Paramanand). 
Finally, I found this video very helpful. It's exactly what I was looking for. Now I have understood how to derive it using the Taylor Theorem.  
A: The Taylor series for a function $f(x)$ is:
$$f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}.$$
We need to calculate $\sin^{(n)}(0)$ for all $n.$
When $n \equiv 0\mod{2},$ $$\sin^{(n)}(0)=0.$$
When $n \equiv 1\mod{4},$ $$\sin^{(n)}(0)=1.$$
When $n \equiv 3\mod{4},$ $$\sin^{(n)}(0)=-1.$$
Using these values, all the even terms disappear from the series, obtaining the formula:
$$\sin x=\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}.$$
A: I faced same problem here. We know that series approximated around $ x= 0$ and it converged but how do we know that it converge to $\sin x $ ?
As limited knowledge in math. I prefered to start with Euler's formular see more in  wikipedia
$$e^{i\theta} =\cos \theta + i\sin \theta $$
Which can be proof as shown in wiki's page by
$$ f(\theta) = e^{-i\theta}(\cos \theta + i \sin \theta)$$
$$ f'(\theta) = e^{-i\theta}(-\sin \theta + i \cos \theta) + -ie^{-i\theta}(\cos \theta + i \sin \theta) = 0$$
And
$$ f(0) = e^{-0i}(\cos 0 + i \sin 0) = 1$$
As $f'(\theta) = 0$ then $f(\theta)$ is constant. Then
$$ f(\theta) = 1$$
$$ e^{-i\theta}(\cos \theta + i \sin \theta) = 1$$
$$ \cos \theta + i \sin \theta = e^{i\theta}$$
Now, we look at power factor of $e^x$
$$ e^x = \sum_{n=0}^{\infty} \frac {x^n} {n!}$$
Then
$$ e^{ix} = 1 + ix + \frac {(ix)^2} {2!} + \frac {(ix)^3} {3!} + \frac {(ix)^4} {4!} + \frac {(ix)^5} {5!}+...+ \frac {(ix)^n} {n!}$$
$$ e^{ix} = 1 + ix - \frac {(x)^2} {2!} - \frac {(ix)^3} {3!} + \frac {(x)^4} {4!} + \frac {(ix)^5} {5!}-...\pm \frac {(ix)^n} {n!}$$
$$ e^{ix} = (1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!}-...) +  i(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!}-...) $$
Just for sin part, you see that
$$sinx = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!}-...$$
Now the only thing you need to proof is power factor of $e$ which easier to find the proof and again, with my limited knowledge of math I will go wih this.
$$ \frac d {dx} e^x = e^x$$
And.
$$ \frac d {dx} e^x = \sum_{n=0}^{\infty} \frac d {dx} \frac {x^n} {n!}$$
$$ \frac d {dx} e^x =  0 + \sum_{n=0}^{\infty} \frac d {dx} \frac {x^n} {n!}$$
$$ \frac d {dx} e^x = \sum_{n=0}^{\infty} \frac {nx^{n-1}} {n!}$$
$$ \frac d {dx} e^x = \sum_{n=0}^{\infty}  \frac {x^{n-1}} {(n-1)!} = \sum_{n=0}^{\infty} \frac {x^n} {n!}$$
As derivative of series equal to itself for all $x$ and For $x = 0,\sum_{n=0}^{\infty} \frac {x^n} {n!} = 1$
which means power series of $e^x$ is valid for all of x
