proof of $\sin(x)$ infinite series I can't seem to find a proof for this
$$\sin( x )= x - \frac{x^3}{3!} + \frac{x^5}{5!} – \frac{x^7}{7!}+\cdots$$
Which euler used to proof the basel problem. 
Would be great if it anyone show us how to this equality
Thanks
 A: Assume $\sin x=a_0+a_1 x^1+a_2 x^2 + a_3 x^3... $
Note that $\sin 0=0\implies a_0=0$
Now differentiate both sides.
$\cos x = a_1+2a_2x+3a_3x^2...$
$\cos 0 =1 \implies a_1=1$
Differentiate again
$-\sin x=2a_2+3!a_3x...$
$\sin 0=0 \implies a_2=0$
Differentiate again
$-\cos x=3! a_3...$
$\cos 0 =1 \implies a_3=-\frac{1}{3!}$
Hence you can find the pattern.
A: The short answer to your question is that that is the Taylor Series for the sine function expanded around $0$. This is a standard early example of a Taylor series, so any intro text detailing Taylor Series (which is to say, most introductory calculus texts) will contain it and a proof.
If you're not interested in learning the whole theory, then you might be interested in something I wrote on my personal blog about understanding Taylor series at a much more basic level. It just so happens to be that my example function is $\sin(x)$, and the post contains a proof of that equality.
A: You can use the fact that: 
$$\sin(x) = \dfrac{e^{ix}-e^{-ix}}{2i}$$ and that:
$$e^{ix} = 1+(ix)+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\cdots$$
$$e^{-ix} = 1+(-ix)+\frac{(-ix)^2}{2!}+\frac{(-ix)^3}{3!}+\cdots$$
So:
$$\begin{align*}\sin(x) &= \dfrac{e^{ix}-e^{-ix}}{2i} \\ &= (1-1)+ \frac{(ix)-(-ix)}{2i}+\frac{1}{2i}\bigg(\frac{-x^2}{2}-\frac{x^2}{2}\bigg)+\frac{1}{2i}\bigg(\frac{-ix^3}{3!}-\frac{ix^3}{3!}\bigg)+\cdots \\ 
&= 0+\frac{2ix}{2i}+0+\frac{-2ix^3}{2i3!}+\cdots \\
&= x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\end{align*}$$
Also using the fact that $i^2 = -1, i^3 = -i, i^4 = i, i^5 = -1,$ etc. 
