Taylor series: $\sin x = x$? Taylor series are used to expand a function to a series of functions that sometimes makes calculations easier. 
The more terms of a series we consider the more precise the solution would be. 
Sometimes a rough approximation, is acceptable, when only first two terms of a series is considered. 
Saying that, if we consider $\sin x = x$ (taking only first term) would that mean, that $\sin x$ is roughly equal to $x$ ? If so what does that mean, how $\sin x$ can be equal to $x$? If not, where am I wrong in my arguments? 
 A: 
The more terms of a series we consider the more precise the solution would be.

Yes, see below, where the black line is $\sin x$, the red is approximating using $x$, the green using $x-x^3/3!$ and the blue with $x-x^3/3!+x^5/5!$ 


Sometimes a rough approximation, is acceptable, when only first two terms of a series is considered.

Yes, for even a single term, it's hard to distinguish $\sin x$ and $x$ when you plot in $[-0.5,0.5]$.

Saying that, if we consider sin(x) = x (taking only first term) would that mean, that sin x is roughly equal to x ? If so what does that mean, how sin x can be equal to x? 

Yes, it means they're approximately equal. Mathematicians and scientists use big O notation to indicate the order or magnitude of the error. E.g.
$$\begin{align}
\sin x &= x + O(x^3)\\
\sin x &= x - \frac{x^3}{3!} + O(x^5)\\
\sin x &= x - \frac{x^3}{3!} + \frac{x^5}{5!} + O(x^7)
\end{align}$$
This means that when you use $x$ for example  to approximate $\sin x$, any errors you get are proportional to $x^3$ at worst, and so long as $x$ is close to zero, $x^3$ will be small and not worth worrying about.
Formally, with Taylor series you can quantify the error, or the remainder, in terms of some multiple of the next term in the series (although this multiple is hard to find explicitly).
A: It's not hard to prove that if $x \neq 0$, then $\sin{x} \neq x$. Graphically one sees
    
             
   
Any sufficiently smooth function, such as $\sin{x}$, is approximately equal to the first term in its Taylor series near where it is centered. The Taylor series for sine (centered about $x=0$) is
$$\sum\limits_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots.
$$
So we have that
$$ \sin{x} \approx x.$$
This approximation is pretty good for small values of $x$. As such, physicsts somtimes use this fact to make their differential equations easier to solve. They call it the small-angle approximation.
By adding more terms, we get a better approximation.
