$\sin x$ approximates $x$ for small angles In physics, particularly in waves, we make use of the fact that for small angles (less than $\pi/12$-ish), the sine function value of an angle is pretty close to the value of the angle itself (in radians of course). Can anyone give a mathematical explanation for why this is?
 A: Hint
Any mathematical function can be, at least locally, approximated by so called Taylor or Mc Laurin expansions.  
To make it as simple as possible, the tangent to a curve is, at the point where it is defined, a local approximation of the curve.
So, write the equation of the tangent to the curve $y=sin(x)$ at $x=0$ and you will obtain, for the tangent line,  $y = 0 + (x-0) = x$. So, close to $x=0$, $sin(x)$ is close to $x$.  
Uisng the same approach, you could show by yourself that, close to $x=0$, $e^x$ is close to $1+x$, that $log(1+x)$ is close to $x$ and so on. For sure, the approximations can be made better and better at the price of more terms. 
A: Hint
Consider the Taylor expansion of $\sin{x}$ about $x = 0$.
Cheers!
A: Alternative solution, if you do not want to deal with series expansion, you could calculate
$$\lim_{x\to0} \frac{\sin x}{x} = 1\quad\text{and/or}\quad \lim_{x\to0}\frac{x}{\sin x}=1$$
Thus $\sin x \sim x$ for $x$ close to $0$.
A: There is a very simple geometrical observation you can make. The arch in a circle with radius 1 equals the angle x (in radians), and for small x the arch is close to sin x. Draw this and you will see!
