Question: How to find the approximation to $\sin(1.58)$ ? By using the Newton's method 

$$x\_{n+1} = x\_{n} - \frac{f(x)}{f'(x)}$$

You always will get $0$. Using this method:

$f(x+\Delta x) \approx f(x) + f'(x)dx$

You get $\sin(1.58) \approx 0.5974$ , but that's not very precise. 

So, is there a method to precisely approximate $\sin(x)$?

Answer: 
Notice that 1.58 is very close to $\pi/2 \approx 1.57079633$

So you can use the Taylor expansion,

$$\sin(x+\delta x) \approx \sin(x) + \cos(x)\delta x - \tfrac{1}{2} \sin(x) \delta x^2$$

with $x=\pi/2$ and $\delta x = 1.58 - \pi/2 \approx 0.00920367321$

This gives

$$\sin(1.58) \approx \sin(\pi/2) + \cos(\pi/2) \delta x - \tfrac{1}{2}\sin(\pi/2) \delta x^2 = 0.999957646$$

compared to the exact value

$$\sin(1.58) = 0.999957646\dots$$

so it is in exact agreement, up to the accuracy provided by Google calculator. The actual error will be of the order $\delta x^4$, or around $10^{-8}$.
**Hint:** You can naturally use [Taylor series](http://en.wikipedia.org/wiki/Taylor_series). It consists in a polynomial approximation of any "reasonable" function around a point. Depending on how many terms you consider in the expansion, you can even obtain a numerical bound of the error occuring by this method. I leave you the care to calculate Taylor series of $sin(x)$ considering it can be a good exercise to become familiar with this method.