How approximate sine using Taylor series
You use argument reduction. You know that sin (2kπ + x) = sin (x), so given an argument y, you calculate k = round (y / 2π), y' = y - 2kπ, and sin (y) = sin (y'), with y between -π and π. 

You can also use sin (π + x) = - sin (x), sin (π/2 + x) = cos (x) to reduce the argument to sin (y') or cos (y') with -π/4 ≤ x ≤ π/4. 

The reason why your Taylor series gives absurd results for large x is that you are adding and subtracting huge terms. Mathematically the series converges just fine, but if you have terms > $10^{20}$, these terms will have rounding errors so large that you cannot expect a good result. Reducing the argument to -π/4 ≤ x ≤ π/4 means you have much smaller terms. (Also see Pseudonym's answer how to get the argument reduction as precise as possible). 

You are adding the terms of the Taylor series from the largest. Each addition has a rounding error, proportional in size to the result of the addition. So adding in this order each addition gives a large error. It is better to start with the smallest term. The two smallest terms are tiny, adding them gives a tiny rounding error. 

And usually we don't actually use the Taylor series. The Taylor series has its largest error at the end of the range. You would use Chebyshev polynomials, which minimise the largest error over the whole interval. And since you don't want any errors at x = 0, you can write sin(x) = x - x * g(x) and minimise the relative error of g(x).