对于罪恶，用泰勒展开可以得到

Sin (x) = x - x^3/3!+ x ^ 5/5 !- x ^ 7/7 !+……(1）

您将继续添加项，直到它们之间的差异低于可接受的容忍水平，或者只是有限的步数(更快，但不太精确)。举个例子:

float sin(float x)
{
  float res=0, pow=x, fact=1;
  for(int i=0; i<5; ++i)
  {
    res+=pow/fact;
    pow*=-1*x*x;
    fact*=(2*(i+1))*(2*(i+1)+1);
  }

  return res;
}

注:(1)适用于小角度的近似值sin(x)=x。对于更大的角度，你需要计算越来越多的项才能得到可接受的结果。 你可以使用while参数并继续，以达到一定的准确性:

double sin (double x){
    int i = 1;
    double cur = x;
    double acc = 1;
    double fact= 1;
    double pow = x;
    while (fabs(acc) > .00000001 &&   i < 100){
        fact *= ((2*i)*(2*i+1));
        pow *= -1 * x*x; 
        acc =  pow / fact;
        cur += acc;
        i++;
    }
    return cur;

}

对于罪恶，用泰勒展开可以得到

Sin (x) = x - x^3/3!+ x ^ 5/5 !- x ^ 7/7 !+……(1）

您将继续添加项，直到它们之间的差异低于可接受的容忍水平，或者只是有限的步数(更快，但不太精确)。举个例子:

float sin(float x)
{
  float res=0, pow=x, fact=1;
  for(int i=0; i<5; ++i)
  {
    res+=pow/fact;
    pow*=-1*x*x;
    fact*=(2*(i+1))*(2*(i+1)+1);
  }

  return res;
}

注:(1)适用于小角度的近似值sin(x)=x。对于更大的角度，你需要计算越来越多的项才能得到可接受的结果。 你可以使用while参数并继续，以达到一定的准确性:

double sin (double x){
    int i = 1;
    double cur = x;
    double acc = 1;
    double fact= 1;
    double pow = x;
    while (fabs(acc) > .00000001 &&   i < 100){
        fact *= ((2*i)*(2*i+1));
        pow *= -1 * x*x; 
        acc =  pow / fact;
        cur += acc;
        i++;
    }
    return cur;

}

盲汉回答的改进版代码

#define EPSILON .0000000000001
// this is smallest effective threshold, at least on my OS (WSL ubuntu 18)
// possibly because factorial part turns 0 at some point
// and it happens faster then series element turns 0;
// validation was made against sin() from <math.h>
double ft_sin(double x)
{
    int k = 2;
    double r = x;
    double acc = 1;
    double den = 1;
    double num = x;

//  precision drops rapidly when x is not close to 0
//  so move x to 0 as close as possible
    while (x > PI)
        x -= PI;
    while (x < -PI)
        x += PI;
    if (x > PI / 2)
        return (ft_sin(PI - x));
    if (x < -PI / 2)
        return (ft_sin(-PI - x));
//  not using fabs for performance reasons
    while (acc > EPSILON || acc < -EPSILON)
    {
        num *= -x * x;
        den *= k * (k + 1);
        acc = num / den;
        r += acc;
        k += 2;
    }
    return (r);
}

无论何时这样一个函数被求值，那么在某种程度上很可能有:

内插的值表(用于快速，不准确的应用程序-例如计算机图形) 收敛于期望值的级数的计算——可能不是泰勒级数，更可能是基于像克伦肖-柯蒂斯这样的奇异正交。

如果没有硬件支持，那么编译器可能会使用后一种方法，只发出汇编代码(没有调试符号)，而不是使用c库——这让您在调试器中跟踪实际代码变得很棘手。

使用泰勒级数，试着找出级数项之间的关系这样你就不用一遍又一遍地计算了

下面是一个关于余窦的例子:

double cosinus(double x, double prec)
{
    double t, s ;
    int p;
    p = 0;
    s = 1.0;
    t = 1.0;
    while(fabs(t/s) > prec)
    {
        p++;
        t = (-t * x * x) / ((2 * p - 1) * (2 * p));
        s += t;
    }
    return s;
}

使用这个，我们可以得到新的和项使用已经使用的和项(我们避免阶乘和x2p)

切比雪夫多项式，正如在另一个答案中提到的，是函数和多项式之间的最大差异尽可能小的多项式。这是一个很好的开始。

在某些情况下，最大误差不是你感兴趣的，而是最大相对误差。例如，对于正弦函数，x = 0附近的误差应该比较大的值小得多;你想要一个小的相对误差。所以你可以计算sinx / x的切比雪夫多项式，然后把这个多项式乘以x。

Next you have to figure out how to evaluate the polynomial. You want to evaluate it in such a way that the intermediate values are small and therefore rounding errors are small. Otherwise the rounding errors might become a lot larger than errors in the polynomial. And with functions like the sine function, if you are careless then it may be possible that the result that you calculate for sin x is greater than the result for sin y even when x < y. So careful choice of the calculation order and calculation of upper bounds for the rounding error are needed.

例如，sinx = x - x^3/6 + x^5 / 120 - x^7 / 5040…如果你天真地计算sinx = x * (1 - x^2/6 + x^4/120 - x^6/5040…)，那么括号中的函数是递减的，如果y是x的下一个大的数字，那么有时siny会小于sinx。相反，计算sinx = x - x^3 * (1/6 - x^2/ 120 + x^4/5040…)，这是不可能发生的。

例如，在计算切比雪夫多项式时，通常需要将系数四舍五入到双倍精度。但是，虽然切比雪夫多项式是最优的，但系数舍入为双精度的切比雪夫多项式并不是具有双精度系数的最优多项式!

For example for sin (x), where you need coefficients for x, x^3, x^5, x^7 etc. you do the following: Calculate the best approximation of sin x with a polynomial (ax + bx^3 + cx^5 + dx^7) with higher than double precision, then round a to double precision, giving A. The difference between a and A would be quite large. Now calculate the best approximation of (sin x - Ax) with a polynomial (b x^3 + cx^5 + dx^7). You get different coefficients, because they adapt to the difference between a and A. Round b to double precision B. Then approximate (sin x - Ax - Bx^3) with a polynomial cx^5 + dx^7 and so on. You will get a polynomial that is almost as good as the original Chebyshev polynomial, but much better than Chebyshev rounded to double precision.

Next you should take into account the rounding errors in the choice of polynomial. You found a polynomial with minimum error in the polynomial ignoring rounding error, but you want to optimise polynomial plus rounding error. Once you have the Chebyshev polynomial, you can calculate bounds for the rounding error. Say f (x) is your function, P (x) is the polynomial, and E (x) is the rounding error. You don't want to optimise | f (x) - P (x) |, you want to optimise | f (x) - P (x) +/- E (x) |. You will get a slightly different polynomial that tries to keep the polynomial errors down where the rounding error is large, and relaxes the polynomial errors a bit where the rounding error is small.

所有这些将使您轻松地获得最多0.55倍于最后一位的舍入误差，其中+，-，*，/的舍入误差最多为0.50倍于最后一位。