没有什么比点击源代码，看看人们是如何在常用的库中实际完成它的了;让我们特别看看一个C库实现。我选择了uLibC。

这是sin函数:

http://git.uclibc.org/uClibc/tree/libm/s_sin.c

看起来它处理了一些特殊情况，然后执行一些参数约简，将输入映射到范围[-pi/4,pi/4]，(将参数分成两部分，一个大的部分和一个尾巴)，然后调用

http://git.uclibc.org/uClibc/tree/libm/k_sin.c

然后作用于这两个部分。 如果没有尾巴，则使用13次多项式生成近似答案。 如果有尾巴，根据sin(x+y) = sin(x) + sin'(x')y的原理，你会得到一个小的修正

我将尝试在一个C程序中回答sin()的情况，该程序用GCC的C编译器在当前的x86处理器(假设是Intel Core 2 Duo)上编译。

在C语言中，标准C库包含了一些常见的数学函数，而这些函数并不包含在语言本身中(例如pow, sin和cos分别表示幂，sin和cos)。它们的头文件包含在math.h中。

现在在GNU/Linux系统上，这些库函数是由glibc (GNU libc或GNU C库)提供的。但是GCC编译器希望您使用-lm编译器标志链接到数学库(libm.so)，以启用这些数学函数的使用。我不确定为什么它不是标准C库的一部分。这些将是浮点函数的软件版本，或“软浮动”。

题外话:将数学函数分开的原因由来已久，据我所知，可能是在共享库可用之前，它仅仅是为了在非常古老的Unix系统中减少可执行程序的大小。

Now the compiler may optimize the standard C library function sin() (provided by libm.so) to be replaced with an call to a native instruction to your CPU/FPU's built-in sin() function, which exists as an FPU instruction (FSIN for x86/x87) on newer processors like the Core 2 series (this is correct pretty much as far back as the i486DX). This would depend on optimization flags passed to the gcc compiler. If the compiler was told to write code that would execute on any i386 or newer processor, it would not make such an optimization. The -mcpu=486 flag would inform the compiler that it was safe to make such an optimization.

现在，如果程序执行sin()函数的软件版本，它将基于CORDIC(坐标旋转数字计算机)或BKM算法，或者更可能是现在通常用于计算此类超越函数的表格或幂级数计算。(Src: http://en.wikipedia.org/wiki/Cordic应用程序)

任何最新的gcc版本(大约2.9倍以来)也提供了内置的sin版本__builtin_sin()，作为优化，它将用于取代对C库版本的标准调用。

我相信这是非常清楚的，但希望给你更多的信息比你期望的，和许多出发点，以了解更多自己。

OK kiddies, time for the pros.... This is one of my biggest complaints with inexperienced software engineers. They come in calculating transcendental functions from scratch (using Taylor's series) as if nobody had ever done these calculations before in their lives. Not true. This is a well defined problem and has been approached thousands of times by very clever software and hardware engineers and has a well defined solution. Basically, most of the transcendental functions use Chebyshev Polynomials to calculate them. As to which polynomials are used depends on the circumstances. First, the bible on this matter is a book called "Computer Approximations" by Hart and Cheney. In that book, you can decide if you have a hardware adder, multiplier, divider, etc, and decide which operations are fastest. e.g. If you had a really fast divider, the fastest way to calculate sine might be P1(x)/P2(x) where P1, P2 are Chebyshev polynomials. Without the fast divider, it might be just P(x), where P has much more terms than P1 or P2....so it'd be slower. So, first step is to determine your hardware and what it can do. Then you choose the appropriate combination of Chebyshev polynomials (is usually of the form cos(ax) = aP(x) for cosine for example, again where P is a Chebyshev polynomial). Then you decide what decimal precision you want. e.g. if you want 7 digits precision, you look that up in the appropriate table in the book I mentioned, and it will give you (for precision = 7.33) a number N = 4 and a polynomial number 3502. N is the order of the polynomial (so it's p4.x^4 + p3.x^3 + p2.x^2 + p1.x + p0), because N=4. Then you look up the actual value of the p4,p3,p2,p1,p0 values in the back of the book under 3502 (they'll be in floating point). Then you implement your algorithm in software in the form: (((p4.x + p3).x + p2).x + p1).x + p0 ....and this is how you'd calculate cosine to 7 decimal places on that hardware.

请注意，在FPU中大多数硬件实现的超越操作通常涉及一些微码和类似的操作(取决于硬件)。 切比雪夫多项式用于大多数先验多项式，但不是全部。例:使用Newton raphson方法的两次迭代，首先使用查询表，使用平方根更快。 同样，《计算机逼近》这本书会告诉你。

If you plan on implmementing these functions, I'd recommend to anyone that they get a copy of that book. It really is the bible for these kinds of algorithms. Note that there are bunches of alternative means for calculating these values like cordics, etc, but these tend to be best for specific algorithms where you only need low precision. To guarantee the precision every time, the chebyshev polynomials are the way to go. Like I said, well defined problem. Has been solved for 50 years now.....and thats how it's done.

Now, that being said, there are techniques whereby the Chebyshev polynomials can be used to get a single precision result with a low degree polynomial (like the example for cosine above). Then, there are other techniques to interpolate between values to increase the accuracy without having to go to a much larger polynomial, such as "Gal's Accurate Tables Method". This latter technique is what the post referring to the ACM literature is referring to. But ultimately, the Chebyshev Polynomials are what are used to get 90% of the way there.

享受。

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。

计算正弦/余弦/正切其实很容易通过代码使用泰勒级数来实现。自己写一个只需5秒钟。

整个过程可以用这个方程来概括:

下面是我为C语言写的一些例程:

double _pow(double a, double b) {
    double c = 1;
    for (int i=0; i<b; i++)
        c *= a;
    return c;
}

double _fact(double x) {
    double ret = 1;
    for (int i=1; i<=x; i++) 
        ret *= i;
    return ret;
}

double _sin(double x) {
    double y = x;
    double s = -1;
    for (int i=3; i<=100; i+=2) {
        y+=s*(_pow(x,i)/_fact(i));
        s *= -1;
    }  
    return y;
}
double _cos(double x) {
    double y = 1;
    double s = -1;
    for (int i=2; i<=100; i+=2) {
        y+=s*(_pow(x,i)/_fact(i));
        s *= -1;
    }  
    return y;
}
double _tan(double x) {
     return (_sin(x)/_cos(x));  
}

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;

}