它如何做到这一点的本质在于杰拉德·惠特利的《应用数值分析》节选:

当你的软件程序要求计算机获取一个值时 或者，你有没有想过它是如何得到 如果它能计算的最强大的函数是多项式? 它不会在表中查找这些并进行插值!相反, 计算机逼近除多项式以外的所有函数 一个多项式，可以精确地给出值。

上面要提到的几点是，一些算法实际上是从表中插值的，尽管只是在前几次迭代中。还要注意它是如何提到计算机利用近似多项式而没有指定哪种类型的近似多项式。正如本文中其他人指出的那样，在这种情况下，切比雪夫多项式比泰勒多项式更有效。

如果你想要一个软件实现，而不是硬件实现，可以在《数值公式》的第5章中找到这个问题的明确答案。我的副本在一个盒子里，所以我不能给出细节，但简短的版本(如果我没记错的话)是你把tan(theta/2)作为你的基本操作，然后从那里计算其他的。计算是用级数近似完成的，但它比泰勒级数收敛得快得多。

抱歉，我没拿到书就想不起来了。

如果您想查看这些函数在C语言中的实际GNU实现，请查看glibc的最新主干。参见GNU C库。

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。

正如许多人指出的那样，它依赖于实现。但就我对你的问题的理解而言，你对数学函数的真正软件实现感兴趣，但只是没有找到一个。如果是这样的话，那么你是这样的:

从http://ftp.gnu.org/gnu/glibc/下载glibc源代码 查看位于解包的glibc根\sysdeps\ieee754\dbl-64文件夹中的文件dosincosc 类似地，您可以找到其余数学库的实现，只需查找具有适当名称的文件

您也可以看看扩展名为.tbl的文件，它们的内容只不过是以二进制形式的不同函数的预计算值的巨大表格。这就是为什么实现如此之快:而不是计算他们使用的任何级数的所有系数，他们只是做一个快速查找，这要快得多。顺便说一下，他们确实用裁缝级数来计算正弦和余弦。

我希望这能有所帮助。

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;

}