Лучший способ поиска в Google - это выяснить, как это делается с помощью некоторого популярного языка, например, путем поиска std::sin implementation, что даст вам реализацию на C ++.
Ниже приведен один изреализации, используемые стандартной математической библиотекой GCC (взятой из здесь ).
/*******************************************************************/
/* An ultimate sin routine. Given an IEEE double machine number x */
/* it computes the correctly rounded (to nearest) value of sin(x) */
/*******************************************************************/
#ifndef IN_SINCOS
double
SECTION
__sin (double x)
{
double t, a, da;
mynumber u;
int4 k, m, n;
double retval = 0;
SET_RESTORE_ROUND_53BIT (FE_TONEAREST);
u.x = x;
m = u.i[HIGH_HALF];
k = 0x7fffffff & m; /* no sign */
if (k < 0x3e500000) /* if x->0 =>sin(x)=x */
{
math_check_force_underflow (x);
retval = x;
}
/*--------------------------- 2^-26<|x|< 0.855469---------------------- */
else if (k < 0x3feb6000)
{
/* Max ULP is 0.548. */
retval = do_sin (x, 0);
} /* else if (k < 0x3feb6000) */
/*----------------------- 0.855469 <|x|<2.426265 ----------------------*/
else if (k < 0x400368fd)
{
t = hp0 - fabs (x);
/* Max ULP is 0.51. */
retval = copysign (do_cos (t, hp1), x);
} /* else if (k < 0x400368fd) */
/*-------------------------- 2.426265<|x|< 105414350 ----------------------*/
else if (k < 0x419921FB)
{
n = reduce_sincos (x, &a, &da);
retval = do_sincos (a, da, n);
} /* else if (k < 0x419921FB ) */
/* --------------------105414350 <|x| <2^1024------------------------------*/
else if (k < 0x7ff00000)
{
n = __branred (x, &a, &da);
retval = do_sincos (a, da, n);
}
/*--------------------- |x| > 2^1024 ----------------------------------*/
else
{
if (k == 0x7ff00000 && u.i[LOW_HALF] == 0)
__set_errno (EDOM);
retval = x / x;
}
return retval;
}
это вызывает
/* Given a number partitioned into X and DX, this function computes the sine of
the number by combining the sin and cos of X (as computed by a variation of
the Taylor series) with the values looked up from the sin/cos table to get
the result. */
static inline double
__always_inline
do_sin (double x, double dx)
{
double xold = x;
/* Max ULP is 0.501 if |x| < 0.126, otherwise ULP is 0.518. */
if (fabs (x) < 0.126)
return TAYLOR_SIN (x * x, x, dx);
mynumber u;
if (x <= 0)
dx = -dx;
u.x = big + fabs (x);
x = fabs (x) - (u.x - big);
double xx, s, sn, ssn, c, cs, ccs, cor;
xx = x * x;
s = x + (dx + x * xx * (sn3 + xx * sn5));
c = x * dx + xx * (cs2 + xx * (cs4 + xx * cs6));
SINCOS_TABLE_LOOKUP (u, sn, ssn, cs, ccs);
cor = (ssn + s * ccs - sn * c) + cs * s;
return copysign (sn + cor, xold);
}
и
/* Given a number partitioned into X and DX, this function computes the cosine
of the number by combining the sin and cos of X (as computed by a variation
of the Taylor series) with the values looked up from the sin/cos table to
get the result. */
static inline double
__always_inline
do_cos (double x, double dx)
{
mynumber u;
if (x < 0)
dx = -dx;
u.x = big + fabs (x);
x = fabs (x) - (u.x - big) + dx;
double xx, s, sn, ssn, c, cs, ccs, cor;
xx = x * x;
s = x + x * xx * (sn3 + xx * sn5);
c = xx * (cs2 + xx * (cs4 + xx * cs6));
SINCOS_TABLE_LOOKUP (u, sn, ssn, cs, ccs);
cor = (ccs - s * ssn - cs * c) - sn * s;
return cs + cor;
}
as
#define SINCOS_TABLE_LOOKUP(u, sn, ssn, cs, ccs) \
({ \
int4 k = u.i[LOW_HALF] << 2; \
sn = __sincostab.x[k]; \
ssn = __sincostab.x[k + 1]; \
cs = __sincostab.x[k + 2]; \
ccs = __sincostab.x[k + 3]; \
})
Таким образом, ответ на ваш вопрос заключается в том, что умная реализация сочетает расширения серии Тейлора с поисками таблиц.
Вот еще один такой подход, используемый Sun :
/* @(#)k_sin.c 1.3 95/01/18 */
/*
* ====================================================
* Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
*
* Developed at SunSoft, a Sun Microsystems, Inc. business.
* Permission to use, copy, modify, and distribute this
* software is freely granted, provided that this notice
* is preserved.
* ====================================================
*/
/* __kernel_sin( x, y, iy)
* kernel sin function on [-pi/4, pi/4], pi/4 ~ 0.7854
* Input x is assumed to be bounded by ~pi/4 in magnitude.
* Input y is the tail of x.
* Input iy indicates whether y is 0. (if iy=0, y assume to be 0).
*
* Algorithm
* 1. Since sin(-x) = -sin(x), we need only to consider positive x.
* 2. if x < 2^-27 (hx<0x3e400000 0), return x with inexact if x!=0.
* 3. sin(x) is approximated by a polynomial of degree 13 on
* [0,pi/4]
* 3 13
* sin(x) ~ x + S1*x + ... + S6*x
* where
*
* |sin(x) 2 4 6 8 10 12 | -58
* |----- - (1+S1*x +S2*x +S3*x +S4*x +S5*x +S6*x )| <= 2
* | x |
*
* 4. sin(x+y) = sin(x) + sin'(x')*y
* ~ sin(x) + (1-x*x/2)*y
* For better accuracy, let
* 3 2 2 2 2
* r = x *(S2+x *(S3+x *(S4+x *(S5+x *S6))))
* then 3 2
* sin(x) = x + (S1*x + (x *(r-y/2)+y))
*/
#include "fdlibm.h"
#ifdef __STDC__
static const double
#else
static double
#endif
half = 5.00000000000000000000e-01, /* 0x3FE00000, 0x00000000 */
S1 = -1.66666666666666324348e-01, /* 0xBFC55555, 0x55555549 */
S2 = 8.33333333332248946124e-03, /* 0x3F811111, 0x1110F8A6 */
S3 = -1.98412698298579493134e-04, /* 0xBF2A01A0, 0x19C161D5 */
S4 = 2.75573137070700676789e-06, /* 0x3EC71DE3, 0x57B1FE7D */
S5 = -2.50507602534068634195e-08, /* 0xBE5AE5E6, 0x8A2B9CEB */
S6 = 1.58969099521155010221e-10; /* 0x3DE5D93A, 0x5ACFD57C */
#ifdef __STDC__
double __kernel_sin(double x, double y, int iy)
#else
double __kernel_sin(x, y, iy)
double x,y; int iy; /* iy=0 if y is zero */
#endif
{
double z,r,v;
int ix;
ix = __HI(x)&0x7fffffff; /* high word of x */
if(ix<0x3e400000) /* |x| < 2**-27 */
{if((int)x==0) return x;} /* generate inexact */
z = x*x;
v = z*x;
r = S2+z*(S3+z*(S4+z*(S5+z*S6)));
if(iy==0) return x+v*(S1+z*r);
else return x-((z*(half*y-v*r)-y)-v*S1);
}
Другие методы включают использование инструкций по сборке, чтобы вычисление производилось аппаратно (хотя это иногда приводило к проблемам ).
Дополнительная информациядоступны на эти ответы .