Sin 和 Cos 对众所周知的角度给出了意想不到的结果 [英] Sin and Cos give unexpected results for well-known angles

问题描述

我确信这是一个非常愚蠢的问题，但是当我将 180 度角传递给 c/c++ 的 cos() 和 sin() 函数时，我似乎收到了不正确的值.我知道它应该是:sin 0.0547 和 cos 0.99但我得到了 3.5897934739308216e-009 的 sin 和 -1.00000 的 cos

I am sure this is a really stupid question, but when I pass an angle of 180 degrees into c/c++'s cos() and sin() functions I appear to receive an incorrect value. I know that it should be: sin of 0.0547 and cos of 0.99 but I get sin of 3.5897934739308216e-009 and cos of -1.00000

我的代码是:

double radians = DegreesToRadians( angle );
double cosValue = cos( radians );
double sinValue = sin( radians );

DegreesToRadians() 是:

DegreesToRadians() is:

double DegreesToRadians( double degrees )
{ 
    return degrees * PI / 180; 
} 

谢谢你:)

推荐答案

C/C++提供sin(a), cos(a), tan(a) 等需要 radian 单位而不是 degrees 的参数的函数.double DegreesToRadians(d) 执行 close 的转换，但转换结果是四舍五入的近似值.机器 M_PI 也很接近，但与数学无理 π 的值不同.

C/C++ provides sin(a), cos(a), tan(a), etc. functions that require a parameter with radian units rather than degrees. double DegreesToRadians(d) performs a conversion that is close but an approximate as the conversion results are rounded. Also machine M_PI is close, but not the same value as the the mathematical irrational π.

带有 180 的 OP 代码传递给 DegreesToRadians(d) 然后传递给 sin()/cos() 给出的结果与预期不同由于四舍五入，double() 的精度有限，并且 PI 的值可能较弱.

OP's code with 180 passed to DegreesToRadians(d) and then to sin()/cos() gives results that differ than expected due to rounding, finite precision of double() and possible a weak value for PI.

一个改进是在调用 trig 函数之前以 degrees 执行参数缩减.下面首先将角度减小到 -45° 到 45° 的范围，然后调用 sin().这将确保 sind(90.0*N) 中 N 的大值 -->-1.0、0.0、1.0..注意:sind(360.0*N +/- 30.0) 可能不完全等于 +/-0.5.需要一些额外的考虑.

An improvement is to perform argument reduction in degrees before calling the trig function. The below reduces the angle first to a -45° to 45° range and then calls sin(). This will insure that large values of N in sind(90.0*N) --> -1.0, 0.0, 1.0. . Note: sind(360.0*N +/- 30.0) may not exactly equal +/-0.5. Some additional considerations needed.

#include <math.h>
#include <stdio.h>

static double d2r(double d) {
  return (d / 180.0) * ((double) M_PI);
}

double sind(double x) {
  if (!isfinite(x)) {
    return sin(x);
  }
  if (x < 0.0) {
    return -sind(-x);
  }
  int quo;
  double x90 = remquo(fabs(x), 90.0, &quo);
  switch (quo % 4) {
    case 0:
      // Use * 1.0 to avoid -0.0
      return sin(d2r(x90)* 1.0);
    case 1:
      return cos(d2r(x90));
    case 2:
      return sin(d2r(-x90) * 1.0);
    case 3:
      return -cos(d2r(x90));
  }
  return 0.0;
}

int main(void) {
  int i;
  for (i = -360; i <= 360; i += 15) {
    printf("sin()  of %.1f degrees is  % .*e
", 1.0 * i, DBL_DECIMAL_DIG - 1,
        sin(d2r(i)));
    printf("sind() of %.1f degrees is  % .*e
", 1.0 * i, DBL_DECIMAL_DIG - 1,
        sind(i));
  }
  return 0;
}

输出

sin()  of -360.0 degrees is   2.4492935982947064e-16
sind() of -360.0 degrees is  -0.0000000000000000e+00  // Exact

sin()  of -345.0 degrees is   2.5881904510252068e-01  // 76-68 = 8 away
//                            2.5881904510252076e-01
sind() of -345.0 degrees is   2.5881904510252074e-01  // 76-74 = 2 away

sin()  of -330.0 degrees is   5.0000000000000044e-01  // 44 away
//  0.5                       5.0000000000000000e-01
sind() of -330.0 degrees is   4.9999999999999994e-01  //  6 away

sin()  of -315.0 degrees is   7.0710678118654768e-01  // 68-52 = 16 away
// square root 0.5 -->        7.0710678118654752e-01
sind() of -315.0 degrees is   7.0710678118654746e-01  // 52-46 = 6 away

sin()  of -300.0 degrees is   8.6602540378443860e-01
sind() of -300.0 degrees is   8.6602540378443871e-01
sin()  of -285.0 degrees is   9.6592582628906842e-01
sind() of -285.0 degrees is   9.6592582628906831e-01
sin()  of -270.0 degrees is   1.0000000000000000e+00  // Exact
sind() of -270.0 degrees is   1.0000000000000000e+00  // Exact
...

这篇关于Sin 和 Cos 对众所周知的角度给出了意想不到的结果的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！