c ++ Sin和Cos [英] c++ Sin and Cos

问题描述

我相信这是一个非常愚蠢的问题，但是当我通过一个180度的角度到c / c ++的cos（）和sin（）函数，我似乎收到一个不正确的值。我知道应该是：
sin 0.0547和cos 0.99
但我得到罪的3.5897934739308216e-009和cos的-1.00000

我的代码是：

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

DegreesToRadians（）是：

  double DegreesToRadians（double degrees）
 {
 return degrees * PI / 180;谢谢：）
 
 
 h2_lin>解决方案
 C / C ++提供 sin（a）， cos（a）， tan（a）等需要具有弧度单位而不是度的参数的函数。  double DegreesToRadians（d）执行关闭的转换，但转换结果四舍五入后为近似值。另外，机器 M_PI 是接近的，但不是与数学非理性π相同的值。
 
 
 
 OP的代码 180 传递给 DegreesToRadians（d），然后到 sin（）/ cos（）由于舍入，有限精度 double（）改进之处是在度之前执行自变量减少。 
调用trig函数。以下将角度首先降低到-45°至45°范围，然后调用 sin（）。这将确保 sind（90.0 * N） - >中的 N  -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。 
  #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：
 //使用* 1.0避免-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（％.1f度的sin（％）是％。* e \\\
 i，DBL_DECIMAL_DIG-1，
 sin（d2r（i）））; 
 printf（％.1f degrees的sind（）为％。* e \\\
，1.0 * i，DBL_DECIMAL_DIG  -  1，
 sind（i） 
} 
 return 0; 
} 
  
输出
 
 
 <$ p -360.0度的$ p>  sin（）是2.4492935982947064e-16 
 -360.0度的sind（）是-0.0000000000000000e + 00 //精确
 
 sin （）-345.0度是2.5881904510252068e-01 // 76-68 = 8分
 // 2.5881904510252076e-01 
 -345.0度的sind（）是2.5881904510252074e-01 // 76-74 = 2 away 
 $ b $ -33.0度的sin（）是5.0000000000000044e-01 // 44 away 
 // 0.5 5.0000000000000000e-01 
 -330.0度的sind 4.9999999999999994e-01 // 6离开
 
 -315.0度的sin（）是7.0710678118654768e-01 // 68-52 = 16离开
 //平方根0.5  - > 7.0710678118654752e-01 
 -315.0度的sind（）是7.0710678118654746e-01 // 52-46 = 6离开
 
 -300.0度的sin（）是8.6602540378443860e-01 $ b -300.0度的$ b sind（）是8.6602540378443871e-01 $ b $ -285.0度的sin（）是9.6592582628906842e-01 
 -285.0度的sind（）是9.6592582628906831e-01 
 -270.0度的sin（）为1.0000000000000000e + 00 //精确
 -270.0度的sind（）为1.0000000000000000e + 00 //精确
 ... 
  




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

My code is:

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

DegreesToRadians() is:

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

Thank you :)

解决方案

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.

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.

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\n", 1.0 * i, DBL_DECIMAL_DIG - 1,
        sin(d2r(i)));
    printf("sind() of %.1f degrees is  % .*e\n", 1.0 * i, DBL_DECIMAL_DIG - 1,
        sind(i));
  }
  return 0;
}

Output

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
...

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