計算COS（X）爲0時x是90度或pi/2弧度在C++

Question

# include <iostream> 
# include <math.h> 
# include <cstdlib> 
using namespace std; 

double cosin_value(double value); 
double sin_value(double value); 
double big_degree (double value); 
double big_radian (double value); 
double x; 
double value; 
double degree; 
double radian; 
const double PI = 3.14159; 

char choice; 
char yes ; 

int main() 
{ 
cout << "Please enter an angle value => "; 
cin >> value; 

cout << "Is the angle in Degree or Radian?" << endl; 
     cout << "\t" << "Type D if it is in Degree" << endl; 
     cout << "\t" << "Type R if it is in Radian" << endl; 
     cout << "Your response => "; 
     cin >> choice; //degree or radian? 

     cout.setf(ios::fixed); 
     cout.setf(ios::showpoint); 
     cout.precision(10); 

    if (choice == 'D' || choice == 'd') 
    { 
     big_degree (value); 
     cout << " " << "sin(x) = " << "\t" << sin_value(degree) << endl; 
     cout << " " << "cos(x) = " << "\t" << cosin_value(degree) << endl; 
     cout << " " << "tan(x) = " << "\t" << sin_value(degree)/cosin_value(degree) << endl; 
    } 
    else if (choice == 'R' || choice == 'r') 
    { 
     cout << " " << "sin(x) = " << "\t" << sin_value(radian) << endl; 
     cout << " " << "cos(x) = " << "\t" << cosin_value(radian) << endl; 
     cout << " " << "tan(x) = " << "\t" << sin_value(radian)/cosin_value(radian) << endl; 
    } 
    return 0; 
} 

// Sine,cosine functions 
// angle -360<value<360 

double sin_value(double value) 
{ 
int count=1; 

double sine, num, dem, sign, term; 
sine=0; 
sign = 1; 
num = value; 
dem = count; 

while (count <= 20) 
{ 
    term = (num/dem); 
    sine = sine + term*sign; 
    num = num*value*value; 
    count = count + 2; 
    dem = dem * count * (count-1); 
    sign = -sign; 

} 
return (sine); 
} 

double cosin_value(double value) 
{ 
int count=0; 
double cosine, num, dem, sign, term; 
cosine=0; 
sign = 1; 

num = 1; 
dem = 1; 

while (count <= 20) 
{ 
    term = (num/dem); 
    cosine = cosine + term*sign; 
    num = num*value*value; 
    count = count + 2; 
    dem = dem * count * (count-1); 
    sign = -sign; 
} 
return (cosine); 
} 

double big_degree (double value) 
{ 
int result; 
const int angle=360; 
if (value >= 360 || value <= -360) 
{ 
    result=value/angle; 
    degree=(value-(result* angle))*PI/180; 
}  
else 
{ 
    degree = (value*PI)/180; 
} 
return (degree); 
} 

double big_radian (double value) 
{ 
int result; 

if (value >= 2*PI || value <= -2*PI) 
{ 
    result=value/(2*PI); 
    radian=(value-(result* 2*PI)); 
}  
else 
{ 
    radian = value; 
} 
return (radian); 
} 

嗨，這是基本上整個程序我使用我用C知道的程度的知識計算三角函數值寫++作爲初學者。爲了更好的觀點，你可以參考以上關於我的代碼的鏈接：codepad.org 從114行開始的行是我創建的函數。那裏有一個問題，當數值爲90度或pi/2弧度時，如何計算我的cosx爲0？ 因爲程序仍然會計算坦克對我來說甚至是90度的值。 讓我們給予價值90度的程序說，它會給我的0.0000013268，而不是0.000000 遺憾的價值，因爲我只是一個初學者，代碼看起來怪異你們。 我感謝你的指導！ 雙big_degree（雙精度值）當值是> = 360或< = -360 *

Answer 1

我不會在腦中爲pi的數字分配任何堆空間，但我確實記得atan(1) == pi/4。

更改PI不斷像這樣：

const double PI = atan(1) * 4; 

以你的代碼，做出這樣的轉變，我得到

Please enter an angle value => 90 
Is the angle in Degree or Radian? 
    Type D if it is in Degree 
    Type R if it is in Radian 
Your response => d 
sin(x) =  1.0000000000 
cos(x) =  0.0000000000 
tan(x) =  15555226593901466.0000000000 

Answer 2

的問題是不是你的代碼裝置。您提供的輸入不夠準確。計算pi/2的正確值，你將得到一個足夠精確的值。

另外，如果你想圓了你可以使用

#rounded關閉值值= Math.Round（#old值，4）

Answer 3

const double PI = 3.14159; 

你做這個定義的更精確，越靠近到0會得到cos PI/2的值！

如果以弧度表示輸入本身，則也適用相同的標準。