cos

返回它的余弦值。

template<class Type> 
   complex<Type> cos( 
      const complex<Type>& _ComplexNum 
   );

参数

• _ComplexNum
  复数的余弦值确定。

返回值

复数是输入的余弦的复数。

备注

定义复杂余弦的标识：

因为 (z) (= 1/2) * (exp (iz) + exp (-iz)

因为 (z) =，因此 (例如 + =) 短棒 (B) 双精度型)，因为 isin) - sinh (B)

示例

// complex_cos.cpp
// compile with: /EHsc
#include <vector>
#include <complex>
#include <iostream>

int main( )
{
   using namespace std;
   double pi = 3.14159265359;
   complex <double> c1 ( 3.0 , 4.0 );
   cout << "Complex number c1 = " << c1 << endl;

   // Values of cosine of a complex number c1
   complex <double> c2 = cos ( c1 );
   cout << "Complex number c2 = cos ( c1 ) = " << c2 << endl;
   double absc2 = abs ( c2 );
   double argc2 = arg ( c2 );
   cout << "The modulus of c2 is: " << absc2 << endl;
   cout << "The argument of c2 is: "<< argc2 << " radians, which is " 
        << argc2 * 180 / pi << " degrees." << endl << endl; 

   // Cosines of the standard angles in the first 
   // two quadrants of the complex plane
   vector <complex <double> > v1;
   vector <complex <double> >::iterator Iter1;
   complex <double> vc1  ( polar (1.0, pi / 6) );
   v1.push_back( cos ( vc1 ) );
   complex <double> vc2  ( polar (1.0, pi / 3) );
   v1.push_back( cos ( vc2 ) );
   complex <double> vc3  ( polar (1.0, pi / 2) );
   v1.push_back( cos ( vc3) );
   complex <double> vc4  ( polar (1.0, 2 * pi / 3) );
   v1.push_back( cos ( vc4 ) );
   complex <double> vc5  ( polar (1.0, 5 * pi / 6) );
   v1.push_back( cos ( vc5 ) );
   complex <double> vc6  ( polar (1.0,  pi ) );
   v1.push_back( cos ( vc6 ) );

   cout << "The complex components cos (vci), where abs (vci) = 1"
        << "\n& arg (vci) = i * pi / 6 of the vector v1 are:\n" ;
   for ( Iter1 = v1.begin( ) ; Iter1 != v1.end( ) ; Iter1++ )
      cout << *Iter1 << endl;
}

要求

复杂页眉： <>

命名空间: std