Напечатать таблицу значений функций sinx и tgx на отрезке [0, 3] с шагом 0.1 - C++ - Ответ 3767506

#include <stdio.h>
#include <math.h>
 
int main()
{
float i;
for(i=0.0; i<3.0;i+=0.1)
{
printf("sin(%.1f): %.5f \n", i, sin(i));
printf("tg(%f): %f %e\n\n", i, sin(i)/cos(i) );
}
 
return 0;
}

#include <stdio.h>
#include <math.h>
 
#define M_PI 3.14159265358979323846
 
float fsin(int n, float xn, float fact);
float fcos(int n, float xn, float fact);
 
int main()
{
float i;
for(i=0.0; i<3.0;i+=0.1)
{
printf("sin(%.1f): %.5f \n", i, fsin(0, i,1.0));
printf("tg(%f): %e\n\n", i, fsin(0, i,1.0)/fcos(0, i,1.0) );
}
 
return 0;
}
 
float fsin(int n, float xn, float fact)
{
if (n == 8)
 return 0.0;
 
int fct = 2*n + 1;
 
float nfact;
 
if (n == 0)
nfact = 1.0;
else
nfact = fact * (float)(fct - 1) * (float)fct;
 
float coef = pow( (float) -1.0, (float) n) / nfact;
float step = 2.0*n+1.0;
 
float result = coef * pow(xn, step);
 
return (result+fsin(n+1, xn, nfact));
 
}
 
float fcos(int n, float xn, float fact)
{
if (n == 8)
 return 0.0;
 
int fct = 2*n;
 
float nfact;
 
if (n == 0)
nfact = 1.0;
else if (n == 1)
nfact = fact * (float)fct;
else
nfact = fact * (float)(fct - 1) * (float)fct;
 
float coef = pow( (float) -1.0, (float) n) / nfact;
float step = 2.0*n;
 
float result = coef * pow(xn, step);
 
return (result+fcos(n+1, xn, nfact));
 
}