How to code a sin progression with c? [closed] - c++

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I'm pretty new in coding and trying to write a sin progression for the uni
As a formula it looks like this:
So I tried to code it... And somehow, it calculates the radian but not the sinus... What is wrong?
#include "stdafx.h"
#include <iostream>
#include <math.h>
using namespace std;
double n, a;
int fakultaet(double a = 2 * n + 1)
{
if (a == 0)
return 1;
else
return (a * fakultaet(a - 1));
}
int _tmain(int argc, _TCHAR* argv[])
{
double sin, y, f; // sin = Sinus, y= angle, n=index
printf("please insert the angle \n");
scanf("%lf", &y);
double x = y * 3.14159265359 * 1 / 180; // x = radian measure
while (n < 5)
{
sin = pow(-1, n) * pow(x, 2 * n + 1) / fakultaet(a);
}
printf("The sinus is %lf\n", sin);
system("Pause");
return 0;
}

Answer 1

You are not incrementing the sin, or the n. Also hardcoding the entry value for the factorial is not good.
It should look more like this:
#include "stdafx.h"
#include <iostream>
#include <math.h>
using namespace std;
int fakultaet(double a)
{
if (a == 0)
return 1;
else
return (a * fakultaet(a - 1));
}
int main()
{
double sin = 0, y, f; // sin = Sinus, y= angle, n=index
printf("please insert the angle \n");
scanf("%lf", &y);
double x = y * 3.14159265359 * 1 / 180; // x = radian measure
for (int n = 0; n < 5; ++n)
{
sin += pow(-1, n) * pow(x, 2 * n + 1) / fakultaet(2 * n + 1);
}
printf("The sinus is %lf\n", sin);
system("Pause");
return 0;
}
I made several edits. I turned the factorial function:
int fakultaet(double a = 2 * n + 1)
into an non hardcoded version:
int fakultaet(double a)
Added the sin initialization:
double sin = 0
Changed your while loop:
while (n < 5)
Into a for which includes the increment that was missing as well:
for (int n = 0; n < 5; ++n)
Also turned the sin calculation:
sin = pow(-1, n) * pow(x, 2 * n + 1) / fakultaet(a);
Into one that sums:
sin += pow(-1, n) * pow(x, 2 * n + 1) / fakultaet(2 * n + 1);

Answer 2

You can get a neater and more efficient result by observing that
adjacent terms in the sum are related. We can express this as a recurrence:
T_n = - T_{n-1} x^2 /((2n+1) (2n))
T_0 = x
This is better since it avoids calls to pow and the need to make any explicit factorial calculations (fakultaet).
The code looks something like:
int main() {
...
double T = x;
double sin = T;
for(int i=1 ; i<5; ++i) {
T *= -(x*x) / ((2*i+1)*(2*i));
sin += T;
}
...
}