用sin(x)≈x－x^3/3!+x^5/5!-……+(-1)^(n-1)*(x^(2n-1))/(2n-1)

用sin(x)≈x－x^3/3!+x^5/5!-……+(-1)^(n-1)*(x^(2n-1))/(2n-1)!的公式求近似值，直到最后一项绝对值小于0.00001为止。设x＝7。答案(0.6569827)
求各位大佬解惑！

#include <math.h>
#include <stdio.h>
double fuc(int m); //修改类型，否则x超过2就跑不了，试过没有问题了
int main()
{
    double x, term, sum;
    int n = 1;
    x = 7;
    term = x;
    sum = x;
    do
    {
        n += 1;
        term = pow((float)-1, (float)(n + 1)) *
               pow((float)x, (float)(2 * n - 1)) / fuc(2 * n - 1);
        sum += term;
    } while (fabs(term) >= 1e-5);
    printf("%lf\n", sum);
    return 0;
}
double fuc(int m)
{
    double t, p;
    for (t = 1, p = 1; t <= m; t++)
        p *= t;
    return p;
}

供参考：

#include <stdio.h>
#include <math.h>
int main()
{
    int    i = 1;
    double s, t, x = 7.0;
    //scanf("%lf", &x);
    s = x; t = x;
    while (fabs(t) >= 1e-5)
    {
        t = -t * x * x / (4 * i * i + 2 * i);
        s += t;
        i++;
    }
    printf("sin(x)≈%.7f\n", s);
    return 0;
}