Java有公式sin(x)的问题

Question

要用Java实现，代码不要太复杂了，就是输入一个x，求这个输入x的sin值。有公式sin(x)=x-x^3/3!+x^5/5!-x^7/7!+x^9/9!-x^11/11!+...(注:x^n表示x的n次方),从键盘输入x,输出相应正弦值,精确到0.000001.

Answer 1

double a=1, b=1, c=0, x=0,n=0,s=0;
Scanner inScanner=new Scanner(System.in);
        x=inScanner.nextInt();
    x = (x*3.1416) / 180;
    s = x;
    for (n = 3;; n += 2)
    {
        a = a*(n - 1)*n;
        b = -b;
        c = Math.pow(x, n) / a;
        s = s + b*c;
        if (c <=0.000001)
        {
            System.out.println(s);

            break;
        }
    }