在Java中找到角度的余弦

Question

我被要求根據以下等式找到cos：

我能夠找到角度的罪過，但是當找到cos時，我得到的數字與正確值有很大不同：

我使用以下代碼查找cos。 ps：我不能使用math.cos

public static double cos(double x, int n){

    // declaring cos and factorial

    double cos = 0.0;


    // this loop determines how long does the loop go so the answer is more accurate
    for (long howlong = 1 ; howlong <=n; howlong++){
        double factorial =1;

    // this will calculate the factorial for even numbers ex/ 2*2 = 4 , 4-2 = 2
    // for the denominator 

    for (int factorialnumber=1; factorialnumber<=2*howlong-2; factorialnumber++){
        factorial = factorial * howlong;    

    }

    // now we need to create the pattern for the + and -
    // so we use % that switches the sign everytime i increments by 1

    if (howlong%2==1){
        cos = cos + (double) (Math.pow(x, 2*howlong-2)/factorial);
    }

    else{
        cos = cos - (double) (Math.pow(x, 2*howlong-2)/factorial);
    }
    }
    return cos;
}

編輯：我發現我的錯誤，因為它是乘以乘號而不是乘數的倍數。

Answer 1

您有兩個錯誤。

（錯誤1）您寫在哪里

factorial = factorial * howlong;

應該是

factorial = factorial * factorialnumber;

（錯誤2）在每次通過外循環的迭代中，您都不會重置自己的階乘。 所以你需要移動線

double factorial =1;

向下幾行，使其位於外循環內。

如果您進行了這兩個更改，則cos(Math.PI / 6, 10)為0.8660254037844386 ，對我來說似乎是正確的。

Answer 2

您的階乘計算錯誤。 嘗試使用以下代碼：

public static double cos(double x, int n) {

    // declaring cos and factorial

    double cos = 0.0;


    // this loop determines how long does the loop go so the answer is more
    // accurate
    for (long howlong = 1; howlong <= n; howlong++) {

        // now we need to create the pattern for the + and -
        // so we use % that switches the sign everytime i increments by 1

        if (howlong % 2 == 1) {
            cos = cos + Math.pow(x, 2 * howlong - 2) / factorial(2 * howlong - 2);
        }

        else {
            cos = cos - Math.pow(x, 2 * howlong - 2) / factorial(2 * howlong - 2);
        }
    }
    return cos;
}

public static long factorial(long n) {
    long result = 1;
    for (int i = 2; i <= n; i++) {
        result *= i;
    }
    return result;
}

Answer 3

您的計算不正確，請更改為

 double value = 1;
            for (int factorialnumber = 1; factorialnumber <= 2 * howlong - 2; factorialnumber++) {
                value = factorialnumber * value;
            }

            factorial = value;
            System.out.println(value + " " + (2 * howlong - 2));