用Java计算正弦和余弦

Question

import java.util.*;
import java.math.*;
public class SinCos{
    public static void main(String args[]){
        Scanner kb=new Scanner(System.in);
        System.out.println("Enter the angle for cosine: ");
        double anglecos=kb.nextDouble();
        System.out.println("Enter the number of expansions required:");
        double n=kb.nextDouble();
        System.out.println("Enter the angle for sine:");
        double anglesin=kb.nextDouble();
        System.out.println("Enter the number of expansions required:");
        double n2=kb.nextDouble();
        System.out.println("Cosine: "+workCos(anglecos,n));
        System.out.println("Sine: " +workSin(anglesin,n2));
    }

    public static double workCos(double angle, double num){
        double ans = 1;
        double ans2=0;

        for(int n = 1;((Math.round(ans * 10000.0) / 10000.0)==ans2)&&n>10; n++) {
            double times=2*n;
            ans2=Math.round(ans * 10000.0) / 10000.0;
            ans += Math.pow(-1, n) * Math.pow(angle, 2*n) / fact(times);
        }

        return ans;
    }

    public static double workSin(double angle, double num){
        double ans = 0;
        double ans2=0;
        double t;
        double t2;

        for(int k=0;(Math.round(ans * 10000.0) / 10000.0)==ans2;k++){
            ans2=Math.round(ans * 10000.0) / 10000.0;
            double times=2*k+1;
            ans += Math.pow(-1, k - 1) * Math.pow(angle, 2*k - 1) / fact(times);


        } return ans;
    }

    public static double fact(double num){
        if(num==0||num==1){
            return 1;
        }
        else{
            return num* fact(num-1);
        } 
    }
}

I have been trying to compute sine and cosine using Taylor's theorem. However, the answers I am given at the end are nowhere near the actual answer.

The last method in my code gives the factorials. The other two methods compute the values using Taylor's theorem. The idea in these two methods is to keep computing until the answers don't vary too much from each other.

Answer 1

I've long forgotten the Taylor's theorem (tho I knew it a while back), so I won't delve deeper into your implementation of it. However, I do see this:

for(int n = 1;((Math.round(ans * 10000.0) / 10000.0)==ans2)&&n>10; n++)

Which makes no sense since this loop will exit immediately because of && n>10 which is not true in the first passing through it. Also check the second for loop since it also does not check the value of int k = 0 declared in the for loop. I'm certain that is the problem, might not be the only one though.

Answer 2

Firstly, I agree with Vucko, you should get rid of n>10 but I assume this is just a mistake.

I'm assuming this part is an attempt to avoid an overflow error.

((Math.round(ans * 10000.0) / 10000.0)==ans2)

This should be changed since you probably want the loop to stop once this is true and keep going if this is false. In a for loop for(A;B;C), the loop will keep going if B is true, hence you should change it to the following.

((Math.round(ans * 10000.0) / 10000.0)!=ans2)

How about manually checking for Nans instead (I cleaned up your code a little while I was at it although it could be a lot more efficient)

public static double workCos(double angle, double num){
    double ans = 0;
    for(int n = 0; n < num; n++) {
        double currAns = Math.pow(-1, n) * Math.pow(angle, 2*n) / fact(2*n);
        if(Double.isNaN(currAns))
            break;
        ans += currAns;
    }
    return ans;
}

Answer 3

对于泰勒级数，输入必须为弧度而不是度。