Approximating sines and cosines using up to five terms of the Taylor series

I have one programming question:


The sine and cosine of \$x\$ can be computed as follows:



\$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \dots\$


\$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \dots\$



Your task is to compute the sine and cosine for given values of \$x\$ (where \$x\$ is in radians) using the above series up to 5 terms.

Input Format


First line will contain \$N\$, the number of test cases. Next \$N\$ lines will contain the input values of \$x\$:



\$1 \le N \le 50\$


\$0 \lt x \lt 10\$



Each value of \$x\$ can contain up to 2 places of decimal in radians.

Output Format


2 \$N\$ lines, corresponding to the \$N\$ input values of \$x\$. For each input, you will output 2 lines. First line will be the sine and the second line will be the cosine of \$x\$. An error margin of \$\pm\$0.001 will be tolerated while evaluating the answers. Please round off your answer to 3 decimal places.

Sample Input

5
2.83
3.24
0.99
2.74
5.04

Sample Output [sic]

0.309
-0.943
-0.089
-0.963
 0.836
 0.549
 0.392
-0.914
 0.195
 2.746

and what I have tried:

public class Trigonometric_Ratios
{  
  int factorial(int number)
  {
    int result = 1;
    for (int i = 1; i 0)
      cosx+=Math.pow(-1,j)*Math.pow(x,c-1)/factorial(c-1);
    }
    sinx=Math.round( sinx * 1000.0 ) / 1000.0;

    cosx=-cosx;
    cosx=Math.round( cosx * 1000.0 ) / 1000.0;
    System.out.println(sinx+" \n"+(cosx));
  }

  public static void main(String [] args)
  {
    int n;
    Trigonometric_Ratios tr=new Trigonometric_Ratios();

    Scanner sc=new Scanner(System.in);

    n=sc.nextInt();

    double a[]=new double[n];
    for(int i=0;i<n;i++)
    {
      a[i]=sc.nextDouble();
    }

    for(int i=0;i<n;i++)
    {
      tr.calc(a[i]);
    }
  }
}

Please review my code. In particular, I am concerned about performance,

Instead of computing Math.pow() from scratch, multiply the previous numerator by x2. (I suspect that Math.pow() uses logarithms so that it can handle the general case.) Also, instead of computing the factorial from scratch, multiply the previous denominator by the next two numbers.

StackExchange Code Review Q#31577, answer score: 2