Find the value of sin(x) using the series up to the given accuracy without using library function - C Language Programming

Find the value of sin(x) using the series up to the given accuracy without using library function.



Click here to open this program in Turbo C++

/**********************************************************************************
 Statement - Find the value of sin(x) using the series up to the given accuracy
 (without using user defined function). Also print sin(x) using library function.
 Programmer - Vineet Choudhary
 Written For - http://developerinsider.co
 **********************************************************************************/

#include <stdio.h>
#include <conio.h>
#include <math.h>
#include <stdlib.h>

void main()
{
    int n, x1;
    float  acc, term, den, x, sinx=0, sinval;
    
    clrscr();
    
    printf("Enter the value of x (in degrees)\n");
    scanf("%f",&x);
    
    x1 = x;
    
    /* Converting degrees to radians*/
    
    x = x*(3.142/180.0);
    sinval = sin(x);
    
    printf("Enter the accuary for the result\n");
    scanf("%f", &acc);
    
    term = x;
    sinx = term;
    n = 1;
    
    do
    {
        den = 2*n*(2*n+1);
        term = -term * x * x / den;
        sinx = sinx + term;
        n = n + 1;
        
    } while(acc <= fabs(sinval - sinx));
    
    printf("Sum of the sine series         = %f\n", sinx);
    printf("Using Library function sin(%d) = %f\n", x1,sin(x));
    
    getch();
    
}         /*End of main() */

/*------------------------------
 Output
 Enter the value of x (in degrees)
 30
 Enter the accuary for the result
 0.000001
 Sum of the sine series         = 0.500059
 Using Library function sin(30) = 0.500059
 
 RUN 2
 Enter the value of x (in degrees)
 45
 Enter the accuary for the result
 0.0001
 Sum of the sine series         = 0.707215
 Using Library function sin(45) = 0.707179
 ---------------------------------------------*/
You've successfully subscribed to Developer Insider
Great! Next, complete checkout for full access to Developer Insider
Welcome back! You've successfully signed in
Success! Your account is fully activated, you now have access to all content.