Program to Calculate the Sine Series in C | C Program

Here's a C program to calculate the sine series with output. This program makes use of C concepts like For loop. The program also uses C's math.h header file and power function pow(i, j).

 Formula of Sine Series: The formula for sine series is -

Sine Series

# include <stdio.h>
# include <conio.h>
# include <math.h> 
void main() 
{ 
 int i, n ; 
 float x, val, sum, t ; 
 clrscr() ; 
 printf("Enter the value for x : ") ; 
 scanf("%f", &x) ; 
 printf("\nEnter the value for n : ") ; 
 scanf("%d", &n) ; 
 val = x ; 
 x = x * 3.14159 / 180 ; 
 t = x ; 
 sum = x ; 
 for(i = 1 ; i < n + 1 ; i++) 
 { 
  t = (t * pow((double) (-1), (double) (2 * i - 1)) * 
   x * x) / (2 * i * (2 * i + 1)) ; 
  sum = sum + t ; 
 } 
 printf("\nSine value of %f is : %8.4f", val, sum) ; 
 getch() ; 
} 

Output of above program is

Enter the value for x : 30

Enter the value for n : 20

Sine value of 30.000000 is :   0.5000

Explanation of above program

Let n = 3, then the sine series generated by above program is as follows:

Initially, x is converted to radian by multiplying it by 3.14159/180 and t and sum are also assigned the value of x. So, after line no 16 of the program, we have:

t = x and sum = x

Now lets take a look at the working of for loop with n = 3:

Exp 1: t = (t * pow((double) (-1), (double) (2 * i - 1)) * x * x) / (2 * i * (2 * i + 1))

Exp 2: sum = sum + t

Iteration 1: i = 1, t = x, sum = x

Putting these values in Exp 1 we get,

t = (x * pow((double) (-1), (double) (2 * 1 - 1)) * x * x) / (2 * 1 * (2 * 1 + 1))

t = x*(-1)*x*x/6 = -x³/3!

Putting value of t in Exp 2 we get,

sum =  x + (-x³/3! ) = x - x³/3! 

Iteration 2: i = 2, t = -x³/3!, sum = x - x³/3!

Putting these values in Exp 1 we get,

t = (-x³/3! * pow((double) (-1), (double) (2 * 2 - 1)) * x * x) / (2 * 2 * (2 * 2 + 1))

t = (-x³/3!)*(-1)*x*x/20 = x⁵/5!

Putting value of t in Exp 2 we get,

sum = x - x³/3! + x⁵/5! 

Iteration 3: i = 3, t = x⁵/5!, sum = x - x³/3! + x⁵/5!

Putting these values in Exp 1 we get,

t = (x⁵/5! * pow((double) (-1), (double) (2 * 3 - 1)) * x * x) / (2 * 3 * (2 * 3 + 1))

t = (x⁵/5!)*(-1)*x*x/42 = -x⁷/7!

Putting value of t in Exp 2 we get,

sum = x - x³/3! + x⁵/5! - x⁷/7!

After iteration 3, i = 4 so the for loop terminates and output is printed.