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 = -x

^{3}/3!
Putting value of t in Exp 2 we get,

sum = x + (-x

^{3}/3! ) = x - x^{3}/3!**Iteration 2: i = 2, t =**

**-x**

^{3}/3!, sum = x - x^{3}/3!
Putting these values in Exp 1 we get,

t = (-x

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

^{3}/3!)*(-1)*x*x/20 = x^{5}/5!
Putting value of t in Exp 2 we get,

sum = x - x

^{3}/3! + x^{5}/5!**Iteration 3: i = 3, t = x**+ x

^{5}/5!, sum = x - x^{3}/3!^{5}/5!
Putting these values in Exp 1 we get,

t = (x

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

^{5}/5!)*(-1)*x*x/42 = -x^{7}/7!
Putting value of t in Exp 2 we get,

sum = x - x

^{3}/3! + x^{5}/5! - x^{7}/7!
After iteration 3, i = 4 so the for loop terminates and output is printed.

what is pow

@Chaitanya Reddy: pow is power function in C. For more information please refer this post - http://cprogramming.language-tutorial.com/2012/01/c-program-to-find-power-of-number-using.html

Why use double

in the brackets?

To avoid precision loss. Since pow((-1), (2 * i - 1)) i.e. without casting to double will result in an integer value. But in the above program, you are multiplying this value to a floating point value i.e. variable "t" and when float/double is multiplied by int type, precision loss occurs.

That is why pow((double) (-1), (double) (2 * i - 1)) equation is used which will result output in double. Double in brackets means that you are casting the value to double.

explain briefly.... cnt understad.. process of it.

@Teja: Explanation has been added in the post.

can we write t=(pow(-1,i)*pow(x,2i-1))/2 * i * (2 * i + 1)) ; insted of t = (t * pow((double) (-1), (double) (2 * i - 1)) *

x * x) / (2 * i * (2 * i + 1)) ; with out initializing t=x

why cant we use -1.0 instead of pow((double) (-1), (double) (2 * i - 1))

THANKS MAN YOU WERE SO HELPFUL!!!