In the last post, I discussed about how would one go about calculating the sum of a finite series using C.

In this post I will apply that method, to find the sum of the Sine series for only a finite number of terms.

**Sine series **is given by:

We will start the numbering the terms from 0. That is, , , ….

It’s easy to see that the ratio between consecutive terms is given by:

Since, we indexed the terms starting from 0, therefore, for the above relation to work, will go from 1 to .

[Hint: To find the general form of the ratio given in the above expression, try writing down t1/t0, t2/t1,…and then you would be able to see the ratio.]

Now, knowing the first() term, the successive terms can be calculated as :

and so on.

Therfore, the C program that calculates the sum of the sin series upto a given number of terms can be written as shown below.

### PROGRAM:

/******************************** ******FINITE SERIES SUM********** Series: sin(x) = x - (x^3/3!) + (x^5/5!) + ..... ********************************/ #include<stdio.h> #include<math.h> main(){ int i,n; double x,t0,t1,R,sum; printf("Enter the value of x:\n"); scanf("%lf",&x); printf("Enter the no. of terms to be summed: "); scanf("%d",&n); //Initialize First Term t0=x; //Make sum equal to the first term sum=t0; printf("n\ttn\t\tSn\n_________________________________"); for(i=1;i<n;i++){ //Find the ratio of the second term to the first term using already known relation R=-(x*x)/(2*i+1)/(2*i); //Calculate the second term t1=R*t0; //find the new sum sum=sum+t1; t0=t1; printf("\n%d\t%f\t%lf\n",i+1,t1,sum); } printf("\nThe sum is: %f",sum); }

The program also prints the value of each term(except the first() term) and sum(partial) upto that term.

### OUTPUT:

The output of the above program for various values of and no. of terms is shown below: