Exercise #3, Functions Given a positive integer n and a floating point number x between 0.0 and 360.0. The objective of this exercise is to write a C++ program to compute the below math equations: y=xxπ/180 S(y, n) = y- y3 y5 y7 v9 y11 + 3! 5! 7! 9! + 11! We decompose the eqations into 5 simplified expressions as follows: 1. y = r(x) = x x π/180 2. p(y,k) = yk 3. f(k) = k! where k is an integer smaller than or equal to n 4. q(y,k) = P(y,k) f(k) 5. S(y,n) =y-q(y,3)+q(y, 5)-q(y, 7) + q(y, 9) -q(y, 11)+...+q(y,n) Method 1 int fact, sign=-1; float y = x*M_PI/180, p, sum =0; for(int i=1;i<=n; i+=2) Write C++ functions to calculate each of the above expressions. In your main, you prompt the user to enter x and n, control the user input, then call function S(x,n) to compute expression 5. Function S(x, n) will call function r(x) to get y, then it calls other functions as needed to complete the computations and display your result at the end. e.g. Value of x: 30 Number of interations n: 12 S(30,12): 0.5 p= 1; fact= 1; for(int j=1; j<=i; j++) +... Two other programmers calculated the above expression using different methods, in Method 1 (see below), the programmer used two loops, in Method 2 the programmer used one single loop. p= p*y; fact=fact*j; } sign=-1*sign; sum += sign*p/fact; ± yn n! if(i==1){ p= y; fact= 1; } else { int fact, sign=-1; float y = x*M_PI/180, p, sum = 0; for(int i=1;i<=n; i+=2) { p= p*y*y; fact= fact* (i-1)*(i); } Method 2 sign=-1*sign; sum += sign*p/fact; } Try to understand the above methods, test them, then compare the three implementations (yours, method1, and method2) to find out: which one is the best, what are the advantages and disadvantages of each method. By the way, when you test the three methods: They all should GIVE THE SAME RESULT. This is one way to verify that your implementation is correct. e.g. S(30,12)= 0.5 S(45,12)= 0.707107 S(60, 12) = 0.866025