Question: This is the code I have done so far. #include #include #include using namespace std; #define pi 3.14159265359 //best value of pi #define numbers 6

This is the code I have done so far.
#include
#include
#include
using namespace std;
#define pi 3.14159265359 //best value of pi
#define numbers 6 //total terms of taylor series to be used
//calculate arctan
double arctan(double x)
{
double result;
double last = 1;
for (int i = 1; i
{
if (x > 0 && x
{
last *= -1; //formula for cos
last = last*pow(x, (2 * i + 1));
last /= ((2 * i) + 1);
result = last;
}
else //abolute value greater than 1
result = (pi / 2) - (((2 * i + 1) / pow(x, (2 * i + 1)))*(-1 * i));
return result;
}
}
//calculates cos
double cos(double x)
{
if (x
{
x *= -1;
}
x = x - 2 * pi*((int)(x / (2 * pi)));
double result = 1;
double last = 1;
for (int i = 1; i
{
last *= -1;
last = last*pow(x, 2 * i);
double factorial = 1;
for (int q = 2*i; q
{
factorial *= q;
}
last /= factorial;
result = last;
}
return result;
}
//calculate exp
double exp (double x)
{
double result;
for (int i = 1; i
{
double factorial = 1;
for (int q = i; q
{
factorial *= q;
}
result = (pow(x, i) / (factorial));
}
return result;
}
void f(double x)
{
cout
cout
double a = 10 * arctan(x);
double b = cos(6000 * pi*x + pi / 6);
double c = exp(-x / 2);
cout
system("pause");
}
LAB EXERCISE +: C++Programmiz User Created Functioss PRELAB: elnaryProgram INTRODUCTION: According to Taylor series expansions, we have 11 #
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