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

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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