Finding an approximation for cos(x) using for loops

Hi there,

I recently got help on creating a script to calculate the value of cos(x) using for loops. The question asks:

"Write a script which will calculate the value of cos(x) (ask the user for the value of x and n):

cos(x)=( from k=1 –> n )∑(-1) ̂(k-1)(x ̂(2(k-1))/(2(k-1))!)

" I had real problems getting the answer and so asked a tutor to help me with it. She managed to get the code working but I failed to understand how it actually works. (her explanation was not very clear)

So, I have sat staring at it for ages and it still doesn't make sense to me.

Here it is:

% Ask user for values x, n
x=input('Please enter a value for x: '); %cos angle
x_rad=x*pi/180;
n=input('Please enter a value for n: '); %nth term
%factorial
fact=1;
sum=0;
for k=1:n
  initial=-1;
  for a=0:(k-1)
    initial=(-1)*initial;
  end
  numerator=1;
  for b=1:2*(k-1)
    numerator=numerator*x_rad;
  end
  for t=1:2*(k-1)
    fact=fact*t;
  end
  total=(initial*numerator)/fact;
  sum=sum+total;
end
fprintf('cos(%d)=%6.6f \n',x,sum);

Can anyone explain how this code is working in the for loop? I am so confused.

Also, any advice on a simpler way to do this?

Thanks so much, Laura

[Walter Roberson]

Which "for" loop ?

[Laura]

All of them. I don't understand why we need nested for loops in the first place and why the end values for each one is different. (e.g. for b=1:2*(k-1))

Answer 1

[Umair Nadeem]

The code has been developed using a more complex approach. You can simply implement the equation you gave above. Here is how to do it.

% Ask user for values x, n
x=input('Please enter a value for x: '); %cos angle

x_rad=x*pi/180;

n=input('Please enter a value for n: '); %nth term

% Initialize resultant variable
sum=0;

for k=1:n
  initial = (-1)^(k-1);
  numerator = x_rad^(2*(k-1));
        denominator = factorial(2*(k-1));
  total=(initial*numerator)/denominator;
  sum=sum+total;
end

fprintf('cos(%d)=%6.6f \n',x,sum);

It will give the same result. You dont have to use nested for loops to calculate the values of ^ (to the power), you can do it right away.

Hope it helps

[Weeraphol Egsaphung]

Hi, I wonder the value of >> fprintf('cos(%d)=%6.6f \n',x,sum); if I want it sin How could I use the code