Question

Using the first three terms of the Maclaurin's series, what is the value of cos(Ï€/2

Answer 1

Do you mean cos(x/2)?

If you do, its easier to start by differentiating the number of levels you need, in your case here, 3.
f = cos(x/2)
This derivative requires the chain rule.
Let u = (x/2), then the chain rule says that we will have
dcos(u)/du * d(x/2)/dx = -sin(u) * 1/2
Substitute back in and you get
f^1 = -1/2sin(x/2)
Do the same thing for f^2 and f^3 and I would recommend f^4 also.

Now the Maclaurin series is the Taylor series at x = 0, so some of these terms will go away.
The first term of the Maclaurin series is f(0),
which in your case is cos(0/2) = cos(0) = 1
The second term goes away as sin(0) = 0. If you have done the other two derivatives you will see that the next term is -1/4cos(0/2) = -1/4 * (x^2/2!)
The third term goes away again so it might be good to get another derivative in here.
The fourth term comes out to be 1/16cos(0/2) = 1/16 * (x^4/4!)

The first three terms of cos(x/2) is:
1 - (1/4)*(x^2/2!) + (1/16)*(x^4/4!)
This is an alternating series and as the limit of x->0 this will come out to be 2.

Hope this helps a little.

Answer 2

\[\cos x = 1 -(x ^{2}/2!)+(x ^{4}/4!)\] where x = \[\pi/2\]