Find a Maclaurin series expansion for each function.
f(x) = cos x
f(x) = 1/(1-x)
f(x) = x sin x
These are pretty standard but if the intent is to "discover" these on your own, just directly apply the formula for the Maclaurin series (a taylor series centred around a = 0):
$\displaystyle f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^{2} + ... + \frac{f^{(n)}(a)}{n!}(x-a)^{n} + ...$
Taking a = 0: $\displaystyle f(x) = f(0) + f'(a)(x) + \frac{f''(a)}{2!}x^{2} + ... + \frac{f^{(n)}(0)}{n!}x^{n} + ... $
For the f(x) = cos x, you would need to find:
$\displaystyle f(0) = \cos (0)$, $\displaystyle f'(0) = -\sin (0)$, $\displaystyle f''(0) = -\cos (0)$ ..
Then generalize a pattern and that should be your Maclaurin series (Of course, if you wanted to prove it you can do it with induction to verify your series).
Also, a google search for maclaurin series would probably give you these results.
Using o_O's above formula we see that
$\displaystyle cos(x)\cos(0)-\sin(0)x-\frac{\cos(0)x^2}{2!}+\frac{\sin(0)x^3}{3!}+\frac{ \cos(0)x^4}{4!}+...=$$\displaystyle 1-\frac{x^2}{2!}+\frac{x^4}{4!}-...=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}$
this is applicable $\displaystyle \forall{x}\in\mathbb{R}$
Because applying the use of the root test for finding intervals of convergence we get
$\displaystyle \lim_{n\to\infty}\bigg|\frac{x^{2n}}{(2n)!}\bigg|^ {\frac{1}{n}}=0<1$
Therefore we run into an identity verifying that cos(x) is analytic across the reals
for
$\displaystyle \frac{1}{1-x}$
Either recognize this as the converse of the sum of an infinite geometric seires or notice that the temrs are given by
$\displaystyle \frac{1}{1-x}=1+x+x^2+x^3+...=\sum_{n=0}^{\infty}x^n$
Using the Root test again we get
$\displaystyle \lim_{n\to\infty}|x^n|^{\frac{1}{n}}=\lim_{n\to\in fty}|x|=|x|<1\Rightarrow{-1<x<1}$
Now seeing that at x=1 and x=-1 we have a divergent series this function is analytic only on $\displaystyle |x|<1$
For the third I will leave it up to you to show the termiwise derivation but
$\displaystyle \sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}$
$\displaystyle \therefore{x\cdot\sin(x)=x\cdot\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+1}}{(2n+1)!}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+2}}{(2n+1)!}}$
and applying the ratio or root test again we see this is also analytic across the reals