Taylor expansion of $\cos^2(\frac{iz}{2})$ 
Expand $\cos^2(\frac{iz}{2})$ around $a=0$

We know that $$\cos t=\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n}}{{2n!}}$$
So $$\cos^2t=[\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n}}{{2n!}}]^2=\sum_{n=0}^{\infty}(-1)^{2n}\frac{t^{4n}}{{4n^2!}}$$
We have $t=\frac{iz}{2}$
$$\sum_{n=0}^{\infty}(-1)^{2n}\frac{(\frac{zi}{2})^{4n}}{{4n^2!}}=\sum_{n=0}^{\infty}(-1)^{2n}\frac{(\frac{zi}{2})^{4n}}{{4n^2!}}=\sum_{n=0}^{\infty}(-1)^{2n}\frac{({zi})^{4n}}{2^{4n}{4n^2!}}=\sum_{n=0}^{\infty}(-1)^{2n}\frac{({z})^{4n}}{2^{4n}{4n^2!}}$$
But the answer in the book is $$1+\frac{1}{2}\sum_{n=1}^{\infty}\frac{({z})^{2n}}{{2n!}}$$.
 A: If $f(x) = \sum_{n=0}^{\infty} a_n x^n$ then 
$$ f^2(x) = \left( \sum_{n=0}^{\infty} a_n x^n \right) \left( \sum_{m=0}^{\infty} a_m x^m \right) = \sum_{k = 0}^{\infty} \left( \sum_{l = 0}^k a_l a_{k - l} \right) x^k \neq \sum_{n=0}^{\infty} a_n^2 x^{2n}.$$
In your case, 
$$ \cos^2(x) = \left( \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n} \right) \left( \sum_{m=0}^{\infty} \frac{(-1)^m}{(2m)!} x^{2m} \right) = \sum_{k=0}^{\infty} \left( \sum_{l=0}^k \frac{(-1)^l}{(2l)!} \frac{(-1)^{k - l}}{(2(k-l))!} \right) x^{2k} = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} \left( \sum_{l=0}^k { 2k \choose 2l} \right) x^{2k} = \sum_{k=0}^{\infty} \frac{(-1)^k 2^{2k-1}}{(2k)!} x^{2k}. $$
Plug in $x = \frac{iz}{2}$ and see what you get.
A: $$ \cos^2(\frac{iz}{2}) = \frac{1}{2}(1+\cos(2*\frac{iz}{2})) = \frac{1}{2}(1+\cos(iz)) = \frac{1}{2}(1+\sum_{n=0}^{+\infty} (-1)^n\frac{(iz)^{2n}}{(2n)!}) = \frac{1}{2}(1+\sum_{n=0}^{+\infty} (-1)^n (i)^{2n}\frac{(z)^{2n}}{(2n)!}) = \frac{1}{2}(1+\sum_{n=0}^{+\infty} (-1)^{2n}\frac{(z)^{2n}}{(2n)!}) = \frac{1}{2}(1+\sum_{n=0}^{+\infty} \frac{(z)^{2n}}{(2n)!}) = \frac{1}{2}(1+1+\sum_{n=1}^{+\infty}\frac{(z)^{2n}}{(2n)!}) = 1+\frac{1}{2}\sum_{n=1}^{+\infty} \frac{(z)^{2n}}{(2n)!} $$
A: We know $ e^{z}= \sum_{n=0}^{\infty} \frac{z^n}{n!}$ and $ \cos z = \frac{e^{iz}+e^{-iz}}{2}$
so $\cos^2 \frac{iz}{2} = \frac{e^z + e^{-z}}{4} + \frac{1}{2}= 1 + \frac{1}{2} \sum_{n=1}^{\infty} \frac{z^{2n}}{2n!}$
A: $\cos^2 \left(\dfrac{i z}{2}\right)=\cosh^2 \left(\dfrac{ z}{2}\right)$
Remember the identity
$\cosh^2 \dfrac{z}{2}=\dfrac{1}{2} (\cosh z +1)$
we know that $\cosh z =\sum _{n=0}^{\infty } \dfrac{z^{2 n}}{n!}$
therefore we have
$$\cos^2 \left(\dfrac{i z}{2}\right)=\sum _{n=0}^{\infty } \dfrac{z^{2 n}}{2 n!}$$
A: Since levap already explained that you failed at squaring the series, I want to show a different approach:
There is the identity
$$
\cos^2(x) = \frac{1}{2}\big(1+\cos(2x)\big)
$$
which is an easy consequence of
$$
\cos(x+y) = \cos(x)\cos(y)-\sin(x)\sin(y)\quad \text{and}\quad \cos^2(x)+\sin^2(x) =1 
$$
Anyway if you use the mentioned identity you only need the taylor series of $\cos$ instead of calculate the taylor series of $\cos^2$. Now you get by using $\cos(2x) = \sum_{n=0}^\infty (-1)^n \frac{(2x)^{2n}}{(2n)!}$
$$
\cos^2(x) = \frac{1}{2} \Big( 1 + \sum_{n=0}^\infty (-1)^n \frac{(2x)^{2n}}{(2n)!}\Big)
$$
Now plug in $x=\frac{\mathrm{i}z}{2}$
$$
\cos^2\Big(\frac{\mathrm{i}z}{2}\Big)
= \frac{1}{2}\Big(1 + \sum_{n=0}^\infty (-1)^n \frac{(2\frac{\mathrm{i}z}{2})^{2n}}{(2n)!}\Big)
=\frac{1}{2} \Big( 1 +\sum_{n=0}^\infty (-1)^n \frac{\mathrm{i}^{2n}z^{2n}}{(2n)!}\Big)
$$
Note that $\mathrm{i}^{2n} = (\mathrm{i}^2)^n = (-1)^n$ and $(-1)^n (-1)^n=1$. Additionally you can take out the first summand and recieve
$$
\cos^2\Big(\frac{\mathrm{i}z}{2}\Big)
=\frac{1}{2}\Big( 2 + \sum_{n=1}^\infty \frac{z^{2n}}{(2n)!} \Big)
=1+\frac{1}{2} \sum_{n=1}^\infty \frac{z^{2n}}{(2n)!}
$$
