<s>[INST] 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 $$1+\frac{1}{2}\sum_{n=1}^{\infty}\frac{({z})^{2n}}{{2n!}}$$.
 [/INST] 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.
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