I have developed the the taylor series of $\cos^{2}x$.
$$\sin x=\sum_{n=0}^{\infty}\frac{(-1)^n\cdot x^{2n+1}}{(2n+1)!}$$
$$\cos x=\sum_{n=0}^{\infty}\frac{(-1)^n\cdot (2n+1)\cdot x^{2n}}{(2n+1)!}=\sum_{n=0}^{\infty}\frac{(-1)^n\cdot x^{2n}}{(2n)!}$$
$$\cos^{2}x=\frac{1}{2}(1+\cos 2x)=\frac{1}{2}+\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^n\cdot x^{2n}}{(2n)!}$$
Now I have to bring the series to a form of $\sum a_nx^n$
Can I used $t=2n\Rightarrow n=\frac{t}{2}$
$$\frac{1}{2}\sum_{\frac{t}{2}=0}^{\infty}\frac{(-1)^{\frac{t}{2}}\cdot x^{t}}{(t)!}$$
So $$a_n=\frac{(-1)^{\frac{t}{2}}}{t!}$$
So $R=\lim_{t\to \infty}|\frac{a_n}{a_{n+1}}|=\frac{(1)}{t!}\cdot\frac{(t+1)!}{1}=\lim_{t\to \infty} (t+1)=\infty$
Are the steps correct? can I say that beacuse $$\sin x=\sum_{n=0}^{\infty}\frac{(-1)^n\cdot x^{2n+1}}{(2n+1)!}$$ had $R=\infty$ so does the new series?
Since $$ \cos(x)=\sum_{n\geq 0}\frac{(-1)^n x^{2n}}{(2n)!} \tag{1}$$ we have: $$ \cos^2(x) = \frac{1+\cos(2x)}{2} = 1+\sum_{n\geq 1}\frac{(-1)^n 2^{2n-1} x^{2n}}{(2n)!}\tag{2} $$ and both $\cos(x)$ and $\cos^2(x)$ are entire functions, with radius of convergence $+\infty$.