Polinomio de Taylor

Derivando

$$\begin{array}{ccc}f(x)=\cos x& f(0)=1& f\left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}\\ f\text{'}(x)=-senx& f\text{'}(0)=0& f\text{'}\left(\frac{\pi }{4}\right)=-\frac{\sqrt{2}}{2}\\ f\text{'}\text{'}(x)=-\cos x& f\text{'}\text{'}(0)=-1& f\text{'}\text{'}\left(\frac{\pi }{4}\right)=-\frac{\sqrt{2}}{2}\\ f\text{'}\text{'}\text{'}(x)=senx& f\text{'}\text{'}\text{'}(0)=0& f\text{'}\text{'}\text{'}\left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}\\ f^{(iv}(x)=\cos x& f^{(iv}(0)=1& f^{(iv}\left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}\\ \mathrm{...}& \mathrm{...}& \mathrm{...}\\ f^{(2n}(x)={(-1)}^n\cos x& f^{(2n}(0)={\left(-1\right)}^n& f^{(2n}\left(\frac{\pi }{4}\right)={(-1)}^n\frac{\sqrt{2}}{2}\\ f^{(2n+1}(x)={(-1)}^{n+1}senx& f^{(2n+1}(0)=0& f^{(2n+1}\left(\frac{\pi }{4}\right)={(-1)}^{n+1}\frac{\sqrt{2}}{2}\end{array}$$ por lo que: $$T_{2n}[\cos x;0]=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\mathrm{...}+\frac{{(-1)}^n{x}^{2n}}{(2n)!}$$ $$\begin{array}{l}{T}_{2n+1}[\cos x;\frac{\pi }{4}]=\frac{\sqrt{2}}{2}(1-\left(x-\frac{\pi }{4}\right)-\frac{1}{2}{\left(x-\frac{\pi }{4}\right)}^2+\mathrm{...}\hfill \\ \mathrm{...}+\frac{{(-1)}^n}{(2n)!}{\left(x-\frac{\pi }{4}\right)}^{2n}+\frac{{(-1)}^{n+1}}{(2n+1)!}{\left(x-\frac{\pi }{4}\right)}^{2n+1})\hfill \end{array}$$