In Exercises 41–48 find the fourth Taylor polynomial $P_4\left(x\right)$for the specified function and the given value of $x_0$.

$Sinx,\mathrm{\pi }$

$Sinx,\mathrm{\pi }$

Since for any function$f$with a derivative of order 4 at $x=\pi$, the fourth Taylor polynomial for $x=\pi$ is given by

$P_4\left(x\right)=f\left(\pi \right)+f^{\text{'}}\left(\pi \right)(x-\pi )+\frac{f^{\text{'}\text{'}}\left(\pi \right)}{2!}(x-\pi {)}^2+\frac{f^{\text{'}\text{'}}\left(\pi \right)}{3!}(x-\pi {)}^3+\frac{f^{\text{'}\text{'}\text{'}\text{'}}\left(\pi \right)}{4!}(x-\pi {)}^4$

Therefore, first find the value of the function along with $f^{\text{'}}\left(x\right),f^{\text{'}\text{'}}\left(x\right),f^{\text{'}\text{'}\text{'}}\left(x\right)$and $f^{\text{'}\text{'}\text{'}\text{'}}\left(x\right)$at $x=\pi$

$$\begin{gathered} Thus,thevalueofthethefunctionstx=\mathrm{\pi }is \\ f\left(\mathrm{\pi }\right)=\sin \left(\mathrm{\pi }\right) \\ =0 \\ Thederivativesofthefunctionf\left(x\right)=\sin \left(x\right)are \\ f\text{'}\left(x\right)=\frac{d}{dx}\left[\sin x\right] \\ =\cos x \\ So,atx=\mathrm{\pi } \\ f\text{'}\left(\mathrm{\pi }\right)=\cos \left(\mathrm{\pi }\right) \\ =-1 \\ Also, \\ f\text{'}\text{'}\left(x\right)=\frac{d}{dx}\left[\cos x\right] \\ =-\sin x \\ So,atx=\mathrm{\pi } \\ f\text{'}\left(\mathrm{\pi }\right)=-\sin \left(\mathrm{\pi }\right) \\ =0 \\ Again, \\ f\text{'}\text{'}\text{'}\left(x\right)=-\frac{d}{dx}\left[\sin x\right] \\ =-\cos x \\ So,atx=\mathrm{\pi } \\ \mathrm{f}\text{'}\text{'}\text{'}\left(\mathrm{\pi }\right)=-\cos \left(\mathrm{\pi }\right) \\ =-(-1) \\ =1 \\ f\text{'}\text{'}\text{'}\text{'}\left(x\right)=-\frac{d}{dx}\left[\cos x\right] \\ =-(-\sin x) \\ =\sin x \\ So,atx=\frac{\mathrm{\pi }}{2} \\ f\text{'}\left(\mathrm{\pi }\right)=\sin \left(\mathrm{\pi }\right) \\ =0 \end{gathered}$$

Therefore, the fourth Taylor polynomial for the function $f\left(x\right)=\sin x$is

$P_4\left(x\right)=0+-1·(x-\pi )+\frac{0}{2!}(x-\pi {)}^2+\frac{1}{3!}(x-\pi {)}^3+\frac{0}{4!}(x-\pi {)}^4$

$=-(x-\pi )+\frac{1}{3!}(x-\pi {)}^3=-(x-\pi )+\frac{1}{6}(x-\pi {)}^3$