91Ó°ÊÓ

Function is $\sin x\text{and}a=\mathrm{Ï€}$.

Taylor's series of about is expressed as follows:

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{+}\frac{{\mathbf{f}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{a}\mathbf{\right)}}{\mathbf{1}\mathbf{!}}\mathbf{(}\mathit{x}\mathbf{-}\mathit{a}\mathbf{)}\mathbf{+}\frac{{\mathbf{f}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{\left(}\mathbf{a}\mathbf{\right)}}{\mathbf{2}\mathbf{!}}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{a}{\mathbf{)}}^{\mathbf{2}}\mathbf{+}\frac{{\mathbf{f}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{\left(}\mathbf{a}\mathbf{\right)}}{\mathbf{3}\mathbf{!}}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{a}{\mathbf{)}}^{\mathbf{3}}\mathbf{+}\mathbf{…}\mathbf{…}\mathbf{.}$

Differentiate the function $\sin x$and substitute$\mathrm{Ï€}$ :

$$\begin{gathered} f^{\text{'}}\left(x\right)=\frac{d\sin x}{dx} \\ {f}^{\text{'}}\left(x\right)=\cos x \\ {f}^{\text{'}}\left(\mathrm{Ï€}\right)=\cos \mathrm{Ï€} \\ {f}^{\text{'}}\left(\mathrm{Ï€}\right)=-1 \end{gathered}$$

Differentiate the function $\sin x$ two times and substitute$\mathrm{Ï€}$ for $x$as follows:

$\begin{array}{rcl}{f}^{\text{'}\text{'}}\left(x\right)& =& \frac{d^2}{d{x}^2}\left(\sin x\right)\\ f^{\text{'}\text{'}}\left(x\right)& =& -\sin x\\ f^{\text{'}\text{'}}\left(\mathrm{Ï€}\right)& =& -\sin \mathrm{Ï€}\\ f^{\text{'}\text{'}}\left(\mathrm{Ï€}\right)& =& 0\\ & & \end{array}$

Differentiate the function $\sin x$three times and substitute $\mathrm{Ï€}$for $x$as follows:

$\begin{array}{rcl}{f}^{\text{'}\text{'}\text{'}}\left(x\right)& =& \frac{d^3}{d{x}^3}\left(\sin x\right)\\ f^{\text{'}\text{'}\text{'}}\left(x\right)& =& -\cos x\\ f^{\text{'}\text{'}\text{'}}\left(\mathrm{Ï€}\right)& =& -\cos \mathrm{Ï€}\\ f^{\text{'}\text{'}\text{'}}\left(\mathrm{Ï€}\right)& =& 1\\ & & \end{array}$

Differentiate the function $\sin x$four times and substitute$\mathrm{Ï€}$ for $x$as follows:

$$\begin{gathered} f^{\text{'}\text{'}\text{'}}\left(x\right)=\frac{d^4}{d{x}^4}\left(\sin x\right) \\ {f}^{\text{'}\text{'}\text{'}}\left(x\right)=\sin x \\ {f}^{\text{'}\text{'}\text{'}}\left(\mathrm{Ï€}\right)=\sin \mathrm{Ï€} \\ {f}^{\text{'}\text{'}\text{'}}\left(\mathrm{Ï€}\right)=0 \end{gathered}$$

Substitute all the values of derivatives of $\mathrm{π}$for $x$in Taylor’s equation as follows:

$$\begin{gathered} \sin x=\sin \mathrm{π}+\frac{-1}{1!}(\mathrm{π}-a)+\frac{0}{2!}(\mathrm{π}-a{)}^2+\frac{1}{3!}(\mathrm{π}-a{)}^3+\frac{0}{4!}(\mathrm{π}-a{)}^4+\frac{-1}{5!}(\mathrm{π}-a{)}^5……. \\ \sin x=-(x-\mathrm{π})+\frac{1}{6}(x-\mathrm{π}{)}^3-\frac{1}{120}(x-\mathrm{π}{)}^5+….. \end{gathered}$$

Thus, the first few terms of the Taylor series$\sin x$at $a=\mathrm{Ï€}$is given as:

$-(x-\mathrm{π})+\frac{1}{6}(x-\mathrm{π}{)}^3-\frac{1}{120}(x-\mathrm{π}{)}^5+…..$