Find the few terms of the Taylor series for the following function about the given points.

$\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathbf{,}\mathit{a}\mathbf{=}\mathit{\pi }$

Function is $\sin x\text{and}a=\pi$.

Taylor's series of about is expressed as follows:

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

Differentiate the function $\sin x$and substitute$\pi$ :

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

Differentiate the function $\sin x$ two times and substitute$\pi$ for $x$as follows:

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

Differentiate the function $\sin x$three times and substitute $\pi$for $x$as follows:

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

Differentiate the function $\sin x$four times and substitute$\pi$ for $x$as follows:

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

Substitute all the values of derivatives of $\pi$for $x$in Taylor’s equation as follows:

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

Thus, the first few terms of the Taylor series$\sin x$at $a=\pi$is given as:

$-(x-\pi )+\frac{1}{6}(x-\pi {)}^3-\frac{1}{120}(x-\pi {)}^5+\dots ..$