91Ó°ÊÓ

We have the given function $f\left(x\right)=\cos \left(2x\right)$with a derivative of order$4$ at $x=\frac{\mathrm{Ï€}}{4}$.

The fourth taylor polynomial for $x=\frac{\mathrm{Ï€}}{4}$is given by,

$\begin{array}{rl}{P}_4\left(x\right)=& f\left(\frac{\mathrm{Ï€}}{4}\right)+f^{\mathrm{′}}\left(\frac{\mathrm{Ï€}}{4}\right)(x−\frac{\mathrm{Ï€}}{4})+\frac{f^{′â\P IJ}\left(\frac{\mathrm{Ï€}}{4}\right)}{2!}{(x−\frac{\mathrm{Ï€}}{4})}^2+\frac{f^{′â\P IJ\mathrm{′}}\left(\frac{\mathrm{Ï€}}{4}\right)}{3!}{(x−\frac{\mathrm{Ï€}}{4})}^3+\frac{f^{′â\P IJ′â\P IJ}\left(\frac{\mathrm{Ï€}}{4}\right)}{4!}{(x−\frac{\mathrm{Ï€}}{4})}^4\end{array}$

Therefore, we have to 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 $x=\frac{\mathrm{Ï€}}{4}$.

The value of the function at $x=\frac{\mathrm{Ï€}}{4}$is,

$\begin{array}{rl}f\left(\frac{\mathrm{Ï€}}{4}\right)& =\cos (2â‹\dots \frac{\mathrm{Ï€}}{4})\\ & =\cos \left(\frac{\mathrm{Ï€}}{2}\right)\\ & =0\end{array}$

The derivatives of the function, $f\left(x\right)=\cos \left(2x\right)$

$\begin{array}{rl}{f}^{\mathrm{′}}\left(x\right)& =\frac{d}{dx}\left[\cos 2x\right]\\ & =−2\sin 2x\end{array}$$\begin{array}{rl}{f}^{\mathrm{′}}\left(x\right)& =\frac{d}{dx}\left[\cos 2x\right]\\ & =−2\sin 2x\end{array}$

So, at $x=\frac{\mathrm{Ï€}}{4}$

$\begin{array}{rl}{f}^{\mathrm{′}}\left(0\right)& =−2\sin (2â‹\dots \frac{\mathrm{Ï€}}{4})\\ & =−2\sin \left(\frac{\mathrm{Ï€}}{2}\right)\\ & =−2â‹\dots 1\\ & =−2\end{array}$

$\begin{array}{rl}{f}^{′â\P IJ}\left(x\right)& =\frac{d}{dx}[−2\sin 2x]\\ & =−2\frac{d}{dx}\left[\sin 2x\right]\\ & =−4\cos 2x\end{array}$

So, at $x=\frac{\mathrm{Ï€}}{4}$

$\begin{array}{rl}{f}^{′â\P IJ}\left(0\right)& =−4\cos (2â‹\dots \frac{\mathrm{Ï€}}{4})\\ & =−4\cos \left(\frac{\mathrm{Ï€}}{2}\right)\\ & =−4â‹\dots 0\\ & =0\end{array}$

$\begin{array}{rl}{f}^{′â\P IJ\mathrm{′}}\left(x\right)& =−4\frac{d}{dx}\left[\cos 2x\right]\\ & =−4(−2\sin 2x)\\ & =8\sin 2x\end{array}$

So, at $x=\frac{\mathrm{Ï€}}{4}$

$\begin{array}{rl}{f}^{′â\P IJ\mathrm{′}}\left(0\right)& =8\sin (2â‹\dots \frac{\mathrm{Ï€}}{4})\\ & =8\sin \left(\frac{\mathrm{Ï€}}{2}\right)\\ & =8â‹\dots 1\\ & =8\end{array}$

$\begin{array}{rl}{f}^{′â\P IJ′â\P IJ}\left(x\right)& =\frac{d}{dx}\left[8\sin 2x\right]\\ & =8\frac{d}{dx}\left[\sin 2x\right]\\ & =16\cos 2x\end{array}$

So, at $x=\frac{\mathrm{Ï€}}{4}$

$\begin{array}{rl}{f}^{′â\P IJ′â\P IJ}\left(0\right)& =16\cos (2â‹\dots \frac{\mathrm{Ï€}}{4})\\ & =16\cos \left(\frac{\mathrm{Ï€}}{2}\right)\\ & =16â‹\dots 0\\ & =0\end{array}$

Hence, the fourth Taylor polynomial of the function $f\left(x\right)=\cos 2x$at$x=\frac{\mathrm{Ï€}}{4}$

$\begin{array}{rl}{P}_4\left(x\right)& =0+−2â‹\dots (x−\frac{\mathrm{Ï€}}{4})+\frac{0}{2!}{(x−\frac{\mathrm{Ï€}}{4})}^2+\frac{8}{3!}{(x−\frac{\mathrm{Ï€}}{4})}^3+\frac{0}{4!}{(x−\frac{\mathrm{Ï€}}{4})}^4\\ & =−2(x−\frac{\mathrm{Ï€}}{4})+\frac{8}{3!}{(x−\frac{\mathrm{Ï€}}{4})}^3\\ & =−2(x−\frac{\mathrm{Ï€}}{4})+\frac{4}{3}{(x−\frac{\mathrm{Ï€}}{4})}^3\end{array}$