# 009C Sample Final 1, Problem 6

Find the Taylor polynomial of degree 4 of ${\displaystyle f(x)=\cos ^{2}x}$ at ${\displaystyle a={\frac {\pi }{4}}}$.

Foundations:
The Taylor polynomial of ${\displaystyle f(x)}$ at ${\displaystyle a}$ is
${\displaystyle \sum _{n=0}^{\infty }c_{n}(x-a)^{n}}$ where ${\displaystyle c_{n}={\frac {f^{(n)}(a)}{n!}}.}$

Solution:

Step 1:
First, we make a table to find the coefficients of the Taylor polynomial.
 ${\displaystyle n}$ ${\displaystyle f^{(n)}(x)}$ ${\displaystyle f^{(n)}(a)}$ ${\displaystyle {\frac {f^{(n)}(a)}{n!}}}$ ${\displaystyle 0}$ ${\displaystyle \cos ^{2}x}$ ${\displaystyle {\frac {1}{2}}}$ ${\displaystyle {\frac {1}{2}}}$ ${\displaystyle 1}$ ${\displaystyle -2\cos x\sin x}$ ${\displaystyle -1}$ ${\displaystyle -1}$ ${\displaystyle 2}$ ${\displaystyle 2\sin ^{2}x-2\cos ^{2}x}$ ${\displaystyle 0}$ ${\displaystyle 0}$ ${\displaystyle 3}$ ${\displaystyle 8\sin x\cos x}$ ${\displaystyle 4}$ ${\displaystyle {\frac {2}{3}}}$ ${\displaystyle 4}$ ${\displaystyle 8\cos ^{2}x-8\sin ^{2}x}$ ${\displaystyle 0}$ ${\displaystyle 0}$
Step 2:
Since ${\displaystyle c_{n}={\frac {f^{(n)}(a)}{n!}},}$ the Taylor polynomial of degree 4 of ${\displaystyle f(x)=\cos ^{2}x}$ is
${\displaystyle T_{4}(x)={\frac {1}{2}}+-1{\bigg (}x-{\frac {\pi }{4}}{\bigg )}+{\frac {2}{3}}{\bigg (}x-{\frac {\pi }{4}}{\bigg )}^{3}.}$
${\displaystyle {\frac {1}{2}}+-1{\bigg (}x-{\frac {\pi }{4}}{\bigg )}+{\frac {2}{3}}{\bigg (}x-{\frac {\pi }{4}}{\bigg )}^{3}}$