Answer
$\dfrac{1}{\sqrt 2}[1-(x-\dfrac{\pi}{4})-\dfrac{1}{2!}(x-\dfrac{\pi}{4})^2+\dfrac{1}{3!}(x-\dfrac{\pi}{4})^3+\dfrac{1}{4!}(x-\dfrac{\pi}{4})^4.......] $
Work Step by Step
Here, we have: $f(x)=\cos x \implies f(\dfrac{\pi}{4})=\dfrac{\sqrt 2}{2} \\ f'(x)=-\sin x \implies f' (\dfrac{\pi}{4})=-\dfrac{\sqrt 2}{2} \\f''(x)=-\cos x \implies f''(\dfrac{\pi}{4})=-\dfrac{\sqrt 2}{2} $
and so on.
The Taylor series of a function $f(x)$ at $x=c$ can be written as:
$T(x)=f(a)+\dfrac{f′(a)}{1!}(x−a)+\dfrac{f′′(a)}{2!}(x−a)^2+......+\dfrac{f^n(a)}{n!}(x−a)^n$
Hence, the required Taylor series is:
$T(x)=\dfrac{\sqrt 2}{2}+\dfrac{-\dfrac{\sqrt 2}{2}}{1!}(x-\dfrac{\pi}{4})+\dfrac{-\dfrac{\sqrt 2}{2}}{2!}(x-\dfrac{\pi}{4})^2+.......=\dfrac{1}{\sqrt 2}[1-(x-\dfrac{\pi}{4})-\dfrac{1}{2!}(x-\dfrac{\pi}{4})^2+\dfrac{1}{3!}(x-\dfrac{\pi}{4})^3+\dfrac{1}{4!}(x-\dfrac{\pi}{4})^4.......] $
