Answer
$\dfrac{\sqrt 2}{2}[1-(x-\dfrac{\pi}{4})-\dfrac{(x-\dfrac{\pi}{4})^2}{2}]$
Work Step by Step
The Taylor approximation for degree $n$ centered at point $a$ can be written as:
$P_n(x)=f(a)+f'(a)(x-a)+\dfrac{1}{2}f''(a)(x-a)^2+.......+\dfrac{1}{n!}f^n(a)(x-a)^n ~~~~........(1)$----
We have: $f(x)=\cos x \implies f(\pi/4)=\dfrac{\sqrt 2}{2}$ with $n=2$ and $a=\pi/4$
Further, $f'(x)=-\sin x \implies f'(\pi/4)=-\dfrac{\sqrt 2}{2} \\ f''(x)=-\cos x \implies f''(\pi/4)=-\dfrac{\sqrt 2}{2}$
Now, plug these values in the equation (1) to obtain:
$P_2(x)=f(\pi/4)+f'(\pi/4)\dfrac{(x-\pi/4)}{1!}+\dfrac{1}{2!}f''(\pi/4)(x-\pi/4)^2\\=\dfrac{\sqrt 2}{2}-\dfrac{\sqrt 2}{2}(x-\dfrac{\pi}{4})-\dfrac{\sqrt 2}{2}\dfrac{(x-\dfrac{\pi}{4})^2}{2!}\\=\dfrac{\sqrt 2}{2}[1-(x-\dfrac{\pi}{4})-\dfrac{(x-\dfrac{\pi}{4})^2}{2}]$
