Answer
$\Sigma_{k=0}^n \dfrac{(-1)^k}{(2k)!}(x-\dfrac{\pi}{2})^{2k}$
Work Step by Step
The nth Taylor polynomial for a function $f$ is
$$p_n(x)= f(x_0) + f'(x_0)(x-x_0) +\dfrac{f''(x_0)}{2!} (x-x_0)^2 +.......+\dfrac{f^n(x_0)}{n!} (x-x_0)^n$$
$f(x)=\cos x$
Differentiate w.r.t x
$f'(x)=-\sin x$
$f''(x)=-\cos x$
$f'''(x)=\sin x$
$f''''(x)=\cos x$
Now $f(\dfrac{\pi}{2})=0$ , $f'(\dfrac{\pi}{2})=-1$ , $f''(\pi/2)=0$, $f'''(\pi/2)=1$, $f''''(\pi/2)=0$
Hence the nth Taylor polynomial for a function is:
$\displaystyle -(x-\dfrac{\pi}{2})^2+\dfrac{1}{6}(x-\dfrac{\pi}{2})^3+ .......+\dfrac{(-1)^n}{(2n)!}(x-\dfrac{\pi}{2})^{2n}=\Sigma_{k=0}^n \dfrac{(-1)^k}{(2k)!}(x-\dfrac{\pi}{2})^{2k}$
