Answer
$\Sigma_{k=0}^{\infty} \dfrac{(-1)^{k+1}}{(2k+1)!} (x-\dfrac{\pi}{2})^{2k+1}$
Work Step by Step
The Maclaurin Series is
$ \displaystyle\sum\limits_{n=0}^{\infty} \frac{f^n(0)x^n} {n!} = f(0) + f'(0)x + \frac{(f''(0)x^2)}{2!} + ..\\
$
$f(x)=\cos (x)$
Differentiate
$f'(x)=-\sin x$
$f''(x)=-\cos x$
$f'''(x)=\sin x$
Now $f(\pi/2)=0$, $f'(\pi/2)=-1$, $f''(\pi/2)=0$, $f'''(\pi/2)=1$
Hence the Maclaurin Series for the function is
$\displaystyle -(x-\dfrac{\pi}{2})+\dfrac{1}{6}(x-\dfrac{\pi}{2})^3+.........+ \dfrac{(-1)^{n+1}}{(2n+1)!} (x-\dfrac{\pi}{2})^{2n+1}+....=\Sigma_{k=0}^{\infty} \dfrac{(-1)^{k+1}}{(2k+1)!} (x-\dfrac{\pi}{2})^{2k+1}$
