Answer
$f(x) = -1+4 \dfrac{(x-(\dfrac{\pi}{4}))^2}{2!}-16 \dfrac{(x-(\dfrac{\pi}{4})^4}{4!}+.... ....\\=\Sigma_{n=0}^{\infty} (-1)^{n+1} (2)^{2n} \dfrac{(x-\dfrac{\pi}{4})^{2n}}{(2n)!}$
Work Step by Step
Differentiate the given function $f(x)$ as follows:
$f'(x)=-2 \cos 2x \implies f'(\dfrac{\pi}{4})=0 \\f''(x)=4 \sin 2x \implies f''(\dfrac{\pi}{4})=4 \\ f'''(x)=8 \cos 2x \implies f'''(\dfrac{\pi}{4})=0$
The Taylor series at $x=\dfrac{\pi}{4}$ can be written as:
$f(x)=f(\dfrac{\pi}{4})+f'(\dfrac{\pi}{4}) (x-(\dfrac{\pi}{4}))+\dfrac{f''(\dfrac{\pi}{4})(x-(\dfrac{\pi}{4})^2 }{2!}+.....\\=-1+4 \dfrac{(x-(\dfrac{\pi}{4}))^2}{2!}-16 \dfrac{(x-(\dfrac{\pi}{4})^4}{4!}+.... ....\\=\Sigma_{n=0}^{\infty} (-1)^{n+1} (2)^{2n} \dfrac{(x-\dfrac{\pi}{4})^{2n}}{(2n)!}$
