已知$f(x)=\sum\limits_{k=1}^{2017}\dfrac{\cos kx}{\cos^k x},$则$f(\dfrac{\pi}{2018})=$_____

分析:
设$g(x)=\sum\limits_{k=1}^{2017}\left(\dfrac{\cos kx}{\cos^k x}+i\dfrac{\sin kx}{\cos^k x}\right)$
$=\sum\limits_{k=1}^{2017}\left(\dfrac{\cos x+i\sin x}{\cos x}\right)^{k}$ 
$ =\dfrac{\frac{\cos x+i\sin x}{\cos x}-(\frac{\cos x+i\sin x}{\cos x})^{2018}}{1-\frac{\cos x+i\sin x}{\cos x}}$ 
$=\dfrac{\cos x+i\sin x-\frac{\cos2018x+i\sin2018x}{\cos^{2017}{x}}}{-i\sin x}$
则$g(\dfrac{\pi}{2018})=-1+i\dfrac{\cos\frac{\pi}{2018}+\cos^{-2017}{\frac{\pi}{2018}}}{\sin\frac{\pi}{2018}}$
比较实部可知$f(\dfrac{\pi}{2018})=-1$

思路参考：MT【34】

有时候实数里不容易解决的问题在复数范围内会变得容易。更大的视野带来更好的思路。

练习: 设$f(x)=\dfrac{\sum\limits_{k=1}^{1009}sin(2k-1)x}{\sum\limits_{k=1}^{1009}cos(2k-1)x},$ 则$f(\dfrac{\pi}{2019})=$_____

提示:$z=cosx+isinx,z+z^3+\cdots+z^{2017}=\dfrac{z(1-(z^2)^{1009}}{1-z^2}$,两边比较辐角的正切值.