已知$f(x)=sumlimits_{k=1}^{2017}dfrac{cos kx}{cos^k x},$则$f(dfrac{pi}{2018})=$_____
分析:
设$g(x)=sumlimits_{k=1}^{2017}left(dfrac{cos kx}{cos^k x}+idfrac{sin kx}{cos^k x}
ight)$
$=sumlimits_{k=1}^{2017}left(dfrac{cos x+isin x}{cos x}
ight)^{k}$
$ =dfrac{frac{cos x+isin x}{cos x}-(frac{cos x+isin x}{cos x})^{2018}}{1-frac{cos x+isin x}{cos x}}$
$=dfrac{cos x+isin x-frac{cos2018x+isin2018x}{cos^{2017}{x}}}{-isin x}$
则$g(dfrac{pi}{2018})=-1+idfrac{cosfrac{pi}{2018}+cos^{-2017}{frac{pi}{2018}}}{sinfrac{pi}{2018}}$
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$f(x)=dfrac{sumlimits_{k=1}^{1009}sin(2k-1)x}{sumlimits_{k=1}^{1009}cos(2k-1)x},$ 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.