两个另类的三角函数

青青子衿

\[\begin{gathered}
  f\left( x \right) = {\cos ^{ - 1}}\left( {\frac{1}{2}\cos \left( {2x} \right)} \right) \\
  g\left( x \right) = - \frac{\pi }{6}\cos \left( {2x} \right) + \frac{\pi }{2} \\
\end{gathered} \]
比较大小

tommywong

$\displaystyle \frac{-1}{2} \le u=\frac{1}{2}cos2x \le \frac{1}{2}$

$\displaystyle y=cos^{-1}u+\frac{\pi}{3}u-\frac{\pi}{2}=0,u=sin\frac{\pi u}{3},u=\frac{-1}{2},0,\frac{1}{2}$

$\displaystyle y'=\frac{-1}{\sqrt{1-u^2}}+\frac{\pi}{3}$