Answer
(a) $ \frac{\sqrt 3}{2}$, $(\frac{\pi}{6},\frac{\sqrt 3}{2})$.
(b) $(\frac{\sqrt 3}{2},\frac{\pi}{6})$.
(c) $(\frac{\pi}{6},2)$.
Work Step by Step
Given $g(x)=cos(x)$, we have:
(a) $g(\frac{\pi}{6})=cos(\frac{\pi}{6})=\frac{\sqrt 3}{2}$ which gives point $(\frac{\pi}{6},\frac{\sqrt 3}{2})$.
(b) As $g^{-1}(x)$ and $g(x)$ are symmetric across $y=x$, we have point on $g^{-1}(x)$ as $(\frac{\sqrt 3}{2},\frac{\pi}{6})$.
(c) For $x=\frac{\pi}{6}$, we have $y=2g(\frac{\pi}{6}-\frac{\pi}{6})=2g(0)=2cos(0)=2$ which gives point $(\frac{\pi}{6},2)$.
