Answer
(a) $ \frac{\sqrt 2}{2}$, $(\frac{\pi}{4},\frac{\sqrt 2}{2})$.
(b) $(\frac{\sqrt 2}{2},\frac{\pi}{4})$.
(c) $(\frac{\pi}{4},-2)$.
Work Step by Step
Given $f(x)=sin(x)$, we have:
(a) $f(\frac{\pi}{4})=sin(\frac{\pi}{4})=\frac{\sqrt 2}{2}$ which gives point $(\frac{\pi}{4},\frac{\sqrt 2}{2})$.
(b) As $f^{-1}(x)$ and $f(x)$ are symmetric across $y=x$, we have point on $f^{-1}(x)$ as $(\frac{\sqrt 2}{2},\frac{\pi}{4})$.
(c) For $x=\frac{\pi}{4}$, we have $y=f(\frac{\pi}{4}+\frac{\pi}{4})-3=f(\frac{\pi}{2})-3=sin(\frac{\pi}{2})-3=-2$ which gives point $(\frac{\pi}{4},-2)$.
