Question

\[sin^{-1}(1)\]
\[sin(1)=\frac\pi 2\]
that would give us \[(1,\frac \pi 2)\]
the inverse would be \[(\frac \pi 2,1)\]

Answer 1

yes

Answer 2

or is it the other way around....kinda confusing

Answer 3

well not really what you wrote
you just should write
\[\sin^{-1}(1)=\frac{\pi}{2}\]

Answer 4

you want \(\sin^{-1}(1)\) that is \(\frac{\pi}{2}\) because \(\sin(\frac{\pi}{2})=1\)

Answer 5

yep makes sense. Thanks again!