Question

\[sin^{-1}0\]
Let's see....
\[sin{-1}x=y\]
\[x=siny\]
\[o=siny\]
where from here?

Answer 1

ask yourself this
"do i know a number (angle) between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\) whose sine is zero? "

Answer 2

you do not need all that stuff above.
like if i ask what is the square root of 25, you say 5 because you know that \(5^2=25\)

Answer 3

if you see \(\sin^{-1}(0)\) you know that is \(0\) because you know \(\sin(0)=0\)

Answer 4

Let's see here
I know that sin of zero is zero. The same holds for the inverse?

Answer 5

and if you see \(\sin(\frac{\sqrt{2}}{2})\) you know it is \(\frac{\pi}{4}\) because
\[\sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}\]

Answer 6

yep

Answer 7

the inverse just bothers me I guess

Answer 8

on the graph of \(\sin(x)\) is \((0,0)\) so if you switch the ordered pairs you still get \((0,0)\)

Answer 9

on the graph of \(\sin(x)\) is the point \((\frac{\pi}{2},\frac{\sqrt{2}}{2})\) so on the graph of the inverse is the point \((\frac{\sqrt{2}}{2},\frac{\pi}{4})\)

Answer 10

Yes! Yep that makes sense. So when we do the inverse we do (a,b) -> (b,a)
Makes sense

Answer 11

right
i made a typo above, ignore it

Answer 12

yeah I noticed. No worries. Thanks though @satellite73

Answer 13

do koala's receive medals too?