Question

\[sin^{-1}(sin\frac{9\pi}8)\]
I thought the limits were
\[\frac{-\pi}2 \le \theta \le \frac{\pi}2\]
so this would be undefined I thought

Answer 1

@satellite73

Answer 2

it is not undefined, it is just not \(\frac{9\pi}{8}\)

Answer 3

why?

Answer 4

what happened to the limits?

Answer 5

\[\frac{-\pi}2 \le \theta \le \frac{\pi}2\] are the limits of the OUTPUT not the input

Answer 6

oh I see

Answer 7

the domain of \(\sin^{-1}(x)\) is \([-1,1]\) what you wrote is the range

Answer 8

oh ok haha

Answer 9

so your real job here is this
find an angle coterminal with \(\frac{9\pi}{8}\) in the interval
\[[-\frac{\pi}{2},\frac{\pi}{2}]\]

Answer 10

coterminal?

Answer 11

ok scratch that

Answer 12

no tell me...please?

Answer 13

find an angle where the sine would be the same
i mean scratch it because what i said is wrong

Answer 14

oh I see

Answer 15

let me draw a picture

Answer 16

ok

Answer 17

there is my picture of \(\frac{9\pi}{8}\) now we want an angle on the other side of the circle with the same \(y\) value (same sine value) so we do this

Answer 18

that point on the unit circle has the same sine value, and the angle is evidently \(-\frac{\pi}{8}\)

Answer 19

and that is your answer

Answer 20

Got it. So I should've probably simplified \[\frac{9\pi}{8}\] first then found the inverse...would that be first?

Answer 21

I meant...would that be wrong?

Answer 22

no there is nothing to "simplify"
the procedure was to locate \(\frac{9\pi}{8}\) on the unit circle, then find another angle on the right hand side of the circle with the same sine

Answer 23

I mean \[\frac{9\pi}8=\frac\pi 8\]

why would this be wrong?

Answer 24

pi/8 is the 'same' as pi/8 + 2pi
but pi/8 is not the 'same' as pi/8 + pi = 9.pi/8

Answer 25

oh I see...

Answer 26

makes sense. Thanks!