Question

Would anyone be able to help me with how to solve this? I don't quite understand how my textbook got its answer.

sin^-1(sin (-7/6))

Answer 1

it is \[\sin^{-1}(\sin(-\frac{7\pi}{6}))\]?

Answer 2

Yes

Answer 3

ok so it look like the answer should be \(-\frac{7\pi}{6}\) since when you compose a function with its inverse, you get the input back

Answer 4

but it is not
that is because the range of arcsine is \([-\frac{\pi}{2},\frac{\pi}{2}]\) so you have to find a number in that interval that has the same sine

Answer 5

So I have to find a value that fits the sine of \[-7\pi/6\], so that would mean I would have to choose \[\pi/6\]?

Answer 6

Since both of their sines equal 1/2?

Answer 7

yes

Answer 8

or not really you can just draw a line across the circle and see what angle it is,
the fact the the output is \(\frac{1}{2}\) is more or less irrelevant

Answer 9

Ok, so I now have \[\sin^{-1} \pi/2\], correct? But how do I find the value of an arcsine with a radial measurement? \[\sin \theta = \pi/2\] or am I thinking wrong?

Answer 10

Sorry, I meant \[\pi/6\]

Answer 11

you are thinking wrong, this is a trick question

Answer 12

\[\frac{\pi}{2}\] is bigger than one, so it is not in the domain of arcsine

Answer 13

OHHHHHH. I forgot about that. My textbook says the answer is \[\pi/6\], but I don't understand WHY it's that.

Answer 14

hmm can you write the exact question
it cannot be \[\sin^{-1}\left(\frac{\pi}{2}\right)\]

Answer 15

i guess what i really mean is that COULD be the question, but if so \(\frac{\pi}{6}\) cannot be the answer to it

Answer 16

Let me check to see if I wrote the question down correctly lol.

Answer 17

I definitely did, so I have no idea.

Answer 18

maybe you were looking at the wrong answer in the answer section
that has happened before

Answer 19

I triple checked and it still says the answer is that. I'll definitely be asking my teacher tomorrow since this makes no sense. But I still am probably looking at something wrong.

Answer 20

it could be a mistake in the book
or rather, if you are sure, it must be one

Answer 21

there is no way on earth that \[\sin(\frac{\pi}{2})=\frac{\pi}{6}\]
no chance

Answer 22

Yeah I'm not getting it. Thanks for the help though. Hopefully I'm not the only one who came across this.

Answer 23

The book is correct. If the original question is \[\sin^{-1}\left( \sin \left( -\frac{ 7 \pi }{ 6 } \right) \right)\]then the correct answer is \(\frac{\pi}{6}\).

Answer 24

How?

Answer 25

lol that is the first one we did!!

Answer 26

I thought there were two separate problems in this thread.
Problem 1: Evaluate arcsin(sin(-7pi/6))
Problem 2: Evaluate arcsin(pi/2)

Answer 27

maybe op confused 1/2 with pi/2

Answer 28

any case sin(-7pi/6)=sin(pi/6)
so arcsin(sin(-7pi/6))=arcsin(sin(pi/6)) where we can just use
arcsin(sin(x))=x if x is between -pi/2 and pi/2 (inclusive)