anonymous
  • anonymous
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))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
it is \[\sin^{-1}(\sin(-\frac{7\pi}{6}))\]?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
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

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More answers

anonymous
  • anonymous
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
anonymous
  • anonymous
|dw:1446778529971:dw|
anonymous
  • anonymous
So I have to find a value that fits the sine of \[-7\pi/6\], so that would mean I would have to choose \[\pi/6\]?
anonymous
  • anonymous
Since both of their sines equal 1/2?
anonymous
  • anonymous
yes
anonymous
  • anonymous
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 |dw:1446778979449:dw|
anonymous
  • anonymous
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?
anonymous
  • anonymous
Sorry, I meant \[\pi/6\]
anonymous
  • anonymous
you are thinking wrong, this is a trick question
anonymous
  • anonymous
\[\frac{\pi}{2}\] is bigger than one, so it is not in the domain of arcsine
anonymous
  • anonymous
OHHHHHH. I forgot about that. My textbook says the answer is \[\pi/6\], but I don't understand WHY it's that.
anonymous
  • anonymous
hmm can you write the exact question it cannot be \[\sin^{-1}\left(\frac{\pi}{2}\right)\]
anonymous
  • anonymous
i guess what i really mean is that COULD be the question, but if so \(\frac{\pi}{6}\) cannot be the answer to it
anonymous
  • anonymous
Let me check to see if I wrote the question down correctly lol.
anonymous
  • anonymous
I definitely did, so I have no idea.
anonymous
  • anonymous
maybe you were looking at the wrong answer in the answer section that has happened before
anonymous
  • anonymous
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.
anonymous
  • anonymous
it could be a mistake in the book or rather, if you are sure, it must be one
anonymous
  • anonymous
there is no way on earth that \[\sin(\frac{\pi}{2})=\frac{\pi}{6}\] no chance
anonymous
  • anonymous
Yeah I'm not getting it. Thanks for the help though. Hopefully I'm not the only one who came across this.
anonymous
  • anonymous
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 \(\frac{\pi}{6}\).
anonymous
  • anonymous
How?
anonymous
  • anonymous
lol that is the first one we did!!
freckles
  • freckles
I thought there were two separate problems in this thread. Problem 1: Evaluate arcsin(sin(-7pi/6)) Problem 2: Evaluate arcsin(pi/2)
freckles
  • freckles
maybe op confused 1/2 with pi/2
freckles
  • freckles
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)

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