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a math question

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\[\cos ^{-1}(\sin \frac{ 11\pi }{ 6 })\]
@satellite73 plz help

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Play with the values for this one... Let \[x = \cos^{-1}(\sin \frac{11\pi}{6})\]and if we get the cosine of both sides, we are left with...\[\cos x = \sin \frac{11\pi}{6}\]perhaps it's simpler from here on in?
the answer is 2pi/3 ??
seems like it
You're right. Hang on... in general, \[\cos(\frac{\pi}{2} - \theta)=\sin(\theta)\] So \[\sin\frac{11\pi}{6}=\cos \left( \frac{\pi}{2}-\frac{11\pi}{6} \right)\] \[=\cos \left( -\frac{8\pi}{6} \right)\] It would appear \[x=-\frac{4\pi}{3}+2k\pi\] where k is any integer. But since we're talking angles, adding 360 degrees, or 2pi, would give the same angle. Consider doing that here, because that negative angle ain't pretty XD \[x = -\frac{4\pi}{3}+2\pi=\frac{2\pi}{3}\]
k thanks but this other one is hard \[\tan(\sin ^{-1}-\frac{ 5 }{ 13 }\]
)
It's a bit complicated, but think of it this way... the SINE of the angle is -5/13, so what is its TANGENT?
here's what you do: sine of what angle will equal 5/13? Then you take that angle and take the tan of it.
but i how do i do that?
answer is 5/12 by the way
|dw:1358823550014:dw|
so... tan of theta is ?
how did u get 12? and one is negative
oops didn't catch the negative
so if its negative then it means the angle is either in the third or fourth quadrant
however my answer would still be correct if the angle was in the 3rd quadrant
omg still confused
Im sorry, i dont know how to work this out.
k :(
did the problems include restrictions?
yes
what were they?
but that's all the questions says
?? so is there or is there no restrictions like 180
no that is all the question says do u know the answer plz i am confused i will just do others if u don't know this
alright... shoot gimme another question
it is k i am doing graphing

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