anonymous
  • anonymous
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Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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jdoe0001
  • jdoe0001
do you know what \(\bf \large cos^{-1}\) stand for?
anonymous
  • anonymous
sec?
jdoe0001
  • jdoe0001
well, is easy to mixed up a reciprocal or inverse with the ARC functions... no quite, \(\bf ^{-1}\) in trig stands for, "WHAT IS THE ANGLE WHOSE VALUE IS THIS"

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jdoe0001
  • jdoe0001
say for example... what's the cosine of \(\bf \cfrac{\pi}{3}\) ?
anonymous
  • anonymous
1/2
jdoe0001
  • jdoe0001
\(\bf cos\left(\frac{\pi}{3}\right) = \cfrac{1}{2}\qquad \qquad cos^{-1}\left(\frac{1}{2}\right) = \cfrac{\pi}{3}\)
jdoe0001
  • jdoe0001
once you get the value for the trig function, the ARC function with the \(\bf ^{-1}\), will give you back the angle that has that value
anonymous
  • anonymous
So tan 3pi/4= -1.. yeah? So then...I...?
jdoe0001
  • jdoe0001
ahemm, well \(\bf tan\left(\cfrac{3\pi}{4}\right) = \textit{some value} \\ tan^{-1}(\textit{some value}) = \cfrac{3\pi}{4}\quad \\ \textit{some value} = tan\left(\cfrac{3\pi}{4}\right) \implies tan^{-1}(\textit{some value}) = tan\left(\cfrac{3\pi}{4}\right)\\ \quad \\\quad \\ tan^{-1}\left(tan\left(\cfrac{3\pi}{4}\right)\right) = \cfrac{3\pi}{4}\)
jdoe0001
  • jdoe0001
hmm, missed something, but anyhow,
jdoe0001
  • jdoe0001
\(\bf tan\left(\cfrac{3\pi}{4}\right) = \textit{some value} \\ tan^{-1}(\textit{some value}) = \cfrac{3\pi}{4}\quad \\ \textit{some value} = tan\left(\cfrac{3\pi}{4}\right) \implies tan^{-1}(\textit{some value}) = tan^{-1}\left(tan\left(\cfrac{3\pi}{4}\right)\right)\\ \quad \\\quad \\ tan^{-1}\left(tan\left(\cfrac{3\pi}{4}\right)\right) = \cfrac{3\pi}{4}\)
anonymous
  • anonymous
Wait so it simplifies out to 3pi/4?
jdoe0001
  • jdoe0001
yes => \(\bf \large{ cos^{-1}(\quad cos(\theta)\quad ) = \theta\\ sin^{-1}(\quad sin(\theta)\quad ) = \theta\\ tan^{-1}(\quad tan(\theta)\quad ) = \theta} \)
anonymous
  • anonymous
So essentially there is no work at all to do/show, its just the inverse?
jdoe0001
  • jdoe0001
yeap, pretty much
anonymous
  • anonymous
so if the problem is cos^-1[(-1)/2], the answer would be (1/2)?
jdoe0001
  • jdoe0001
did you mean \(\bf cos^{-1}\left(cos\left(-\frac{1}{2}\right)\right) \quad ?\)
anonymous
  • anonymous
No the numerator is only negative
jdoe0001
  • jdoe0001
... ok... then \(\bf cos^{-1}\left(\frac{-1}{2}\right)\)
anonymous
  • anonymous
yes
jdoe0001
  • jdoe0001
it's a bit different... one sec
anonymous
  • anonymous
And isn't 3pi/4 out of range so shouldnt it be -pi/4?
jdoe0001
  • jdoe0001
\(\bf cos^{-1}\left(\frac{-1}{2}\right)\\ \frac{-1}{2} \textit{is the cosine value obtained from an angle, say }\theta\\ \textit{so we can say that } cos(\theta) = \frac{-1}{2}\) if you look at your Unit Circle, it'll be the angles, whose cosine is \(\bf \cfrac{-1}{2}\)
anonymous
  • anonymous
I understand that, but on the previous question isnt it -pi/4 answer?
jdoe0001
  • jdoe0001
hmm, no, why not, well, the inverse functions have a range from where they return their angle value for the \(\bf cos^{-1}\) the range is \(\bf 0 \ to \ \pi\)
jdoe0001
  • jdoe0001
-pi/4 will be in the 4th quadrant, out of range from it
anonymous
  • anonymous
if its tan^-1(tan) why use the cos restrictions?
anonymous
  • anonymous
isnt tan^-1 range -pi/2 to pi/2?
anonymous
  • anonymous
so 3pi/4 -> -pi/4?
jdoe0001
  • jdoe0001
yes, you're correct, those are the restrictions for inverse tangent what's the difference? with -1/2 we don't have an angle we can look at, as source what angle it came from, it could have been from who know what revolution or cycle thus the inverse cosine restrictions apply when you have the case of \(\bf cos^{-1}(cos(\textit{some angle here}))\) we have a source to see for the angle, thus we can safely return it, without being restricted
anonymous
  • anonymous
So for the problem tan^-1[tan(3pi/4)] the restrictions make the final answer neg pi/4 because of that
jdoe0001
  • jdoe0001
if you apply the restrictions, yes
anonymous
  • anonymous
so how do i dtermine the angle based off the unit circle for the cos^-1(-1/2)? I just use 1/2?
jdoe0001
  • jdoe0001
yes, and since we dunno the source right off, the range restrictions will apply when you get the angle from the inverse function usually Unit Circles have the cos/sin pair for the angles
anonymous
  • anonymous
so thats pi/3?
jdoe0001
  • jdoe0001
hmmm thought you said ...-1/2 that'd be \(\bf \cfrac{2\pi}{3}\)
anonymous
  • anonymous
making the answer sq rt 3/2?
anonymous
  • anonymous
or 120 degrees?
jdoe0001
  • jdoe0001
\(\bf sin\left(\cfrac{2\pi}{3}\right) = \cfrac{\sqrt{3}}{2}\)
jdoe0001
  • jdoe0001
cosine of \(\bf \cfrac{2\pi}{3}\) is -1/2
anonymous
  • anonymous
So cos^-1[(-1)/2] = 2pi/3 or 120 degrees?
jdoe0001
  • jdoe0001
yeap \(\bf cos^{-1}\left(\cfrac{-1}{2}\right) = \cfrac{2\pi}{3} = 120^o\)
anonymous
  • anonymous
Wow thanks. A lot. are you any good with identities?
jdoe0001
  • jdoe0001
depends, you may want to post anew :)
anonymous
  • anonymous
k

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