anonymous
  • anonymous
sin inverse[sin 6pi/7]. I'm suppose to find the exact value.
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
that is just 6pi/7
anonymous
  • anonymous
that is true for all inverse functions f inverse f(x)=x
anonymous
  • anonymous
so the same thing applies for this: cos inverse[cos pi/2] is the the exact value pi/2 ?

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anonymous
  • anonymous
yes
anonymous
  • anonymous
thanks!
anonymous
  • anonymous
this is what inverse functions do
anonymous
  • anonymous
what if its negative like in this case: tan inverse[tan(-pi/4)] ?
watchmath
  • watchmath
You need to be careful here, the inverse sine function has range (-pi/2,pi/2). So the answer can't be 6pi/7.
anonymous
  • anonymous
would you take the absolute value of "(-pi/4)" so it would be "pi/4"
anonymous
  • anonymous
watchmath, i'm confused.
anonymous
  • anonymous
"f inverse f(x)=x"
anonymous
  • anonymous
oh true, did not think of that
watchmath
  • watchmath
What I am trying to say that \(\sin^{-1}(\sin x)=x\) only for \(x\in[-\pi/2,\pi/2]\)
watchmath
  • watchmath
So first you need to find an angle x so that sin(x)=sin(6pi/7). And we know that sin(pi/7)=sin(6pi/7) So \(\sin^{-1}(\sin 6\pi/7)=\sin^{-1}(\sin (\pi/7))=\pi/7.\)
anonymous
  • anonymous
so the exact value is pi/7?
watchmath
  • watchmath
Yes For the tangent, we have \[\tan^{-1}(\tan x)=x\]only for \(x\in (-\pi/2,\pi/2)\)

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