anonymous
  • anonymous
i know this is easy but, i just forget how to do it.
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\cos(\arcsin(\frac{ \sqrt{3} }{ 2 }))\]
anonymous
  • anonymous
\[\sin^{-1}\frac{ \sqrt{3} }{ 2 }\]This is an angle measurement. Like if you had \[\sin \theta = \frac{ \sqrt{3} }{ 2 }\]You would find theta by taking the inverse. Use your knowledge of the unit circle to find the value of the y coordinate (sine) at sqrt(3)/2. Then once you have this, you can find cos of whatever it was
anonymous
  • anonymous
okay so arc sin is actually sin^-1

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anonymous
  • anonymous
Ya they are used interchangeably.
anonymous
  • anonymous
so it's sin of 30 degrees?
myininaya
  • myininaya
\[\text{ Let } u=\arcsin(\frac{\sqrt{3}}{2})\] => \[ \sin(u)=\frac{\sqrt{3}}{2}\] sin( )=opp/hyp So we can do this: |dw:1352310919231:dw| Use Pythagorean thm to find the adjacent side to u. Then you can find cos(u) which is the same as what your question is asking since we \[\text{ let } u=\arcsin(\frac{\sqrt{3}}{2})\] \[\cos(u)=\cos(\arcsin(\frac{\sqrt{3}}{2}))=\frac{\text{ adjacent side to u}}{hyp}\]
anonymous
  • anonymous
right so it's just root 3 over 2
myininaya
  • myininaya
Nope sin(u) is that what is cos(u)?
anonymous
  • anonymous
or it's at 60 degrees and 120
anonymous
  • anonymous
Based on the unit circle, sqrt(3)/2 (the y value) is at 60 deg. So it would be cos 60deg
anonymous
  • anonymous
60 and 120, right?
anonymous
  • anonymous
right it is also at 120. good catch
anonymous
  • anonymous
so i write it as cos(60) ??
anonymous
  • anonymous
so my answer should be 1/2 and -1/2?
myininaya
  • myininaya
but arcsin( ) only has range between -90 and 90.
anonymous
  • anonymous
Thats what I got !!
myininaya
  • myininaya
So you will only have one answer.
anonymous
  • anonymous
Explain that plx @myininaya . I don't remember that.
anonymous
  • anonymous
i thought that had to be specified
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Inverse_trigonometric_functions He's right. The only answer would be 1/2
myininaya
  • myininaya
In order to make sin( ) one to one, we must restrict the domain of sin( ) in order to do this. We chose the restriction [-90 degrees, 90 degrees] So now that are inverse for sin( ) exists, we call it arcsin( ). And arcsin( ) only has range [-90 degrees, 90 degrees]
anonymous
  • anonymous
i can only do this type of problem if it's one to one?
myininaya
  • myininaya
I'm a her.
anonymous
  • anonymous
Sry I meant nothing by it
anonymous
  • anonymous
so if i wanted to do sec(arctan(2)) i do the same thing?
myininaya
  • myininaya
I like making a pretty right triangle. :)
anonymous
  • anonymous
huh?
myininaya
  • myininaya
\[\text{ Letting } u=\arctan(2) \] => tan(u)=2 tan(u)=2/1 tan(u)=opp/adj We let opp side of u be 2 We let adj side of u be 1 Let the labeling begin.|dw:1352311632755:dw| Now find the hyp by using the Pythagorean thm.
myininaya
  • myininaya
After that, find sec(u) and you are done.
anonymous
  • anonymous
there is no hypotnus tho :s
anonymous
  • anonymous
oh i gott find it lol
myininaya
  • myininaya
Yes there is. You find it by using the Pythagorean thm.
anonymous
  • anonymous
square root of 5 is the hyp.
anonymous
  • anonymous
???
myininaya
  • myininaya
Yes:)
anonymous
  • anonymous
okay so now i take sec(squareroot 5)
myininaya
  • myininaya
so what is sec(u)
anonymous
  • anonymous
and do i restrict my tan?
myininaya
  • myininaya
|dw:1352311961504:dw|
myininaya
  • myininaya
sec(u)=hyp/adj
anonymous
  • anonymous
so its just root 5
myininaya
  • myininaya
Yep. :)
anonymous
  • anonymous
Study her method because it was way easier than my explanation.
anonymous
  • anonymous
and i state my domain and range for just sec right?
myininaya
  • myininaya
No it was just looking for sec(arctan(2)) which is just \[\sqrt{5}\]

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