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bmelykBest ResponseYou've already chosen the best response.0
\[\cos(\arcsin(\frac{ \sqrt{3} }{ 2 }))\]
 one year ago

ChmEBest ResponseYou've already chosen the best response.1
\[\sin^{1}\frac{ \sqrt{3} }{ 2 }\]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
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
okay so arc sin is actually sin^1
 one year ago

ChmEBest ResponseYou've already chosen the best response.1
Ya they are used interchangeably.
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
so it's sin of 30 degrees?
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
\[\text{ Let } u=\arcsin(\frac{\sqrt{3}}{2})\] => \[ \sin(u)=\frac{\sqrt{3}}{2}\] 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 \[\text{ let } u=\arcsin(\frac{\sqrt{3}}{2})\] \[\cos(u)=\cos(\arcsin(\frac{\sqrt{3}}{2}))=\frac{\text{ adjacent side to u}}{hyp}\]
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
right so it's just root 3 over 2
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
Nope sin(u) is that what is cos(u)?
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
or it's at 60 degrees and 120
 one year ago

ChmEBest ResponseYou've already chosen the best response.1
Based on the unit circle, sqrt(3)/2 (the y value) is at 60 deg. So it would be cos 60deg
 one year ago

ChmEBest ResponseYou've already chosen the best response.1
right it is also at 120. good catch
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
so i write it as cos(60) ??
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
so my answer should be 1/2 and 1/2?
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
but arcsin( ) only has range between 90 and 90.
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
So you will only have one answer.
 one year ago

ChmEBest ResponseYou've already chosen the best response.1
Explain that plx @myininaya . I don't remember that.
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
i thought that had to be specified
 one year ago

ChmEBest ResponseYou've already chosen the best response.1
http://en.wikipedia.org/wiki/Inverse_trigonometric_functions He's right. The only answer would be 1/2
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
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]
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
i can only do this type of problem if it's one to one?
 one year ago

ChmEBest ResponseYou've already chosen the best response.1
Sry I meant nothing by it
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
so if i wanted to do sec(arctan(2)) i do the same thing?
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
I like making a pretty right triangle. :)
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
\[\text{ Letting } u=\arctan(2) \] => 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.
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
After that, find sec(u) and you are done.
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
there is no hypotnus tho :s
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
Yes there is. You find it by using the Pythagorean thm.
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
square root of 5 is the hyp.
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
okay so now i take sec(squareroot 5)
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
and do i restrict my tan?
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
dw:1352311961504:dw
 one year ago

ChmEBest ResponseYou've already chosen the best response.1
Study her method because it was way easier than my explanation.
 one year ago

bmelykBest ResponseYou've already chosen the best response.0
and i state my domain and range for just sec right?
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
No it was just looking for sec(arctan(2)) which is just \[\sqrt{5}\]
 one year ago
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