bmelyk
i know this is easy but, i just forget how to do it.
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bmelyk
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\[\cos(\arcsin(\frac{ \sqrt{3} }{ 2 }))\]
ChmE
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\[\sin^{-1}\frac{ \sqrt{3} }{ 2 }\]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
bmelyk
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okay so arc sin is actually sin^-1
ChmE
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Ya they are used interchangeably.
bmelyk
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so it's sin of 30 degrees?
myininaya
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\[\text{ Let } u=\arcsin(\frac{\sqrt{3}}{2})\]
=>
\[ \sin(u)=\frac{\sqrt{3}}{2}\]
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
\[\text{ let } u=\arcsin(\frac{\sqrt{3}}{2})\]
\[\cos(u)=\cos(\arcsin(\frac{\sqrt{3}}{2}))=\frac{\text{ adjacent side to u}}{hyp}\]
bmelyk
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right so it's just root 3 over 2
myininaya
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Nope sin(u) is that
what is cos(u)?
bmelyk
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or it's at 60 degrees and 120
ChmE
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Based on the unit circle, sqrt(3)/2 (the y value) is at 60 deg.
So it would be cos 60deg
bmelyk
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60 and 120, right?
ChmE
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right it is also at 120. good catch
bmelyk
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so i write it as cos(60) ??
bmelyk
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so my answer should be 1/2 and -1/2?
myininaya
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but arcsin( ) only has range between -90 and 90.
ChmE
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Thats what I got !!
myininaya
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So you will only have one answer.
ChmE
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Explain that plx @myininaya . I don't remember that.
bmelyk
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i thought that had to be specified
myininaya
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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]
bmelyk
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i can only do this type of problem if it's one to one?
myininaya
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I'm a her.
ChmE
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Sry I meant nothing by it
bmelyk
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so if i wanted to do sec(arctan(2)) i do the same thing?
myininaya
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I like making a pretty right triangle. :)
bmelyk
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huh?
myininaya
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\[\text{ Letting } u=\arctan(2) \]
=>
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
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After that, find sec(u) and you are done.
bmelyk
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there is no hypotnus tho :s
bmelyk
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oh i gott find it lol
myininaya
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Yes there is.
You find it by using the Pythagorean thm.
bmelyk
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square root of 5 is the hyp.
bmelyk
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???
myininaya
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Yes:)
bmelyk
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okay so now i take sec(squareroot 5)
myininaya
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so what is sec(u)
bmelyk
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and do i restrict my tan?
myininaya
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|dw:1352311961504:dw|
myininaya
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sec(u)=hyp/adj
bmelyk
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so its just root 5
myininaya
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Yep. :)
ChmE
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Study her method because it was way easier than my explanation.
bmelyk
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and i state my domain and range for just sec right?
myininaya
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No it was just looking for sec(arctan(2))
which is just \[\sqrt{5}\]