i know this is easy but, i just forget how to do it.

- anonymous

i know this is easy but, i just forget how to do it.

- jamiebookeater

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- anonymous

\[\cos(\arcsin(\frac{ \sqrt{3} }{ 2 }))\]

- 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

okay so arc sin is actually sin^-1

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## More answers

- anonymous

Ya they are used interchangeably.

- anonymous

so it's sin of 30 degrees?

- 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

right so it's just root 3 over 2

- myininaya

Nope sin(u) is that
what is cos(u)?

- anonymous

or it's at 60 degrees and 120

- anonymous

Based on the unit circle, sqrt(3)/2 (the y value) is at 60 deg.
So it would be cos 60deg

- anonymous

60 and 120, right?

- anonymous

right it is also at 120. good catch

- anonymous

so i write it as cos(60) ??

- anonymous

so my answer should be 1/2 and -1/2?

- myininaya

but arcsin( ) only has range between -90 and 90.

- anonymous

Thats what I got !!

- myininaya

So you will only have one answer.

- anonymous

Explain that plx @myininaya . I don't remember that.

- anonymous

i thought that had to be specified

- anonymous

http://en.wikipedia.org/wiki/Inverse_trigonometric_functions
He's right. The only answer would be 1/2

- 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

i can only do this type of problem if it's one to one?

- myininaya

I'm a her.

- anonymous

Sry I meant nothing by it

- anonymous

so if i wanted to do sec(arctan(2)) i do the same thing?

- myininaya

I like making a pretty right triangle. :)

- anonymous

huh?

- 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

After that, find sec(u) and you are done.

- anonymous

there is no hypotnus tho :s

- anonymous

oh i gott find it lol

- myininaya

Yes there is.
You find it by using the Pythagorean thm.

- anonymous

square root of 5 is the hyp.

- anonymous

???

- myininaya

Yes:)

- anonymous

okay so now i take sec(squareroot 5)

- myininaya

so what is sec(u)

- anonymous

and do i restrict my tan?

- myininaya

|dw:1352311961504:dw|

- myininaya

sec(u)=hyp/adj

- anonymous

so its just root 5

- myininaya

Yep. :)

- anonymous

Study her method because it was way easier than my explanation.

- anonymous

and i state my domain and range for just sec right?

- myininaya

No it was just looking for sec(arctan(2))
which is just \[\sqrt{5}\]

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