Question

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

Answer 1

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

Answer 2

\[\sin^{-1}\frac{ \sqrt{3} }{ 2 }\]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

Answer 3

okay so arc sin is actually sin^-1

Answer 4

Ya they are used interchangeably.

Answer 5

so it's sin of 30 degrees?

Answer 6

\[\text{ Let } u=\arcsin(\frac{\sqrt{3}}{2})\]
=>
\[ \sin(u)=\frac{\sqrt{3}}{2}\]
sin( )=opp/hyp

So we can do this:



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
\[\text{ let } u=\arcsin(\frac{\sqrt{3}}{2})\]

\[\cos(u)=\cos(\arcsin(\frac{\sqrt{3}}{2}))=\frac{\text{ adjacent side to u}}{hyp}\]

Answer 7

right so it's just root 3 over 2

Answer 8

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

Answer 9

or it's at 60 degrees and 120

Answer 10

Based on the unit circle, sqrt(3)/2 (the y value) is at 60 deg.

So it would be cos 60deg

Answer 11

60 and 120, right?

Answer 12

right it is also at 120. good catch

Answer 13

so i write it as cos(60) ??

Answer 14

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

Answer 15

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

Answer 16

Thats what I got !!

Answer 17

So you will only have one answer.

Answer 18

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

Answer 19

i thought that had to be specified

Answer 20

http://en.wikipedia.org/wiki/Inverse_trigonometric_functions

He's right. The only answer would be 1/2

Answer 21

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]

Answer 22

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

Answer 23

I'm a her.

Answer 24

Sry I meant nothing by it

Answer 25

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

Answer 26

I like making a pretty right triangle. :)

Answer 27

huh?

Answer 28

\[\text{ Letting } u=\arctan(2) \]
=>
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.
Now find the hyp by using the Pythagorean thm.

Answer 29

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

Answer 30

there is no hypotnus tho :s

Answer 31

oh i gott find it lol

Answer 32

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

Answer 33

square root of 5 is the hyp.

Answer 34

???

Answer 35

Yes:)

Answer 36

okay so now i take sec(squareroot 5)

Answer 37

so what is sec(u)

Answer 38

and do i restrict my tan?