## bmelyk 2 years ago i know this is easy but, i just forget how to do it.

1. bmelyk

$\cos(\arcsin(\frac{ \sqrt{3} }{ 2 }))$

2. ChmE

$\sin^{-1}\frac{ \sqrt{3} }{ 2 }$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

3. bmelyk

okay so arc sin is actually sin^-1

4. ChmE

Ya they are used interchangeably.

5. bmelyk

so it's sin of 30 degrees?

6. 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 $\text{ let } u=\arcsin(\frac{\sqrt{3}}{2})$ $\cos(u)=\cos(\arcsin(\frac{\sqrt{3}}{2}))=\frac{\text{ adjacent side to u}}{hyp}$

7. bmelyk

right so it's just root 3 over 2

8. myininaya

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

9. bmelyk

or it's at 60 degrees and 120

10. ChmE

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

11. bmelyk

60 and 120, right?

12. ChmE

right it is also at 120. good catch

13. bmelyk

so i write it as cos(60) ??

14. bmelyk

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

15. myininaya

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

16. ChmE

Thats what I got !!

17. myininaya

So you will only have one answer.

18. ChmE

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

19. bmelyk

i thought that had to be specified

20. ChmE

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

21. 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]

22. bmelyk

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

23. myininaya

I'm a her.

24. ChmE

Sry I meant nothing by it

25. bmelyk

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

26. myininaya

I like making a pretty right triangle. :)

27. bmelyk

huh?

28. 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.

29. myininaya

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

30. bmelyk

there is no hypotnus tho :s

31. bmelyk

oh i gott find it lol

32. myininaya

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

33. bmelyk

square root of 5 is the hyp.

34. bmelyk

???

35. myininaya

Yes:)

36. bmelyk

okay so now i take sec(squareroot 5)

37. myininaya

so what is sec(u)

38. bmelyk

and do i restrict my tan?

39. myininaya

|dw:1352311961504:dw|

40. myininaya

41. bmelyk

so its just root 5

42. myininaya

Yep. :)

43. ChmE

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

44. bmelyk

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

45. myininaya

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