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bmelyk

  • 3 years ago

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

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  1. bmelyk
    • 3 years ago
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    \[\cos(\arcsin(\frac{ \sqrt{3} }{ 2 }))\]

  2. ChmE
    • 3 years ago
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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

  3. bmelyk
    • 3 years ago
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    okay so arc sin is actually sin^-1

  4. ChmE
    • 3 years ago
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    Ya they are used interchangeably.

  5. bmelyk
    • 3 years ago
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    so it's sin of 30 degrees?

  6. myininaya
    • 3 years ago
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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 \[\text{ let } u=\arcsin(\frac{\sqrt{3}}{2})\] \[\cos(u)=\cos(\arcsin(\frac{\sqrt{3}}{2}))=\frac{\text{ adjacent side to u}}{hyp}\]

  7. bmelyk
    • 3 years ago
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    right so it's just root 3 over 2

  8. myininaya
    • 3 years ago
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    Nope sin(u) is that what is cos(u)?

  9. bmelyk
    • 3 years ago
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    or it's at 60 degrees and 120

  10. ChmE
    • 3 years ago
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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

  11. bmelyk
    • 3 years ago
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    60 and 120, right?

  12. ChmE
    • 3 years ago
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    right it is also at 120. good catch

  13. bmelyk
    • 3 years ago
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    so i write it as cos(60) ??

  14. bmelyk
    • 3 years ago
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    so my answer should be 1/2 and -1/2?

  15. myininaya
    • 3 years ago
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    but arcsin( ) only has range between -90 and 90.

  16. ChmE
    • 3 years ago
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    Thats what I got !!

  17. myininaya
    • 3 years ago
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    So you will only have one answer.

  18. ChmE
    • 3 years ago
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    Explain that plx @myininaya . I don't remember that.

  19. bmelyk
    • 3 years ago
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    i thought that had to be specified

  20. ChmE
    • 3 years ago
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    http://en.wikipedia.org/wiki/Inverse_trigonometric_functions He's right. The only answer would be 1/2

  21. myininaya
    • 3 years ago
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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]

  22. bmelyk
    • 3 years ago
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    i can only do this type of problem if it's one to one?

  23. myininaya
    • 3 years ago
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    I'm a her.

  24. ChmE
    • 3 years ago
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    Sry I meant nothing by it

  25. bmelyk
    • 3 years ago
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    so if i wanted to do sec(arctan(2)) i do the same thing?

  26. myininaya
    • 3 years ago
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    I like making a pretty right triangle. :)

  27. bmelyk
    • 3 years ago
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    huh?

  28. myininaya
    • 3 years ago
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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.

  29. myininaya
    • 3 years ago
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    After that, find sec(u) and you are done.

  30. bmelyk
    • 3 years ago
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    there is no hypotnus tho :s

  31. bmelyk
    • 3 years ago
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    oh i gott find it lol

  32. myininaya
    • 3 years ago
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    Yes there is. You find it by using the Pythagorean thm.

  33. bmelyk
    • 3 years ago
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    square root of 5 is the hyp.

  34. bmelyk
    • 3 years ago
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    ???

  35. myininaya
    • 3 years ago
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    Yes:)

  36. bmelyk
    • 3 years ago
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    okay so now i take sec(squareroot 5)

  37. myininaya
    • 3 years ago
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    so what is sec(u)

  38. bmelyk
    • 3 years ago
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    and do i restrict my tan?

  39. myininaya
    • 3 years ago
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    |dw:1352311961504:dw|

  40. myininaya
    • 3 years ago
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    sec(u)=hyp/adj

  41. bmelyk
    • 3 years ago
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    so its just root 5

  42. myininaya
    • 3 years ago
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    Yep. :)

  43. ChmE
    • 3 years ago
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    Study her method because it was way easier than my explanation.

  44. bmelyk
    • 3 years ago
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    and i state my domain and range for just sec right?

  45. myininaya
    • 3 years ago
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    No it was just looking for sec(arctan(2)) which is just \[\sqrt{5}\]

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