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anonymous

  • 5 years ago

Find the exact value of the following expression (multiple choice)

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  1. anonymous
    • 5 years ago
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    http://tigger.uic.edu/~calkafka/spring2011math121practiceexam3.pdf - number 17

  2. anonymous
    • 5 years ago
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    method or answers?

  3. anonymous
    • 5 years ago
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    method. i'll figure out the answer from the method..that way, i learn the method and figure the answer

  4. anonymous
    • 5 years ago
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    or calculator?

  5. anonymous
    • 5 years ago
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    k

  6. anonymous
    • 5 years ago
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    number 1. think of a number (angle) between 0 and pi whose cosine is \[\frac{\sqrt{3}}{2}\]

  7. anonymous
    • 5 years ago
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    look at trig cheat sheet here http://tutorial.math.lamar.edu/cheat_table.aspx if you do not remember

  8. mathteacher1729
    • 5 years ago
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    Drawing pictures is SUPER helpful. The trig sheet Satellite mentioned is also very nice. :)

  9. anonymous
    • 5 years ago
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    oh i only nueed help on number 17. thanks though

  10. anonymous
    • 5 years ago
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    aaahhhhhhhh

  11. mathteacher1729
    • 5 years ago
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    Still,l a pic would help. Lemmie sketch out one... brb.

  12. anonymous
    • 5 years ago
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    addition angle formula: \[sin(a+b)=sin(a)cos(b)+cos(a)sin(b)\]

  13. mathteacher1729
    • 5 years ago
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    You won't even have to do that, I think, cuz you have to do the inverse sin & cos stuff inside the main sin argument.

  14. anonymous
    • 5 years ago
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    thats what i thought..at math teacher

  15. anonymous
    • 5 years ago
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    i just dont know how to go about solving it

  16. anonymous
    • 5 years ago
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    typeset is goofy but i assume it means \[sin(sin^{-1}(\frac{2}{3}) + cos^{-1}(\frac{1}{3}))\]

  17. anonymous
    • 5 years ago
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    yuppers. thats what it means

  18. anonymous
    • 5 years ago
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    put \[a=sin^{-1}(\frac{2}{3}), b=cos^{-1}(\frac{1}{3})\]

  19. anonymous
    • 5 years ago
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    use formula above. you aready know \[sin(sin^{-1}(\frac{2}{3}))=\frac{2}{3}\]

  20. anonymous
    • 5 years ago
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    you need \[cos^{-1}(\frac{2}{3})=\frac{\sqrt{5}}{3}\] by pythagoras

  21. anonymous
    • 5 years ago
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    oops typo

  22. anonymous
    • 5 years ago
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    you need \[cos(sin^{-1}(\frac{2}{3}))=\frac{\sqrt{5}}{3}\]

  23. anonymous
    • 5 years ago
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    by pythagoras.

  24. mathteacher1729
    • 5 years ago
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    Here is the first part.

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  25. anonymous
    • 5 years ago
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    ahh the pythagorean thm

  26. anonymous
    • 5 years ago
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    what math teacher said. you only need to find \[sin(cos^{-1}(\frac{1}{3}))=\frac{\sqrt{2}}{3}\]

  27. anonymous
    • 5 years ago
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    now you have everything you need to plug in to the "addition angle" formula

  28. anonymous
    • 5 years ago
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    but i think you must use the formula now. you have all 4 numbers that you need. \[sin(a)=\frac{2}{3}\] \[cos(a)=\frac{\sqrt{5}}{3}\]

  29. anonymous
    • 5 years ago
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    \[sin(b)=\frac{\sqrt{2}}{3}\] \[cos(b)=\frac{1}{3}\]

  30. anonymous
    • 5 years ago
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    write out the formula, substitute the numbers, and be done.

  31. anonymous
    • 5 years ago
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    right. gimme a sec.

  32. anonymous
    • 5 years ago
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    whoooooooooooops

  33. anonymous
    • 5 years ago
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    \[sin(b)=\frac{\sqrt{8}}{3}\]

  34. anonymous
    • 5 years ago
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    fraid both math teacher and i made a mistake.

  35. anonymous
    • 5 years ago
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    its okay. i'm still working on it. thanks

  36. anonymous
    • 5 years ago
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    if you look at the picture math teachers sent unfortunately the second triangle is wrong. adjacent side is 1, hypotenuse is 3, opposite side should be \[{\sqrt{8}}=2\sqrt{2}\]

  37. anonymous
    • 5 years ago
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    get \[\frac{2}{3}\times \frac{\sqrt{5}}{3}+\frac{1}{3}\times \frac{2\sqrt{2}}{3}\]

  38. anonymous
    • 5 years ago
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    is the answer for number 17 "A"? thats what i'm getting. http://tigger.uic.edu/~calkafka/spring2011math121practiceexam3.pdf

  39. mathteacher1729
    • 5 years ago
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    Oh my gosh, I can't believe I did that. :( here is the corrected version.

  40. anonymous
    • 5 years ago
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    dont worry mathteacher. satellite clarified. i'm debating on whether its A or B for number 17 in the link i posted

  41. anonymous
    • 5 years ago
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    A yes!

  42. anonymous
    • 5 years ago
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    SWEET! thanky you so much!

  43. anonymous
    • 5 years ago
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    welcome

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