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StokesParadox Group Title

@zepdrix

  • one year ago
  • one year ago

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  1. zepdrix Group Title
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    Hmm darn, I have plenty of resources for Calculus... I can't think of any for trig though haha.

    • one year ago
  2. StokesParadox Group Title
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    I'm looking at yesterdays problems...one seconds.... we should probably start with inverses of trig functions..one second

    • one year ago
  3. StokesParadox Group Title
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    http://tutorial.math.lamar.edu/Extras/AlgebraTrigReview/SolveTrigEqn.aspx

    • one year ago
  4. StokesParadox Group Title
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    do these look difficult enough (for me)

    • one year ago
  5. zepdrix Group Title
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    Hmm they look pretty simple. I wouldn't say they're way easier than what we were working on yesterday. They might be worth spending some time on c:

    • one year ago
  6. zepdrix Group Title
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    Take a stab at #3

    • one year ago
  7. StokesParadox Group Title
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    yeah they look too easy....I'll find some harder ones \[sin(5x)=-\frac{\sqrt 3}{2}\] |dw:1362793858479:dw| am I right so far?

    • one year ago
  8. zepdrix Group Title
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    Solve for x! :)

    • one year ago
  9. zepdrix Group Title
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    I mean, yes you're right so far!

    • one year ago
  10. zepdrix Group Title
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    Keep in mind we want to rotate in the positive direction. So make sure your angle corresponds to that.

    • one year ago
  11. StokesParadox Group Title
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    \[sin^{-1}(-\sqrt{3}/2)=5x\] uhm I'm stuck \[\frac {5\pi} 3\]

    • one year ago
  12. zepdrix Group Title
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    \[\large 5\pi/3=5x\] So you found that your angle, \(\large 5x\) was equal to the special angle \(\large 5\pi/3\). Looks good so far.

    • one year ago
  13. StokesParadox Group Title
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    oh so that is right!

    • one year ago
  14. StokesParadox Group Title
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    \[x=\frac {\pi}{3}\]

    • one year ago
  15. zepdrix Group Title
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    Yay good job

    • one year ago
  16. StokesParadox Group Title
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    wait, that's it? what did we just discover?

    • one year ago
  17. StokesParadox Group Title
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    wait wait wait....

    • one year ago
  18. zepdrix Group Title
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    There is some angle \(\large 5x\) that corresponded to the given equation. And we were ultimately trying to find 1/5 of that angle. No waiting! >:O lol

    • one year ago
  19. StokesParadox Group Title
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    LOL \[sin(5x)=-\frac{\sqrt 3}{2}\] why are we trying to find 1/5 of that angle do you mean: "there is some angle x"

    • one year ago
  20. zepdrix Group Title
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    Our angle is the thing inside of the trig function. Think of it maybe as like, \(\large 5x=\theta\). We want to solve for part of that angle. But for the purposes of relating this to special angles, it's good to think of the entire contents as the angle. :)

    • one year ago
  21. StokesParadox Group Title
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    I am.....so why are we trying to find 1/5 of that angle? Oh you're just saying we solve for x which is 1/5th of that angle....I think I get it

    • one year ago
  22. StokesParadox Group Title
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    si?

    • one year ago
  23. zepdrix Group Title
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    si c: I'm just being a little technical I guess.

    • one year ago
  24. StokesParadox Group Title
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    It makes sense :)

    • one year ago
  25. StokesParadox Group Title
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    should we do a more difficult problem? what are these problems called \[sec\left(cos^{-1}\frac 12\right)\] I feel sooooo d u m b LOL

    • one year ago
  26. zepdrix Group Title
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    lol poor gal c:

    • one year ago
  27. zepdrix Group Title
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    Ok so we're dealing with an inverse even function on the inside. So our angle will be in the 1st or second quadrant, yes? Hmmm +1/2. Is that a length to the right, or left? :D

    • one year ago
  28. StokesParadox Group Title
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    I will feel smarter by the end of the weekend....hopefully.... When I Know How To Do These Darn Problems o.O

    • one year ago
  29. StokesParadox Group Title
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    right

    • one year ago
  30. StokesParadox Group Title
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    the length to the right

    • one year ago
  31. StokesParadox Group Title
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    I found some nice aerobics music!!!!

    • one year ago
  32. zepdrix Group Title
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    lol that was random XD

    • one year ago
  33. StokesParadox Group Title
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    Thanks :)

    • one year ago
  34. StokesParadox Group Title
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    I'm planning to work out in 15 mins

    • one year ago
  35. StokesParadox Group Title
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    I just need to find a way to download it

    • one year ago
  36. zepdrix Group Title
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    Oh i see c:

    • one year ago
  37. StokesParadox Group Title
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    ....so these problems are called inverse functions within trig functions? i"m trying to find some problems on google.

    • one year ago
  38. zepdrix Group Title
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    Sorry I'm not quite sure what they're called :( Maybe something Composition of Trig functions...

    • one year ago
  39. zepdrix Group Title
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    http://www.ck12.org/book/CK-12-Trigonometry-Concepts/r1/section/4.6/ Near the bottom there is a \(\large \text{Practice}\) section. Some good problems there.

    • one year ago
  40. StokesParadox Group Title
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    • one year ago
  41. StokesParadox Group Title
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    which one should a take a stab at?

    • one year ago
  42. zepdrix Group Title
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    5

    • one year ago
  43. StokesParadox Group Title
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    \[tan(cos^{-1}1)\] \[cos\alpha =1\] at zero and \(2\pi\)

    • one year ago
  44. zepdrix Group Title
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    \[\large \tan(0)\] K looks good so far.

    • one year ago
  45. StokesParadox Group Title
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    hmmmmm..... 0? final answer

    • one year ago
  46. zepdrix Group Title
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    Yes good!

    • one year ago
  47. StokesParadox Group Title
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    yayyayayayayay!!!!! ok I'm gonna work out now. THanks!!!

    • one year ago
  48. zepdrix Group Title
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    Cya c:

    • one year ago
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