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anonymous

  • one year ago

Help for a medal? I'm really struggling with limits. (I'll post the problem below)

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  1. anonymous
    • one year ago
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  2. jim_thompson5910
    • one year ago
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    Did you make a random guess at 1/4 ?

  3. anonymous
    • one year ago
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    Yes, I know that it was the right answer because the system told me it was. Luckily I wasn't graded on it.

  4. anonymous
    • one year ago
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    I just tried every option until it said correct, since all my answers were wrong.

  5. jim_thompson5910
    • one year ago
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    I see

  6. jim_thompson5910
    • one year ago
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    |dw:1441858889495:dw|

  7. jim_thompson5910
    • one year ago
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    what is sin^2 equal to in terms of cos^2 ?

  8. anonymous
    • one year ago
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    I personally don't have a clue. I know that cos(0) = 1 and sin(0) = 0 but that's about it

  9. jim_thompson5910
    • one year ago
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    you've learned that cos^2 + sin^2 = 1 right?

  10. anonymous
    • one year ago
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    I probably have in the past.

  11. jim_thompson5910
    • one year ago
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    look on page 2 of the pdf http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf look at the pythagorean identities

  12. jim_thompson5910
    • one year ago
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    look familiar?

  13. anonymous
    • one year ago
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    Oh wow! I do recognize a lot of this stuff. Thanks for the pdf.

  14. jim_thompson5910
    • one year ago
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    yeah it's handy to keep it as a reference. Print it out if you can. Anyways, solving for sin^2 gives \[\Large \sin^2(\theta) = 1-\cos^2(\theta)\]

  15. jim_thompson5910
    • one year ago
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    |dw:1441859333706:dw| I'm getting everything in terms of cos(theta)

  16. jim_thompson5910
    • one year ago
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    what would \(\Large 1-\cos^2(\theta)\) factor to?

  17. anonymous
    • one year ago
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    \[\sin^2(\theta)\] ??

  18. jim_thompson5910
    • one year ago
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    hint: think of a^2 - b^2

  19. anonymous
    • one year ago
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    1?

  20. jim_thompson5910
    • one year ago
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    another hint: a^2 - b^2 = (a-b)(a+b)

  21. anonymous
    • one year ago
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    I'm sorry, but I am completely confused. Math used to be my best subject, though that doesn't seem to be the case anymore...

  22. jim_thompson5910
    • one year ago
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    if you had x^2 - 5^2, what would that factor to?

  23. anonymous
    • one year ago
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    (x+5)(x-5)

  24. jim_thompson5910
    • one year ago
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    so like that idea, we factor 1 - cos^2 into (1+cos)*(1-cos)

  25. jim_thompson5910
    • one year ago
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    |dw:1441860277846:dw|

  26. anonymous
    • one year ago
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    Oh okay. I see now.

  27. jim_thompson5910
    • one year ago
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    I'm sure you see what cancels?

  28. anonymous
    • one year ago
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    Yes the two (1-cos theta) ones cancel

  29. jim_thompson5910
    • one year ago
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    yep |dw:1441860261204:dw|

  30. jim_thompson5910
    • one year ago
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    we're now left with \[\Large \frac{1}{2(1+\cos(\theta))}\]

  31. jim_thompson5910
    • one year ago
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    at this point, plugging in theta = 0 does not lead to a division by zero error (as it did before in the previous expressions). So you can now plug in theta = 0 and simplify

  32. anonymous
    • one year ago
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    then it would be... 1 / 2(1+1) 1 / 4

  33. jim_thompson5910
    • one year ago
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    yep

  34. anonymous
    • one year ago
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    You're amazing! Thank you so much!

  35. jim_thompson5910
    • one year ago
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    no problem

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