anonymous
  • anonymous
Help for a medal? I'm really struggling with limits. (I'll post the problem below)
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
1 Attachment
jim_thompson5910
  • jim_thompson5910
Did you make a random guess at 1/4 ?
anonymous
  • anonymous
Yes, I know that it was the right answer because the system told me it was. Luckily I wasn't graded on it.

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anonymous
  • anonymous
I just tried every option until it said correct, since all my answers were wrong.
jim_thompson5910
  • jim_thompson5910
I see
jim_thompson5910
  • jim_thompson5910
|dw:1441858889495:dw|
jim_thompson5910
  • jim_thompson5910
what is sin^2 equal to in terms of cos^2 ?
anonymous
  • anonymous
I personally don't have a clue. I know that cos(0) = 1 and sin(0) = 0 but that's about it
jim_thompson5910
  • jim_thompson5910
you've learned that cos^2 + sin^2 = 1 right?
anonymous
  • anonymous
I probably have in the past.
jim_thompson5910
  • jim_thompson5910
look on page 2 of the pdf http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf look at the pythagorean identities
jim_thompson5910
  • jim_thompson5910
look familiar?
anonymous
  • anonymous
Oh wow! I do recognize a lot of this stuff. Thanks for the pdf.
jim_thompson5910
  • jim_thompson5910
yeah it's handy to keep it as a reference. Print it out if you can. Anyways, solving for sin^2 gives \[\Large \sin^2(\theta) = 1-\cos^2(\theta)\]
jim_thompson5910
  • jim_thompson5910
|dw:1441859333706:dw| I'm getting everything in terms of cos(theta)
jim_thompson5910
  • jim_thompson5910
what would \(\Large 1-\cos^2(\theta)\) factor to?
anonymous
  • anonymous
\[\sin^2(\theta)\] ??
jim_thompson5910
  • jim_thompson5910
hint: think of a^2 - b^2
anonymous
  • anonymous
1?
jim_thompson5910
  • jim_thompson5910
another hint: a^2 - b^2 = (a-b)(a+b)
anonymous
  • anonymous
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...
jim_thompson5910
  • jim_thompson5910
if you had x^2 - 5^2, what would that factor to?
anonymous
  • anonymous
(x+5)(x-5)
jim_thompson5910
  • jim_thompson5910
so like that idea, we factor 1 - cos^2 into (1+cos)*(1-cos)
jim_thompson5910
  • jim_thompson5910
|dw:1441860277846:dw|
anonymous
  • anonymous
Oh okay. I see now.
jim_thompson5910
  • jim_thompson5910
I'm sure you see what cancels?
anonymous
  • anonymous
Yes the two (1-cos theta) ones cancel
jim_thompson5910
  • jim_thompson5910
yep |dw:1441860261204:dw|
jim_thompson5910
  • jim_thompson5910
we're now left with \[\Large \frac{1}{2(1+\cos(\theta))}\]
jim_thompson5910
  • jim_thompson5910
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
anonymous
  • anonymous
then it would be... 1 / 2(1+1) 1 / 4
jim_thompson5910
  • jim_thompson5910
yep
anonymous
  • anonymous
You're amazing! Thank you so much!
jim_thompson5910
  • jim_thompson5910
no problem

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