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

This Question is Closed

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2Did you make a random guess at 1/4 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, I know that it was the right answer because the system told me it was. Luckily I wasn't graded on it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I just tried every option until it said correct, since all my answers were wrong.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441858889495:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2what is sin^2 equal to in terms of cos^2 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I personally don't have a clue. I know that cos(0) = 1 and sin(0) = 0 but that's about it

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2you've learned that cos^2 + sin^2 = 1 right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I probably have in the past.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2look on page 2 of the pdf http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf look at the pythagorean identities

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2look familiar?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh wow! I do recognize a lot of this stuff. Thanks for the pdf.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2yeah 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
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441859333706:dw I'm getting everything in terms of cos(theta)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2what would \(\Large 1\cos^2(\theta)\) factor to?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sin^2(\theta)\] ??

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2hint: think of a^2  b^2

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2another hint: a^2  b^2 = (ab)(a+b)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'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
 one year ago
Best ResponseYou've already chosen the best response.2if you had x^2  5^2, what would that factor to?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2so like that idea, we factor 1  cos^2 into (1+cos)*(1cos)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441860277846:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2I'm sure you see what cancels?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes the two (1cos theta) ones cancel

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2yep dw:1441860261204:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2we're now left with \[\Large \frac{1}{2(1+\cos(\theta))}\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2at 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
 one year ago
Best ResponseYou've already chosen the best response.0then it would be... 1 / 2(1+1) 1 / 4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You're amazing! Thank you so much!
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