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

- anonymous

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

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- chestercat

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- anonymous

##### 1 Attachment

- jim_thompson5910

Did you make a random guess at 1/4 ?

- 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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## More answers

- anonymous

I just tried every option until it said correct, since all my answers were wrong.

- jim_thompson5910

I see

- jim_thompson5910

|dw:1441858889495:dw|

- jim_thompson5910

what is sin^2 equal to in terms of cos^2 ?

- 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

you've learned that cos^2 + sin^2 = 1 right?

- anonymous

I probably have in the past.

- 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

look familiar?

- anonymous

Oh wow! I do recognize a lot of this stuff.
Thanks for the pdf.

- 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

|dw:1441859333706:dw|
I'm getting everything in terms of cos(theta)

- jim_thompson5910

what would \(\Large 1-\cos^2(\theta)\) factor to?

- anonymous

\[\sin^2(\theta)\] ??

- jim_thompson5910

hint: think of a^2 - b^2

- anonymous

1?

- jim_thompson5910

another hint: a^2 - b^2 = (a-b)(a+b)

- 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

if you had x^2 - 5^2, what would that factor to?

- anonymous

(x+5)(x-5)

- jim_thompson5910

so like that idea, we factor 1 - cos^2 into (1+cos)*(1-cos)

- jim_thompson5910

|dw:1441860277846:dw|

- anonymous

Oh okay. I see now.

- jim_thompson5910

I'm sure you see what cancels?

- anonymous

Yes the two (1-cos theta) ones cancel

- jim_thompson5910

yep
|dw:1441860261204:dw|

- jim_thompson5910

we're now left with
\[\Large \frac{1}{2(1+\cos(\theta))}\]

- 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

then it would be...
1 / 2(1+1)
1 / 4

- jim_thompson5910

yep

- anonymous

You're amazing! Thank you so much!

- jim_thompson5910

no problem

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