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anonymous

  • one year ago

I don't believe that I solved this problem correctly... (will post below)

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  1. anonymous
    • one year ago
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    \[\frac{ \cos2\theta }{ \cos \theta -\sin \theta}\]

  2. anonymous
    • one year ago
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    I know that \[\cos2 \theta = 1-2\sin^2\] so I changed that to the numerator, and then plugged in my limit into the thetas, which is pi/4

  3. anonymous
    • one year ago
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    I ended up with 0/0 meaning that the limit didn't exist, but I have a feeling I did this wrong.

  4. anonymous
    • one year ago
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    Try the identity\[\cos 2\theta = \cos^2\theta - \sin^2\theta\]This can be factored and then the fraction can be simplified

  5. anonymous
    • one year ago
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    Let me clarify... pi/4 = c not the limit. Sorry about that.

  6. Mertsj
    • one year ago
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    \[\frac{\cos ^2\theta-\sin ^2\theta}{\cos \theta-\sin \theta}=\frac{(\cos \theta-\sin \theta)(\cos \theta+\sin \theta)}{\cos \theta-\sin \theta}\]

  7. anonymous
    • one year ago
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    Then I would just cancel out the like terms in the numerator and the denominator, plug in my c values and solve?

  8. anonymous
    • one year ago
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    That's right

  9. anonymous
    • one year ago
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    Oh okay, thank you. :) I guess I'll medal ospreytriple since they responded first? But I'll fan both of you. :)

  10. anonymous
    • one year ago
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    not to butt in but if you get \[\frac{0}{0}\] it does NOT mean the limit doesn't exist it means you have to do more work

  11. anonymous
    • one year ago
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    Great to know! I'll keep that in mind next time I get it as an answer.

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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