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I can't solve either of these.

  1. Feb 11, 2007 #1
    1. The problem statement, all variables and given/known data
    1. DIRECTIONS: Use half-angle formulas to simplify the expression.

    negative >>- *all square root*(1+cos4x/1-cos4x)

    2. Verify the identity.

    (sinx +/- siny)/(sinx + cosy) = Tan {(x +/- y)/(2)}


    2. Relevant equations
    1. All half angle formulas.
    2. Trig formulas. there is a bunch

    3. The attempt at a solution
    1. i don't think its even possible
    2. i don't see how this one is possible either
     
    Last edited by a moderator: Feb 11, 2007
  2. jcsd
  3. Feb 11, 2007 #2

    cristo

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    Have you really tried these?

    1. [tex]- \sqrt{\frac{1+cos(4x)}{1-cos(4x)}}[/tex]

    What have you tried? What is the double angle formula for cos(4x) in terms of cos(2x), sin(2x)?

    2.[tex]\frac{sinx \pm siny}{sinx+cosy}=tan\left(\frac{x \pm y}{2}\right)[/tex]

    Again, what have you tried? It might be easier to start with the right hand side, and write tan in terms of sin and cos, then work to the left.
     
    Last edited: Feb 11, 2007
  4. Feb 11, 2007 #3
    None of the half-angle formulas seemed to correspond to number 1. I tried using the half angle formula for number 2, but was thrown off by the "+/-" and don't know what to do.
     
  5. Feb 11, 2007 #4

    cristo

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    1. Make the substitution y=2x. Can you simplify cos(2y)?

    2. Follow my hint, and write tan in terms of sin and cos. There are standard identities for [itex]sin(A \pm B)[/itex] and [itex]cos(A \pm B)[/itex]. What identities do you know?

    Show some work!
     
  6. Feb 11, 2007 #5
    so like sin/cos (x +- y)/2
     
  7. Feb 11, 2007 #6

    cristo

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    You mean [tex]tan\left(\frac{x \pm y}{2}\right)=\frac{sin\left(\frac{x \pm y}{2}\right)}{cos\left(\frac{x \pm y}{2}\right)}[/tex]

    Now, do you know the identies I hinted at above?
     
  8. Feb 11, 2007 #7
    They look like sum and difference identities. And how do you do that thing with the problem so that it looks good.
     
  9. Feb 11, 2007 #8
    and how did you go from this:

    [​IMG]

    to this:

    [​IMG]


    It looks like part of sum to product but its not.
     
  10. Feb 11, 2007 #9

    cristo

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    I didn't! Following my hint in post #2 I wrote tan in terms of sin and cos. Try and work from this to obtain the LHS of the identity.

    I wrote my equations in LaTex; to see the code, click on the image.
     
  11. Feb 11, 2007 #10
    oh ok i see what you did you wrote what tan of that equals in sin over cosine and then that equals the left side. ok ok ok. now were on the same page i was confused. now........ hmm.....

    so then do i now have to use these?

    [​IMG]
     
  12. Feb 12, 2007 #11

    cristo

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    I would use the identities [itex]\sin(A\pm B)=\sin A \cos B\pm \cos A\sin B[/itex] and the corresponding identity for cosine, and let A = x/2 and B = y/2.
     
  13. Feb 14, 2007 #12
    It just so happens that it doesn't really matter anyways. Cause my teacher said that this problems isn't possible to do anyway. Thanks for your time.
     
  14. Feb 14, 2007 #13

    cristo

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    Why would you teacher set questions that cannot be solved? I'm sure number 1 can be simplified into trigonometric terms of argument x. I've not worked through number two, but I don't see why it would be set if it cannot be solved!
     
  15. Feb 15, 2007 #14
    He accidentally gave us the wrong problem from the book. He said do problem 100 but had done 99 instead. Even when one of the genius kids in our class pointed out to him that it couldn't be done he told him he must have made a mistake cause the book wouldn't give a bad problem. Then he told us the next day that he spent an hour on it and couldn't do it so he gave up. Thats for number 2.

    For number 1 he said the directions were bad cause you can't do it with half angle formulas
     
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