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I am Completely Clueless in Calculus

  1. Sep 17, 2004 #1
    Today i was given some review questions from pre-calc, but i forgot how to even start the problems. One problem looks like this:
    ln lsinxl=(ln l1-cos2x)-ln2)


    i know that i have to prove that each side is equal, but i don't know where to begin...any suggestions?
     
  2. jcsd
  3. Sep 17, 2004 #2

    Pyrrhus

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    [tex] \ln |\sin(x)| = \ln |1-cos2x| - \ln(2) [/tex]

    Apply Logarithm properties, (Review them)

    [tex] \ln |\sin(x)| = \ln |\frac{1-cos2x}{2}| [/tex]

    Here's another hint Look up Half-Angle identities.
     
  4. Sep 17, 2004 #3

    mmmm...I love Trig Identities.
     
  5. Sep 17, 2004 #4
    Greetings friend,

    Here is a step by step solution for your inquiry:

    ln |sin(x)| = ln | (1-cos2x) / 2 | (after applying neccessary log laws)
    you raise base e to some exponets:
    e^(ln|sin(x)|) = e^(ln|(1-cos2x)/2)
    sin(x) = (1-cos2x)/2 (notice 1-cos2x/2 is another way of saying sin^2(x))

    so they are not equal... unless you made a type and meant ln|sin^2(x)| originally.
     
  6. Sep 17, 2004 #5
    PrudensOptimus is correct. There must be something wrong with your equality, unless the question was if you were to prove or disprove it. If the equality should be true, then

    [tex]

    ln |sin^2 x| = ln | 1 - cos2x | - ln (2)

    [/tex]
     
  7. Sep 19, 2004 #6
    That's for all the help guys!!!
    You are all lifesavers!!!
     
  8. Sep 19, 2004 #7

    Pyrrhus

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    It's glad to be of help!, and Welcome to PF!, i hope you enjoy your stay :rofl:
     
  9. Sep 19, 2004 #8
    This is the second best site in the world, after google of course.
     
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