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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


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    Homework Helper

    [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


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

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


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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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