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Help with trig identities

  1. Feb 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Show that (sin^4 x + (sin^2 x * cos^2 x)) / (cos^2 x - 1) == -1
    2. Relevant equations
    Sin^2 x + cos^2 x == 1

    3. The attempt at a solution
    (sin ^4 x + (sin^2 x * cos^2 x)) / (cos^2 x - 1)
    = ((sin^2 x)(sin^2 x) + (sin^2 x * cos^2 x)) / (cos^2 x - 1)
    =((sin^2 x)(1 - cos^2 x) + (sin^2x * cos^2 x)) / (cos^2 x -1 )
    = (sin^2 x - (sin^2 x * cos^2 x) + (sin^2 x * cos^2 x)) / (cos^2 x - 1 )
    = (sin^2 x) / (cos^2 x - 1 )
    = (1 - cos^2 x ) / (cos^2 x -1)
    = -1

    I think this is correct, but when I looked up the answer in the back of the textbook it showed completely different working using different substitutions. Did I make any mistakes? or are there two or more solutions to this problem?, if this is the case would I be marked down in an exam for using this method?.
     
  2. jcsd
  3. Feb 18, 2015 #2

    SteamKing

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    In working trig identity problems, there may be more than one valid substitution which can be used to obtain a simplification, especially with complicated or lengthy expressions.

    If this were an exam exercise, no, you should not be penalized for using a valid method of simplification, even if it differs from a method preferred by the instructor.
     
  4. Feb 18, 2015 #3
    Thanks, that puts my mind at ease.
     
  5. Feb 22, 2015 #4

    HallsofIvy

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    I, for example, would, seeing that "[itex]sin^4(x)[/itex]" change everything else to "sin(x)".
    Since [itex]cos^2(x)= 1- sin^2(x)[/itex], the numerator is [itex]sin^4(x)+ sin^2(x)(1- sin^2(x))= sin^4()+ sin^2(x)- sin^4(x)= sin^2(x)[/itex]. Is that what your text book does?
     

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