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Trig: Basic Identities Problem

  1. Mar 26, 2004 #1


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    Hi, today I was writing my Identities final test and got stuck on one question which I couldn't solve.

    The question was: If [tex]sinB=-\frac{5}{13}[/tex] find the exact value of [tex]tan2B[/tex].

    I've got this far, correct me if I went wrong anywhere: [tex]tan2B=\frac{sin2B}{cos2B}=\frac{2sinBcosB}{2cos^2B-1}[/tex]

    Then, when I tryied to solve further I came to a dead end, even though I tried many different ways.

    Thanx in advance!
  2. jcsd
  3. Mar 26, 2004 #2


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    There should probably be 2 answers.

    First of all, with that triangle, the hypotinuse is 13, x is +12 and -12, y is -5, and B is either in the third or fourth quadrant.

    Fill in the equation using x as 12, then fill in the equation using -12.
    Last edited: Mar 26, 2004
  4. Mar 26, 2004 #3


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    Tnx for your help, the answers I've got are [tex]\frac{120}{119}[/tex] and [tex]\frac{-120}{119}[/tex]

    Once I get my test back I'll post if the answers match.
    Last edited: Mar 27, 2004
  5. Mar 27, 2004 #4
    If sin(B) = -5/13, then sin^2(B) = 25/169. Using the Pythagorean identity, we have that cos^2(B) = 1 - sin^2(B) = 1 - 25/169 = 144/169, which gives cos(B) = +/- 12/13. Plug those values into the formula for tan(2B) and voila...
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