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Trigonometric Identity

  1. Jul 7, 2008 #1
    1. The problem statement, all variables and given/known data

    I'm attempting to prove that

    1 - sin^2 t /(1 + cos t) - cos^2/(1+tan t) = cos t sin t

    2. The attempt at a solution

    I've tried various approaches. The most promising has the LHS reduced to:

    (sin t cos t (1 + cos t + sin t cos t))/((1 + cos t)(cos t + sin t)).

    I've also shown numerically that the LHS resembles the RHS so I don't think there was a typo in the original problem.

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jul 7, 2008 #2
    Note that:

    \frac{\sin^2(t)}{1+\cos(t)} = \frac{(1+\cos(t))(1-\cos(t))}{1+\cos(t)} = 1-\cos(t) [/tex]

    You think you can proceed from this point on?
  4. Jul 8, 2008 #3
    The first two terms simplify to cos t, but I'm still having the same issue wrt further simplification. One version has me stuck at:

    cos t (cos t + sin t - cos^2 t)/(cos t + sin t)

    I'd appreciate another hint. Thanks.
  5. Jul 8, 2008 #4


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    Homework Helper

    Is this the trigo identity you're trying to prove:

    [tex]1 - \frac{\sin^2 t}{1 + \cos t} - \frac{\cos^2 t}{1+\tan t} = \sin t \cos t[/tex]

    If so, then it doesn't appear to hold for [itex]t=\frac{\pi}{4}[/itex].
  6. Jul 8, 2008 #5

    So it appears. Let me go back to the person who set the original problem. Sorry!
  7. Jul 8, 2008 #6
    It turns out that it was a typo. It should read:
    [tex]1 - \frac{\sin^2 t}{1 + \cot t} - \frac{\cos^2 t}{1+\tan t} = \sin t \cos t[/tex]

    Solution is straightforward.
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