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Math Help - problem in addition formulae and double angle formulae

  1. #1
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    problem in addition formulae and double angle formulae

    please help me with this question:

    given tan2\theta=\frac{4}{3}, without using calculator,find the values of sin\theta.
    Last edited by mastermin346; July 5th 2010 at 05:49 AM.
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  2. #2
    Senior Member eumyang's Avatar
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    Uh... come again? Maybe I'm not seeing it, but is there more to this problem? It this part of a bigger problem?

    EDIT: Do you mean this:
    given sin 2\theta=\frac{4}{3}, without using calculator,find the values of sin\theta.
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  3. #3
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    yes..that is the question.

    opp..sory..the question is tan 2\theta..i have edit it
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  4. #4
    Senior Member eumyang's Avatar
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    I'm getting 2 answers:
    sin\,\theta= \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}
    I started by using the tangent double identity:
    \frac{2tan\,\theta}{1-tan^2\,\theta}=\frac{4}{3}
    Cross-multiplied:
    3\left(2tan\,\theta \right)=4\left(1-tan^2\,\theta \right)
    ... solved for tan\,\theta, then rewrote in terms of sin\,\theta.

    EDIT: I originally thought that both answers I wrote above were plus-minus, but I forgot that tan\,2\theta is positive, which means that the angle 2\theta has to be in Quadrants I or III. So \theta is between 0 and 45 or between 90 and 135, and for those values of \theta, sine is positive.
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  5. #5
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    You should know that

    \tan{2\theta} = \frac{2\tan{\theta}}{1 - \tan^2{\theta}}.


    So if \tan{2\theta} = \frac{4}{3} then

    \frac{2\tan{\theta}}{1 - \tan^2{\theta}} = \frac{4}{3}

    3(2\tan{\theta}) = 4(1 - \tan^2{\theta})

    6\tan{\theta} = 4 - 4\tan^2{\theta}

    4\tan^2{\theta} + 6\tan{\theta} - 4 = 0

    2\tan^2{\theta} + 3\tan{\theta} - 2 = 0

    2\tan^2{\theta} - \tan{\theta} + 4\tan{\theta} - 2 = 0

    \tan{\theta}(2\tan{\theta} - 1) + 2(2\tan{\theta} - 1) = 0

    (2\tan{\theta} - 1)(\tan{\theta} + 2) = 0

    2\tan{\theta} - 1 = 0 or \tan{\theta} + 2 = 0

    \tan{\theta} = \frac{1}{2} or \tan{\theta} = -2.


    Case 1:

    \tan{\theta} = \frac{1}{2}.

    This means you are dealing with a right-angle triangle of side lengths O = 1 and A = 2.

    By Pythagoras, that means that H = \sqrt{5}.

    Since \sin{\theta} = \frac{O}{H} that means

    \sin{\theta} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}.


    Apply a similar process for Case 2.
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