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Thread: 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 $\displaystyle tan2\theta=\frac{4}{3},$ without using calculator,find the values of $\displaystyle sin\theta$.
    Last edited by mastermin346; Jul 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 $\displaystyle sin 2\theta=\frac{4}{3},$ without using calculator,find the values of $\displaystyle sin\theta$.
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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:
    $\displaystyle sin\,\theta= \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}$
    I started by using the tangent double identity:
    $\displaystyle \frac{2tan\,\theta}{1-tan^2\,\theta}=\frac{4}{3}$
    Cross-multiplied:
    $\displaystyle 3\left(2tan\,\theta \right)=4\left(1-tan^2\,\theta \right)$
    ... solved for $\displaystyle tan\,\theta$, then rewrote in terms of $\displaystyle sin\,\theta$.

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

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


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

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

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

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

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

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

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

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

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

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

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


    Case 1:

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

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

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

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

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


    Apply a similar process for Case 2.
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