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Thread: How to find sinθ given sin2θ?

  1. #1
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    How to find sinθ given sin2θ?

    I have spent hours trying to figure out this problem. I know how to get sin2θ, given sinθ, but not the other way around:

    Find sinθ, given that sin2θ = -2/3, (3pi/2) < 2θ < 2pi.

    The answer is sq(18 - 6sq(5)) / 6.

    Thanks for any help.
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  2. #2
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    First of all, if $\displaystyle \frac{3\pi}{2} < 2\theta < 2\pi$, then $\displaystyle \frac{3\pi}{4} < \theta < \pi$.

    Since this is in Quadrant 2, you know $\displaystyle 0 < \sin{\theta} < 1$.


    You have $\displaystyle \sin{2\theta} = -\frac{2}{3}$

    $\displaystyle 2\sin{\theta}\cos{\theta} = - \frac{2}{3}$

    $\displaystyle 2\sin{\theta}\sqrt{1 - \sin^2{\theta}} = -\frac{2}{3}$

    $\displaystyle (2\sin{\theta}\sqrt{1 - \sin^2{\theta}})^2 = \left(-\frac{2}{3}\right)^2$

    $\displaystyle 4\sin^2{\theta}(1 - \sin^2{\theta}) = \frac{4}{9}$

    $\displaystyle \sin^2{\theta}(1 - \sin^2{\theta}) = \frac{1}{9}$

    $\displaystyle \sin^2{\theta} - \sin^4{\theta} = \frac{1}{9}$

    $\displaystyle \sin^4{\theta} - \sin^2{\theta} + \frac{1}{9} = 0$

    $\displaystyle X^2 - X + \frac{1}{9} = 0$ where $\displaystyle X = \sin^2{\theta}$

    $\displaystyle X^2 - X + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 + \frac{1}{9} = 0$

    $\displaystyle \left(X - \frac{1}{2}\right)^2 - \frac{1}{4} + \frac{1}{9} = 0$

    $\displaystyle \left(X - \frac{1}{2}\right)^2 - \frac{5}{36} = 0$

    $\displaystyle \left(X - \frac{1}{2}\right)^2 = \frac{5}{36}$

    $\displaystyle X - \frac{1}{2} = \frac{\pm \sqrt{5}}{6}$

    $\displaystyle X = \frac{1}{2} \pm \frac{\sqrt{5}}{6}$

    $\displaystyle X = \frac{3 \pm \sqrt{5}}{6}$

    $\displaystyle \sin^2{\theta} = \frac{3 \pm \sqrt{5}}{6}$

    $\displaystyle \sin{\theta} = \frac{\sqrt{3 \pm \sqrt{5}}}{\sqrt{6}}$

    $\displaystyle \sin{\theta} = \frac{\sqrt{6}\sqrt{3 \pm \sqrt{5}}}{6}$

    $\displaystyle \sin{\theta} = \frac{\sqrt{6(3 \pm \sqrt{5})}}{6}$

    $\displaystyle \sin{\theta} = \frac{\sqrt{18 \pm 6\sqrt{5}}}{6}$.


    The only solution that lies in $\displaystyle 0 < \sin{\theta} < 1$ is

    $\displaystyle \sin{\theta} = \frac{\sqrt{18 - 6\sqrt{5}}}{6}$.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    First of all, if $\displaystyle \frac{3\pi}{2} < 2\theta < 2\pi$, then $\displaystyle \frac{3\pi}{4} < \theta < \pi$.

    Since this is in Quadrant 2, you know $\displaystyle 0 < \sin{\theta} < 1$.


    You have $\displaystyle \sin{2\theta} = -\frac{2}{3}$

    $\displaystyle 2\sin{\theta}\cos{\theta} = - \frac{2}{3}$

    $\displaystyle 2\sin{\theta}\sqrt{1 - \sin^2{\theta}} = -\frac{2}{3}$
    What sign is $\displaystyle \cos(\theta)$ in quadrant 2?

    CB
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    Ouch. I need to work on my algebra. It wouldn't have occurred to me to do it that way at all.

    It makes sense I think, but in: , is there a reason (-1/2) was used, as opposed to any other number?
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    What sign is $\displaystyle \cos(\theta)$ in quadrant 2?

    CB
    $\displaystyle ^{-}ve$.

    But $\displaystyle \sin{\theta}$ is $\displaystyle ^{+}ve$, so $\displaystyle 2\sin{\theta}\cos{\theta}$ is $\displaystyle ^{-}ve$.

    I suppose I should have written $\displaystyle -2\sin{\theta}\sqrt{1 - \sin^2{\theta}}$, but it doesn't really matter once you square both sides.
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  6. #6
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    Quote Originally Posted by michael710 View Post
    Ouch. I need to work on my algebra. It wouldn't have occurred to me to do it that way at all.

    It makes sense I think, but in: , is there a reason (-1/2) was used, as opposed to any other number?
    Yes, when you complete the square in a quadratic, you add and subtract half of the square of the $\displaystyle b$ value. In this case the $\displaystyle b$ value is $\displaystyle -1$.
    Last edited by Prove It; Jul 13th 2010 at 09:25 PM.
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  7. #7
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    I'm not even sure I've ever covered completing the square, but the concept makes sense, and that's the only place I questioned. I'll drill this into my brain until I can do it myself.

    Thanks! And thanks for such a quick response.
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