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Thread: double angle question

  1. #1
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    double angle question

    How do you solve:

    sin(2θ) + sin(4θ) = 0

    on the interval of [0 degrees, 360 degrees)
    Last edited by topsquark; May 20th 2008 at 04:11 PM. Reason: Restored original question
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by stasis View Post
    How do you solve:

    sin(2θ) + sin(4θ) = 0

    on the interval of [0 degrees, 360 degrees)
    Write : $\displaystyle \sin 4 \theta=2 \sin 2\theta \cdot \cos 2\theta$

    Notice that $\displaystyle \cos A=\sqrt{1-\sin^2 A} \implies \cos 2\theta=\sqrt{1-\sin^2 2 \theta}$

    Substitute : $\displaystyle y=\sin 2 \theta$


    $\displaystyle \implies \sin 2 \theta+2 \sin 2 \theta \sqrt{1-\sin^2 2 \theta}=y+2y \cdot \sqrt{1-y^2}$

    So you want to solve for y in $\displaystyle y+2y \cdot \sqrt{1-y^2}=0 \longleftrightarrow y(1+2\sqrt{1-y^2})=0$

    Does it help ?
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by stasis View Post
    How do you solve:

    sin(2θ) + sin(4θ) = 0

    on the interval of [0 degrees, 360 degrees)
    Note that $\displaystyle \sin(2u)=2\sin(u) \cos(u)$. Thus, $\displaystyle \sin(4\theta)=2\sin(2\theta) \cos(2\theta)$. Substituting this into the equation, we get:

    $\displaystyle \sin(2\theta)+2\sin(2\theta)\cos(2\theta)=0\implie s \sin(2\theta)\left(1+2\cos(2\theta)\right)=0$

    Thus we get two equations:

    $\displaystyle \sin(2\theta)=0 \ \text{and} \ \cos(2\theta)=-\frac{1}{2}$. You can take it from here...
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