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Thread: sin cos equation

  1. #1
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    sin cos equation

    can someone solve the equation
    cos4x=sin2x in the easiest way possible using trigometric formula. i keep going around in circles and getting no where.
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  2. #2
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    Re: sin cos equation

    \cos^{2}2x-\sin^{2}2x-\sin2x=0
    1-\sin^{2}2x-\sin^{2}2x-\sin2x=0
    2\sin^{2}2x+\sin2x-1=0,\ \sin2x=t
    2t^2+t-1=0
    \vdots
    t_1=\frac12,\ t_2=-1
    \ldots
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  3. #3
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    Re: sin cos equation

    which formula converts cos4x into cos^2 2x -sin^2 2x
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  4. #4
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    Re: sin cos equation

    Quote Originally Posted by markosheehan View Post
    which formula converts cos4x into cos^2 2x -sin^2 2x
    The double angle formula for cosine: $\cos (2u) = \cos^2 u-\sin^2 u$. If $u = 2x$, you get $\cos(4x) = \cos^2(2x)-\sin^2(2x)$.
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  5. #5
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    Re: sin cos equation

    cos(4x)=sin(2x)
    $1 - 2\sin^2(2x) = \sin(2x)$

    $0 = 2\sin^2(2x) + \sin(2x) - 1$

    $0 = [2\sin(2x)-1][\sin(2x)+1]$

    $\sin(2x) = \dfrac{1}{2}$

    $2x = \dfrac{\pi}{6} \pm 2k\pi \implies x = \dfrac{\pi}{12} \pm k\pi \text{ where } k \in \mathbb{Z}$

    $2x = \dfrac{5\pi}{6} \pm 2k\pi \implies x = \dfrac{5\pi}{12} \pm k\pi \text{ where } k \in \mathbb{Z}$

    $\sin(2x) = -1$

    $2x = \dfrac{3\pi}{2} \pm k \cdot 2\pi \implies x = \dfrac{3\pi}{4} \pm k \cdot \pi \text{ where } k \in \mathbb{Z}$
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