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Math Help - Prove that 1+ cos 2θ + cos 4θ+ cos 6θ = 4 cos θ cos 2θ cos 3θ

  1. #1
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    Prove that 1+ cos 2θ + cos 4θ+ cos 6θ = 4 cos θ cos 2θ cos 3θ

    Yet again another hard question I don't know how to do!

    Prove that 1+ cos 2θ + cos 4θ+ cos 6θ = 4 cos θ cos 2θ cos 3θ
    Hence, find all the angles between 0 degrees and 180 degrees inclusive, which satisfy the equation(below)

    cos 2θ+ cos 4θ+ cos 6θ= -1

    Thank you all in advance
    Last edited by mr fantastic; December 6th 2011 at 03:49 AM. Reason: Re-titled.
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  2. #2
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    Re: Trigonometrical question!!!!!!!!!!!

    Quote Originally Posted by zResistance View Post
    Yet again another hard question I don't know how to do!

    Prove that 1+ cos 2θ + cos 4θ+ cos 6θ = 4 cos θ cos 2θ cos 3θ
    Hence, find all the angles between 0 degrees and 180 degrees inclusive, which satisfy the equation(below)

    cos 2θ+ cos 4θ+ cos 6θ= -1

    Thank you all in advance
    \displaystyle \begin{align*} 1 + \cos{(2\theta)} + \cos{(4\theta)} + \cos{(6\theta)} &\equiv 1 + 2\cos^2{(\theta)} - 1 + 2\cos^2{(2\theta)} - 1 + 2\cos^2{(3\theta)} - 1 \\ &\equiv 2\cos^2{(\theta)} + 2\cos^2{(2\theta)} + 2\cos^2{(3\theta)} - 2 \\ &\equiv 2\cos^2{(\theta)} + 2\left[2\cos^2{(\theta)} - 1\right]^2 + 2\left[4\cos^3{(\theta)} - 3\cos{(\theta)}\right]^2 - 2 \\ &\equiv 2\cos^2{(\theta)} + 2\left[ 4\cos^4{(\theta)} - 4\cos^2{(\theta)} + 1 \right]  + 2\left[16\cos^6{(\theta)} - 24\cos^4{(\theta)} + 9\cos^2{(\theta)}\right] - 2 \\ &\equiv 2\cos^2{(\theta)} + 8\cos^4{(\theta)} - 8\cos^2{(\theta)} + 2 + 32\cos^6{(\theta)} - 48\cos^4{(\theta)} + 18\cos^2{(\theta)} - 2 \\ &\equiv 12\cos^2{(\theta)} - 40\cos^4{(\theta)} + 32\cos^6{(\theta)} \end{align*}


    \displaystyle \begin{align*} 4\cos{(\theta)}\cos{(2\theta)}\cos{(3\theta)} &\equiv 4\cos{(\theta)}\left[2\cos^2{(\theta)} - 1\right]\left[4\cos^3{(\theta)} - 3\cos{(\theta)}\right] \\ &\equiv \left[8\cos^3{(\theta)} - 4\cos{(\theta)}\right]\left[ 4\cos^3{(\theta)} - 3\cos{(\theta)}\right]  \\ &\equiv 32\cos^6{(\theta)} - 24\cos^4{(\theta)} - 16\cos^4{(\theta)} + 12\cos^2{(\theta)} \\ &\equiv 12\cos^2{(\theta)} - 40\cos^4{(\theta)} + 32\cos^6{(\theta)} \end{align*}


    So the identity holds
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  3. #3
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    Re: Trigonometrical question!!!!!!!!!!!

    Quote Originally Posted by zResistance View Post
    Yet again another hard question I don't know how to do!

    Prove that 1+ cos 2θ + cos 4θ+ cos 6θ = 4 cos θ cos 2θ cos 3θ
    Hence, find all the angles between 0 degrees and 180 degrees inclusive, which satisfy the equation(below)

    cos 2θ+ cos 4θ+ cos 6θ= -1

    Thank you all in advance

    \begin{array}{l}4 \cdot \cos \theta  \cdot \cos 2\theta  \cdot \cos 3\theta  = 4 \cdot \frac{{\cos \theta  + \cos 3\theta }}{2} \cdot \cos 3\theta  = 2 \cdot (\cos \theta  \cdot \cos 3\theta  + \cos 3\theta  \cdot \cos 3\theta ) = \\= 2 \cdot (\frac{{\cos 2\theta  + \cos 4\theta }}{2} + \frac{{\cos 0 + \cos 6\theta }}{2}) = 1 + \cos 2\theta  + \cos 4\theta  + \cos 6\theta \end{array}
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    Re: Trigonometrical question!!!!!!!!!!!

    WOW! thanks alot
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