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

**zResistance** · December 5th 2011, 11:12 PM

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

**Prove It** · December 5th 2011, 11:48 PM

Originally Posted by zResistance

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

**princeps** · December 5th 2011, 11:55 PM

Originally Posted by zResistance

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

$\begin{array}{l}4 \cdot \cos \theta \cdot \cos 2\theta \cdot \cos 3\theta = 4 \cdot \frac{{\cos \theta + \cos 3\theta }}{2} \cdot \cos 3\theta = 2 \cdot (\cos \theta \cdot \cos 3\theta + \cos 3\theta \cdot \cos 3\theta ) = \\= 2 \cdot (\frac{{\cos 2\theta + \cos 4\theta }}{2} + \frac{{\cos 0 + \cos 6\theta }}{2}) = 1 + \cos 2\theta + \cos 4\theta + \cos 6\theta \end{array}$

**zResistance** · December 5th 2011, 11:56 PM

WOW! thanks alot

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

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