Famous five

To prove that \(\theta=\frac{\pi}{5}\) satisfies \(\frac{1}{2}+\cos{(2\theta)}=\cos{\theta}\):

2sin⁡θ(12+cos⁡(2θ))=2sin⁡θcos⁡θ2\sin{\theta}(\frac{1}{2}+\cos{(2\theta)})=2\sin{\theta} \cos{\theta}

sin⁡θ+2sin⁡θcos⁡(2θ)=2sin⁡θcos⁡θ\sin{\theta}+ 2\sin{\theta} \cos{(2\theta)}=2\sin{\theta} \cos{\theta}

sin⁡(2−1)θ+2sin⁡θcos⁡(2θ)=sin⁡(2θ)\sin{(2-1)\theta}+2\sin{\theta} \cos{(2\theta)}=\sin{(2\theta)}

sin⁡(2θ)cos⁡θ−sin⁡θcos⁡(2θ)+2sin⁡θcos⁡(2θ)=sin⁡(2θ)\sin{(2\theta)}\cos{\theta}-\sin{\theta}\cos{(2\theta)}+2\sin{\theta} \cos{(2\theta)}=\sin{(2\theta)}

sin⁡(2θ)cos⁡θ+sin⁡θcos⁡(2θ)=sin⁡(2θ)\sin{(2\theta)}\cos{\theta}+\sin{\theta}\cos{(2\theta)}=\sin{(2\theta)}

sin⁡(2+1)θ=sin⁡(2θ)\sin{(2+1)\theta}=\sin{(2\theta)}

sin⁡(3θ)=sin⁡(2θ)\sin{(3\theta)}=\sin{(2\theta)}

sin⁡(3θ)=sin⁡(π−2θ)\sin{(3\theta)}=\sin{(\pi-2\theta)}

3θ=π−2θ3\theta=\pi-2\theta

(3+2)θ=π(3+2)\theta=\pi

5θ=π5\theta=\pi

θ=π5\theta=\frac{\pi}{5}

QED

#Geometry

Note by Noel Lo
3 years, 10 months ago

No vote yet
1 vote

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Comments

More generally ∑k=0n−1cos⁡(2k+12n+1π)=12\displaystyle \sum_{k=0}^{n-1} \cos \left( \frac {2k+1}{2n+1}\pi \right) = \frac 12 (see proof). This means cos⁡π5+cos⁡3π5=cos⁡π5−cos⁡2π5=12\cos \frac \pi 5 + \cos \frac {3 \pi} 5 = \cos \frac \pi 5 - \cos \frac {2 \pi} 5 = \frac 12

Chew-Seong Cheong - 3 years, 10 months ago
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