Proof: $\cos\dfrac{\pi}{7}=\dfrac{1}{6}+\dfrac{\sqrt{7}}{6}$

[anemone]

Prove that $\cos\dfrac{\pi}{7}=\dfrac{1}{6}+\dfrac{\sqrt{7}}{6}\left(\cos\left(\dfrac{1}{3}\arccos\dfrac{1}{2\sqrt{7}}\right)+\sqrt{3}\left(\dfrac{1}{3}\cos\dfrac{1}{2\sqrt{7}}\right)\right)$.

 

[jvicens]

Dear forum members,

I would like to present a proof for the identity $\cos\dfrac{\pi}{7}=\dfrac{1}{6}+\dfrac{\sqrt{7}}{6}\left(\cos\left(\dfrac{1}{3}\arccos\dfrac{1}{2\sqrt{7}}\right)+\sqrt{3}\left(\dfrac{1}{3}\cos\dfrac{1}{2\sqrt{7}}\right)\right)$.

First, let us consider the following trigonometric identities:

1. $\cos(3x)=4\cos^3x-3\cos x$
2. $\cos(2x)=2\cos^2x-1$
3. $\cos\left(\dfrac{1}{2}x\right)=\sqrt{\dfrac{1+\cos x}{2}}$

Using these identities, we can rewrite the left side of the given identity as:

$\cos\dfrac{\pi}{7}=\cos\left(\dfrac{3\pi}{7}-\dfrac{2\pi}{7}\right)=\cos\left(\dfrac{3\pi}{7}\right)\cos\left(\dfrac{2\pi}{7}\right)+\sin\left(\dfrac{3\pi}{7}\right)\sin\left(\dfrac{2\pi}{7}\right)$

Next, we can use the angle sum formula for cosine to rewrite $\cos\left(\dfrac{3\pi}{7}\right)$ and $\cos\left(\dfrac{2\pi}{7}\right)$ as:

$\cos\left(\dfrac{3\pi}{7}\right)=\cos\left(\dfrac{\pi}{7}+\dfrac{2\pi}{7}\right)=\cos\dfrac{\pi}{7}\cos\dfrac{2\pi}{7}-\sin\dfrac{\pi}{7}\sin\dfrac{2\pi}{7}$

$\cos\left(\dfrac{2\pi}{7}\right)=\cos\left(\dfrac{\pi}{7}+\dfrac{\pi}{7}\right)=\cos^2\dfrac{\pi}{7}-\sin^2\dfrac{\pi}{7}$

Plugging these values into the original equation, we get: