0

I am trying to prove that

$$ \sqrt{7} \cot\frac{\pi}{7}-4 \sin\frac{3 \pi}{14}=3$$

My attempt is to set $x=\frac{\pi}{7}$ and the above relation becomes (using some trigonometry):

$$ 4 \sin^3\frac{x}{2}- \frac{\sqrt {7}}{2} \left(\tan\frac{x}{2}- \cot\frac{x}{2}\right) -12 \sin\frac{x}{2} \cos^2\frac{x}{2}=3$$

My goal was to set $y=\sin\frac{x}{2}$ and make some calculations, but the relation is getting more complex and I don't know if I am right

Michael Rozenberg
  • 194,933
George
  • 750

2 Answers2

3

We need to prove that: $$\sqrt7\cos\frac{\pi}{7}=4\sin\frac{\pi}{7}\sin\frac{3\pi}{14}+3\sin\frac{\pi}{7}$$ or $$\sqrt7\cos\frac{\pi}{7}=2\cos\frac{\pi}{14}-2\cos\frac{5\pi}{14}+3\sin\frac{\pi}{7}$$ or $$\sqrt7\cos\frac{\pi}{7}=2\sin\frac{3\pi}{7}+\sin\frac{\pi}{7}$$ or $$\frac{7}{2}\left(1+\cos\frac{2\pi}{7}\right)=2-2\cos\frac{6\pi}{7}+2\cos\frac{2\pi}{7}-2\cos\frac{4\pi}{7}+\frac{1}{2}\left(1-\cos\frac{2\pi}{7}\right)$$ or $$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2},$$ which is true because $$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}}{2\sin\frac{\pi}{7}}=$$ $$=\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}}{2\sin\frac{\pi}{7}}=-\frac{1}{2}.$$

Michael Rozenberg
  • 194,933
  • I rewrite your first equation to wolfram and gives a False answer: https://www.wolframalpha.com/input/?i=%E2%88%9A7++cot%E2%81%A1%28%CF%80%2F7%29%3D4+sin%E2%81%A1%28%CF%80%2F7%29+sin%E2%81%A1%283%CF%80%2F14%29%2B3+sin%E2%81%A1%28%CF%80%2F7%29 So, the rest of them – George Feb 19 '20 at 14:01
  • 1
    @George Your equality is wrong, of course. See please better my post – Michael Rozenberg Feb 19 '20 at 21:24
0

$\sqrt{7}\cot\dfrac{\pi}{7} -4 \sin\dfrac{3 \pi}{14}=3$

Since $\dfrac{3\pi}{14}=\dfrac{\pi}{2}-\dfrac{2\pi}{7}$ we have $$\sqrt{7} \cot\dfrac{\pi}{7}-4 \cos\dfrac{2\pi}{7}=3$$ and because of $\cot x=\sqrt{\dfrac{1+\cos 2x}{1-\cos 2x}}$, putting $X=\cos 2x$, one has the equation $$8X^3+4X^2-4X-1=0$$ which have three real roots $X_1=\cos 2x_1\approx0.623489$ which corresponds to $x_1=\dfrac{\pi}{7}$.

The other two roots $X_2\approx-0.90097$ and $X_3\approx -0.22257$ could correspond to other two solutions of the problem if the corresponding angles such that $\cos x_2=-0.90097$ and $\cos x_3=-0.22257$ satisfy $\dfrac{3x_i}{2}=\dfrac{\pi}{2}-2x$. Maybe but this is not asked

Piquito
  • 29,594