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If $\dfrac{2\sqrt{19}}{3}\cos{\left(\dfrac{1}{3}\arccos{\dfrac{7}{\sqrt{76}}}\right)}-\dfrac{1}{3}$ can be simpified to $2\left(\cos{\dfrac{4\pi}{19}}+\cos{\dfrac{6\pi}{19}}+\cos{\dfrac{10\pi}{19}}\right)$.

How to simplify $\dfrac{4\sqrt{7}}{3}\cos{\left(\dfrac{1}{3}\arccos{\dfrac{1}{\sqrt{28}}}\right)}+\dfrac{1}{3}$ ?

edit : Now, I have get the answer : $$\dfrac{4\sqrt{7}}{3}\cos{\left(\dfrac{1}{3}\arccos{\dfrac{1}{\sqrt{28}}}\right)}+\dfrac{1}{3}=2\left(\cos{\dfrac{\pi}{7}}+\cos{\dfrac{2\pi}{7}}+\cos{\dfrac{3\pi}{7}}\right)$$

How to prove it?

Bless
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1 Answers1

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Call $\theta=\frac{1}{3}\arccos \frac{1}{\sqrt {28}}$. We can use the formula $\cos (3\theta)=4\cos^3\theta -3\cos \theta$ to deduce $$\frac{1}{\sqrt {28}}=4\cos^3\theta -3\cos \theta. \tag{1}$$ The question is asking us to evaluate the quantity $$ y=\frac{4 \sqrt 7}{3}\cos \theta+\frac{1}{3}$$ with $\cos \theta$ satisfying $(1)$. Applying the substitution $\cos \theta=\frac{3y-1}{4 \sqrt 7}$we get $$y^3-y^2-9y=-1. \tag{2}$$ The last step is to show that $$y=2\left(\cos{\dfrac{\pi}{7}}+\cos{\dfrac{2\pi}{7}}+\cos{\dfrac{3\pi}{7}}\right)$$ satisfies $(2)$. This can be done by a straightforward computation, using Werner's formula to deal with the product of cosines. In details we get

$$y^2=2\Big(\cos{\dfrac{2\pi}{7}}+\cos{\dfrac{4\pi}{7}}+2\cos{\dfrac{2\pi}{7}}-\cos{\dfrac{6\pi}{7}}\Big)+6$$ and $$y^3=y^2\cdot y=24\cos{\frac{\pi}{7}}+24\cos{\frac{2\pi}{7}}+18\cos{\frac{3\pi}{7}}+4\cos{\frac{4\pi}{7}}+2\cos{\frac{5\pi}{7}}+2\cos{\frac{6\pi}{7}}+6.$$ Now we can compute \begin{align} y^3-y^2-9y &=2\cos{\frac{\pi}{7}}+4\cos{\frac{2\pi}{7}}+2\cos{\frac{4\pi}{7}}+2\cos{\frac{5\pi}{7}}+4\cos{\frac{6\pi}{7}} \\ &= 2\cos{\frac{2\pi}{7}}-2\cos{\frac{3\pi}{7}}+ 2\cos{\frac{6\pi}{7}}=-1 \end{align} from which follows $$y^3-y^2-9y=-1.$$ The last cosines equality is a well known result and it's the only non trivial step.

mrprottolo
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