Simplifying $\frac{2\sqrt{19}}{3}\cos{\left(\frac{1}{3}\arccos{\frac{7}{\sqrt{76}}}\right)}-\frac{1}{3}$

Asked Jul 02 '16 at 14:51

Active Sep 08 '19 at 07:05

Viewed 101 times

Using Vieta's Trignometric Solutions, I am getting $$\dfrac{2\sqrt{19}}{3}\cos{\left(\dfrac{1}{3}\arccos{\dfrac{7}{\sqrt{76}}}\right)}-\dfrac{1}{3}\tag{1}$$ And it is also equal to $$2\left(\cos\left(\frac {4\pi}{19}\right)+\cos\left(\frac {6\pi}{19}\right)+\cos\left(\frac {10\pi}{19}\right)\right)\tag{2}$$

My Question: How would you get from $(1)$ to $(2)$, and is there a way to generalize it? Anything helps!

edited Sep 08 '19 at 07:05

StubbornAtom

17,052

asked Jul 02 '16 at 14:51

Frank

5,984

Hm... they aren't equal – Frank Jul 02 '16 at 15:14
I edited the question. Now, they are :) – Peter Jul 02 '16 at 15:15
1

It would help to know the equation from which you got $(1)$ – Peter Jul 02 '16 at 15:15
$y^3+y^2-6y-7$. – Frank Jul 02 '16 at 15:16
OK, we have to show that $(2)$ is a root of the cubic. Not sure, whether there is another way to come from $(1)$ to $(2)$ – Peter Jul 02 '16 at 15:17
Oh... But $(2)$ is already a root of $x^3+x^2-6x-7$. I just want to know if there is a way to get from $(1)$ to $(2)$... It should be possible, right? – Frank Jul 02 '16 at 15:19
Let us continue this discussion in chat. – Peter Jul 02 '16 at 15:19
see here google will help you – Dr. Sonnhard Graubner Jul 02 '16 at 15:24
http://math.stackexchange.com/questions/31352/proving-2-cos-frac4-pi19-cos-frac6-pi19-cos-frac10-pi19 – Dr. Sonnhard Graubner Jul 02 '16 at 15:24
In your previous question you got $(2)\mapsto(1)$, and I do not think there is an easy way to spot $(1)\mapsto (2)$. – Jack D'Aurizio Jul 02 '16 at 16:32

Simplifying $\frac{2\sqrt{19}}{3}\cos{\left(\frac{1}{3}\arccos{\frac{7}{\sqrt{76}}}\right)}-\frac{1}{3}$

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