MHB Cos (π/7) is a root to a cubic equation

[anemone]

Prove that $\cos \dfrac{\pi}{7}$ is a root of the equation $8x^3-4x^2-4x+1=0$.

 

[kaliprasad]

Prove that $\cos \dfrac{\pi}{7}$ is a root of the equation $8x^3-4x^2-4x+1=0$.

Spoiler

Let us put $x = \cos \frac{\pi}{7}$ and see if expression is zero
we have $8x^3-4x^2-4x+1=8\cos^3 \frac{\pi}{7} - 4 \cos^2 \frac{\pi}{7} + 4 \cos \frac{\pi}{7} + 1$
$= 2(4\cos^3 \frac{\pi}{7}) - 2 (2 \cos^2 \frac{\pi}{7}) - 4 \cos \frac{\pi}{7} + 1$
$= 2 (3 cos \frac{\pi}{7} + cos \frac{3\pi}{7}) - 2 ( cos \frac{2\pi}{7} + 1) + 4 \cos \frac{\pi}{7} + 1$ using $4cos^3 x = 3 \cos\, x + cos 3x$ and $2\cos^2x = cos 2x + 1$
$= 2(cos \frac{3\pi}{7} - cos \frac{2\pi}{7} + cos \frac{\pi}{7} + \frac{1}{2})$
$= 2(cos \frac{3\pi}{7} + cos \frac{5\pi}{7} + cos \frac{\pi}{7} + \frac{1}{2})$
$= 2(cos \frac{5\pi}{7} + cos \frac{3\pi}{7} + cos \frac{\pi}{7} + \frac{1}{2})$
$= 2(-\frac{1}{2} + \frac{1}{2})= 0$

Note: I have used $(cos \frac{5\pi}{7}) + cos \frac{3\pi}{7} + cos \frac{\pi}{7} = - \frac{1}{2}$ and I can prove it if required

 

[anemone]

Well done kaliprasad and thanks for participating!

Spoiler

Note: I have used $(cos \frac{5\pi}{7}) + cos \frac{3\pi}{7} + cos \frac{\pi}{7} = - \frac{1}{2}$ and I can prove it if required

Spoiler

Yes, that is the trick I believe one has to use to crack this problem but I think the readers would appreciate it if you show how we obtained $\cos \frac{5\pi}{7} + \cos \frac{3\pi}{7} + \cos \frac{\pi}{7} = -\frac{1}{2}$...(Happy)

 

[Greg]

Prove that $\cos \dfrac{\pi}{7}$ is a root of the equation $8x^3-4x^2-4x+1=0$.

Spoiler

If you'll admit the identity $\cos\dfrac{\pi}{7}\cos\dfrac{2\pi}{7}\cos\dfrac{3\pi}{7}=\dfrac18$ then here is an alternative:

$$8x^3-4x^2-4x+1=0$$

$$4x(2x^2-1-x)=-1$$

$$4\cos\dfrac{\pi}{7}\left(\cos\dfrac{2\pi}{7}-\cos\dfrac{\pi}{7}\right)=-1$$

$$-8\cos\dfrac{\pi}{7}\sin\dfrac{3\pi}{14}\sin\dfrac{\pi}{14}=-1$$

$$-8\cos\dfrac{\pi}{7}\cos\dfrac{2\pi}{7}\cos\dfrac{3\pi}{7}=-1$$

$$-1=-1$$

 

[kaliprasad]

Spoiler

Let us put $x = \cos \frac{\pi}{7}$ and see if expression is zero
we have $8x^3-4x^2-4x+1=8\cos^3 \frac{\pi}{7} - 4 \cos^2 \frac{\pi}{7} + 4 \cos \frac{\pi}{7} + 1$
$= 2(4\cos^3 \frac{\pi}{7}) - 2 (2 \cos^2 \frac{\pi}{7}) - 4 \cos \frac{\pi}{7} + 1$
$= 2 (3 cos \frac{\pi}{7} + cos \frac{3\pi}{7}) - 2 ( cos \frac{2\pi}{7} + 1) + 4 \cos \frac{\pi}{7} + 1$ using $4cos^3 x = 3 \cos\, x + cos 3x$ and $2\cos^2x = cos 2x + 1$
$= 2(cos \frac{3\pi}{7} - cos \frac{2\pi}{7} + cos \frac{\pi}{7} + \frac{1}{2})$
$= 2(cos \frac{3\pi}{7} + cos \frac{5\pi}{7} + cos \frac{\pi}{7} + \frac{1}{2})$
$= 2(cos \frac{5\pi}{7} + cos \frac{3\pi}{7} + cos \frac{\pi}{7} + \frac{1}{2})$
$= 2(-\frac{1}{2} + \frac{1}{2})= 0$

Note: I have used $(cos \frac{5\pi}{7}) + cos \frac{3\pi}{7} + cos \frac{\pi}{7} = - \frac{1}{2}$ and I can prove it if required

Sorry : there were 2 serious errors . Now I correct the same

Spoiler

Let us put $x = \cos \frac{\pi}{7}$ and see if expression is zero
we have $8x^3-4x^2-4x+1=8\cos^3 \frac{\pi}{7} - 4 \cos^2 \frac{\pi}{7} + 4 \cos \frac{\pi}{7} + 1$
$= 2(4\cos^3 \frac{\pi}{7}) - 2 (2 \cos^2 \frac{\pi}{7}) - 4 \cos \frac{\pi}{7} + 1$
$= 2 (3 cos \frac{\pi}{7} + cos \frac{3\pi}{7}) - 2 ( cos \frac{2\pi}{7} + 1) + 4 \cos \frac{\pi}{7} + 1$ using $4cos^3 x = 3 \cos\, x + cos 3x$ and $2\cos^2x = cos 2x + 1$
$= 2(cos \frac{3\pi}{7} - cos \frac{2\pi}{7} + cos \frac{\pi}{7} - \frac{1}{2})$
$= 2(cos \frac{3\pi}{7} + cos \frac{5\pi}{7} + cos \frac{\pi}{7} - \frac{1}{2})$
$= 2(cos \frac{5\pi}{7} + cos \frac{3\pi}{7} + cos \frac{\pi}{7} - \frac{1}{2})$
$= 2(\frac{1}{2} - \frac{1}{2})= 0$

Spoiler

I had mentioned $(cos \frac{5\pi}{7}) + cos \frac{3\pi}{7} + cos \frac{\pi}{7} = - \frac{1}{2}$ but it should be
$(cos \frac{5\pi}{7}) + cos \frac{3\pi}{7} + cos \frac{\pi}{7} = \frac{1}{2}$

Now for the proof:
let $z = \cos \frac{\pi}{7} + i \sin \frac{\pi}{7}$
$z^7 = \cos \pi + i \sin \pi = - 1$
so $z^7+1 = 0$
($z+1) (z^6-z^5+z^4 - z^3+z^2 -z + 1) = 0$
as z is not - 1 so
$z^6-z^5+z^4 - z^3+z^2 -z + 1 = 0$
so $z^6-z^5+z^4 - z^3+z^2 -z= -1$
so $z+z^3+ z^5 = 1 + (z^2+z^4+z^6)$
equating the real part
$\cos \frac{\pi}{7} + \cos \frac{3\pi}{7} + \cos \frac{5\pi}{7} = 1 + \cos \frac{2\pi}{7}+ \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7})$
but $\cos\frac{6\pi}{7}= - \cos(\pi-\frac{\pi}{7}) = - \cos \frac{\pi}{7}$
$\cos \frac{4\pi}{7}= - \cos \frac{3\pi}{7}$
$\cos \frac{2\pi}{7} = - \cos \frac{5\pi}{7}$
so $\cos \frac{\pi}{7} + \cos \frac{3\pi}{7} + \cos \frac{5\pi}{7} = 1 - \cos \frac{2\pi}{7} - \cos \frac{4\pi}{7} - \cos \frac{6\pi}{7})$
so $2( cos \frac{\pi}{7} + \cos \frac{3\pi}{7} + \cos \frac{5\pi}{7}) = 1$
or $2( cos \frac{\pi}{7} + \cos \frac{3\pi}{7} + \cos \frac{5\pi}{7}) = \frac{1}{2}$

 

[anemone]

Hi greg1313! Your answer is great and thanks for participating!(Cool)

Hi kaliprasad, I admit that I didn't go through your whole solution and missed something obviously wrong. Sorry!

In this problem, I used the following identity for the proof:

Spoiler

$$\cos \dfrac{\pi}{7}-\cos \dfrac{2\pi}{7}+\cos \dfrac{3\pi}{7}=\frac{1}{2}.$$