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

  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Cos Cubic Root
AI Thread Summary
The discussion centers on proving that cos(π/7) is a root of the cubic equation 8x^3 - 4x^2 - 4x + 1 = 0. Participants share their solutions and corrections, with some acknowledging errors in their initial approaches. The use of specific mathematical identities is mentioned as part of the proof process. Overall, the conversation highlights collaborative problem-solving and the importance of accuracy in mathematical proofs. The focus remains on validating the root of the equation through rigorous proof techniques.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Prove that $\cos \dfrac{\pi}{7}$ is a root of the equation $8x^3-4x^2-4x+1=0$.
 
Mathematics news on Phys.org
anemone said:
Prove that $\cos \dfrac{\pi}{7}$ is a root of the equation $8x^3-4x^2-4x+1=0$.

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
 
Well done kaliprasad and thanks for participating!

kaliprasad said:
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

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

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 said:
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

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 $2\cos^2x = cos 2x + 1$
$= 2(cos \frac{3\pi}{7} - cos \frac{2\pi}{7} + cos \frac{\pi}{7} - \frac{1}{2})$
$= 2(cos \frac{3\pi}{7} + cos \frac{5\pi}{7} + cos \frac{\pi}{7} - \frac{1}{2})$
$= 2(cos \frac{5\pi}{7} + cos \frac{3\pi}{7} + cos \frac{\pi}{7} - \frac{1}{2})$
$= 2(\frac{1}{2} - \frac{1}{2})= 0$

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 $2( cos \frac{\pi}{7} + \cos \frac{3\pi}{7} + \cos \frac{5\pi}{7}) = \frac{1}{2}$
 
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:
$$\cos \dfrac{\pi}{7}-\cos \dfrac{2\pi}{7}+\cos \dfrac{3\pi}{7}=\frac{1}{2}.$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.

Similar threads

Back
Top