MHB Can we express cos(pi/11) using radicals?

  • Thread starter Thread starter mathbalarka
  • Start date Start date
  • Tags Tags
    Exercise
AI Thread Summary
The discussion revolves around expressing cos(π/11) using radicals, stemming from a solvable quintic polynomial. The polynomial presented is 32x^5 - 16x^4 - 32x^3 + 12x^2 + 6x - 1 = 0, which has a cyclic Galois group of order 5, indicating that its roots can be expressed using elementary functions. The challenge is to find a radical expression for cos(π/11), which is suggested to be more difficult. The conversation highlights the intersection of polynomial roots and trigonometric functions in mathematical exploration. Ultimately, the discussion invites further exploration into the relationship between quintic equations and trigonometric identities.
mathbalarka
Messages
452
Reaction score
0
This is the question I found long ago in another math forum. I thought that it would be a good sweat for everyone in order to find the answer :

Find all the closed from of the roots of this solvable quintic

$$32 x^5 - 16 x^4 - 32 x^3 + 12 x^2 + 6x - 1 = 0$$
 
Mathematics news on Phys.org
Here is my solution :

$$32 x^5 - 16 x^4 - 32 x^3 + 12 x^2 + 6x - 1 = 0$$

$$\Rightarrow (x + 1)(32 x^5 - 16 x^4 - 32 x^3 + 12 x^2 + 6x - 1)^2 = 0$$

Expanding this gives

$$1024 x^{11} - 2816 x^9 + 2816 x^7 - 1232 x^5 + 220 x^3 -11 x + 1 = 0$$

$$\Rightarrow [1024 x^{11} - 2816 x^9 + 2816 x^7 - 1232 x^5 + 220 x^3 -11 x] + 1 = 0$$

Looking at the expression closed with 3-rd bracket carefully, we see that it's one of the Chebyshev polynomials. Hence, letting $$x = \cos(\theta)$$, we get

$$\cos(11 \theta) + 1 = 0$$

So, the roots of the equation above are $$\theta = \frac{(2n + 1) \pi}{11}$$.

Hence, the (uncertified) roots of the quintic of interest are $$x = \cos \left ( \frac{(2n + 1) \pi}{11} \right )$$.

By a quick check, we see that only n = 0, 1, 2, 3, 4 works.

Hence, the roots are $$\cos \left ( \frac{\pi}{11} \right ), \cos \left ( \frac{3 \pi}{11} \right ) , \cos \left ( \frac{5 \pi}{11} \right ) , \cos \left ( \frac{7 \pi}{11} \right ) , \cos \left ( \frac{9 \pi}{11} \right )$$

Balarka
.
 
Interestingly enough, the polynomial of the general interest has a solvable galois group (i.e, cyclic group of order 5) implying that all the roots over there can be expressed by finite number of elementary functions.

So, my question is how can we express at least, say, $$\cos (\pi / 11)$$ in terms of radicals?

Maybe this one is a challenge too and probably a harder one!
 
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...
Back
Top