0

Find a monic, irreducible polynomial with a root at $\cos (π/5)$

Craig Winston
  • 1
  • 7
    $x-\cos{(\pi/5)}$ – Peter Foreman Apr 27 '19 at 09:32
  • The number $\cos(\pi/5)$ is not an algebraic integer, so such a monic polynomial does not exist with integer coefficients. Either rational coefficients must be allowed or $\cos(\pi/5)$ has to be multiplied by $2$ (that product is an algebraic integer). Please clarify which direction to take. – Oscar Lanzi Apr 27 '19 at 09:38
  • You might want to see https://math.stackexchange.com/questions/136978/showing-that-cos-left-frac-pi5-right-frac12-phi. – Minus One-Twelfth Apr 27 '19 at 09:51
  • Or https://math.stackexchange.com/questions/113466/show-4-cos2-frac-pi5-2-cos-frac-pi5-1-0 – Minus One-Twelfth Apr 27 '19 at 09:58

2 Answers2

3

The question is not precise but I think you want to find polynomial with integer coefficient. Assuming that:

$\frac{\pi}{5}$ is a solution of trig equation $\cos(5x)=-1$. Now you expand $\cos{5x}$ in terms of powers of $\cos x$. You can use Euler’s formula and show this : $$\cos{5x}=16\cos^5{x}-20\cos^3x+5\cos x $$ After that put $x=\frac{\pi}{5}$ and $t=\cos x$.
Finally note that $16t^5-20t^3+5t+1=(t+1)(4t^2-2t-1)^2$ and take it from here

DINEDINE
  • 6,081
1

If $5x=(2n+1)\pi,5\nmid(2n+1)\implies\cos x\ne-1$

$$\cos(3x)=-\cos2x$$

Expand both sides to form a cubic equation is n $\cos x$

Divide both sides by $\cos x+1$ as $\cos x\ne-1$

lab bhattacharjee
  • 274,582