Fallen Angel Messages 202 Reaction score 0 Thread starter Feb 19, 2015 #1 Is $\sin\left(10^{\circ}\right)$ rational or not? Prove it.
chisigma Gold Member MHB Messages 1,627 Reaction score 0 Feb 19, 2015 #2 Fallen Angel said: Is $\sin\left(10^{\circ}\right)$ rational or not? Prove it. [sp]From the relation... $\displaystyle \sin \alpha = 3\ \sin \frac{\alpha}{3} - 4\ \sin^{3} \frac{\alpha}{3}\ (1)$ ... it follows that $\displaystyle x= \sin \frac{\pi}{18}$ must be root of the cubic equation... $\displaystyle 8\ x^{3} - 6\ x + 1 =0\ (2)$ If x is rational, then You can write $\displaystyle x = \frac{a}{b}$, being a and b integers... but in this case for (2) it must be... $\displaystyle b = 4\ \sqrt{3\ a - \frac{1}{2}}\ (3)$ ... and that's impossible for a and b integers... the conclusion is that x isn't rational...[/sp] Kind regards $\chi$ $\sigma$
Fallen Angel said: Is $\sin\left(10^{\circ}\right)$ rational or not? Prove it. [sp]From the relation... $\displaystyle \sin \alpha = 3\ \sin \frac{\alpha}{3} - 4\ \sin^{3} \frac{\alpha}{3}\ (1)$ ... it follows that $\displaystyle x= \sin \frac{\pi}{18}$ must be root of the cubic equation... $\displaystyle 8\ x^{3} - 6\ x + 1 =0\ (2)$ If x is rational, then You can write $\displaystyle x = \frac{a}{b}$, being a and b integers... but in this case for (2) it must be... $\displaystyle b = 4\ \sqrt{3\ a - \frac{1}{2}}\ (3)$ ... and that's impossible for a and b integers... the conclusion is that x isn't rational...[/sp] Kind regards $\chi$ $\sigma$
Fallen Angel Messages 202 Reaction score 0 Feb 20, 2015 #3 Good work chisigma, my solution was almost the same. Spoiler From $8x^3-6x+1=0$, let $y=2 sin \ 10º$, then $y^3-3y+1=0$ and this polynomial has no rational roots (because of Gauss Lemma).
Good work chisigma, my solution was almost the same. Spoiler From $8x^3-6x+1=0$, let $y=2 sin \ 10º$, then $y^3-3y+1=0$ and this polynomial has no rational roots (because of Gauss Lemma).