So, this is the problem I am working on.

Show that $\sin 10^\circ$ is irrational.

The solution to the problem is $$1/2 = \sin 30^\circ = 3 \sin 10^\circ - 4\sin^3 10^\circ .$$ Let $$x = 2\sin 10^\circ.$$

Then we have, $$x^3 - 3x + 1 = 0.$$ And, we have to work on this to find out the roots. But, what I don't understand is that why do I have to subtract $4\sin^3 10^\circ$ from $3\sin 10^\circ$. And, how did they come up with $x^3 - 3x+1 = 0?$ I am confused. Can someone please explain this in details and is there any other way we can do this problem?

  • 1
    $\begingroup$ Sorry my bad. $1/2 = sin 30^\circ$ $\endgroup$ – user136422 Apr 11 '14 at 17:33
  • $\begingroup$ @user136442 Do you know that $\sin(3\alpha)=3\sin(\alpha)-4(\sin(\alpha))^3$, for all $\alpha \in \mathbb R$? $\endgroup$ – Git Gud Apr 11 '14 at 17:36
  • $\begingroup$ Sorry. Must've forgotten. $\endgroup$ – user136422 Apr 11 '14 at 17:40
  • $\begingroup$ @user136442 It is likely, by reading the proposed solution, that that identity is a prerequisite for the problem. I suggest you assume it as true for the purpose of this problem. $\endgroup$ – Git Gud Apr 11 '14 at 17:42
  • $\begingroup$ This question is related to, but in no way is a duplicate of, this question. $\endgroup$ – robjohn Apr 11 '14 at 18:55

identity: $\sin(3a)=3\sin(a)-4\sin^3(a)$ By using this identity,

$$1/2 = \sin 30^\circ = 3 \sin 10^\circ - 4\sin^3 10^\circ$$

$$1=2\sin 30^\circ = 6 \sin 10^\circ - 8\sin^3 10^\circ$$ Then if you set $x=2\sin(10)$ you will get $$x^3 - 3x+1 = 0.$$

  • $\begingroup$ By the way I assumed that you reached the conlusion. If not, set $x=t-1$ in above equation then you will get $t^3-3t^2+3$ by Einstain creteria, it has no rational roots. $\endgroup$ – mesel Apr 11 '14 at 17:55
  • $\begingroup$ Yes, I didn't find any integer root here. $\endgroup$ – user136422 Apr 11 '14 at 18:06
  • $\begingroup$ @user136422: it also works and easier. $\endgroup$ – mesel Apr 11 '14 at 18:10
  • $\begingroup$ Yep.Sometimes I just need some handholding, that's all :-) $\endgroup$ – user136422 Apr 11 '14 at 18:12
  • 1
    $\begingroup$ @user136422: me, too :) $\endgroup$ – mesel Apr 11 '14 at 18:13

mesel's answer about why $2\sin(10^\circ)$ satisfies $x^3-3x+1=0$ is very good.

Let's answer the question about why $x^3-3x+1=0$ implies $x$ is irrational. In this answer, it is shown that if $x^3-3x+1=0$ has a rational root, then that root must be an integer.

Suppose that $|x|\ge2$, then dividing by $x^3$ yields $$ \begin{align} 1 &=\left|\,\frac3{x^2}-\frac1{x^3}\,\right|\\ &\le\frac34+\frac18\\ &=\frac78 \end{align} $$ Thus, there can be no solutions for $|x|\ge2$.

Simply checking $\{-1,0,1\}$, we see there are no integer solutions. Therefore, any solution to $x^3-3x+1=0$ is irrational.

  • $\begingroup$ Really make sense. $\endgroup$ – user136422 Apr 11 '14 at 18:35
  • $\begingroup$ Why not just say that by the rational root theorem, if there is a rational root it must be $\pm 1$? $\endgroup$ – Ovi Dec 2 '16 at 15:26
  • $\begingroup$ Because there are many paths that lead to the same destination. I chose a different path. $\endgroup$ – robjohn Dec 2 '16 at 15:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.