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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?

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    $\begingroup$ 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
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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 $$x^3 - 3x+1 = 0.$$

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  • $\begingroup$ 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
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    $\begingroup$ @user136422: me, too :) $\endgroup$ – mesel Apr 11 '14 at 18:13
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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.

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  • $\begingroup$ 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

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