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

Question

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?

@user136442 Do you know that $\sin(3\alpha)=3\sin(\alpha)-4(\sin(\alpha))^3$, for all $\alpha \in \mathbb R$? — Git Gud, Apr 11 '14 at 17:36
@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. — Git Gud, Apr 11 '14 at 17:42
This question is related to, but in no way is a duplicate of, this question. — robjohn, Apr 11 '14 at 18:55
You can prove this in one line using Niven's Theorem:
https://en.wikipedia.org/wiki/Niven%27s_theorem — , May 18 '21 at 07:23

score 8 · Accepted Answer · edited Apr 11 '14 at 17:42

8

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

edited Apr 11 '14 at 17:42

Git Gud

31,706

answered Apr 11 '14 at 17:40

mesel

15,125

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. – mesel Apr 11 '14 at 17:55
Yes, I didn't find any integer root here. – user136422 Apr 11 '14 at 18:06
@user136422: it also works and easier. – mesel Apr 11 '14 at 18:10
Yep.Sometimes I just need some handholding, that's all :-) – user136422 Apr 11 '14 at 18:12
1

@user136422: me, too :) – mesel Apr 11 '14 at 18:13

score 3 · Answer 2 · edited Apr 13 '17 at 12:21

3

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.

edited Apr 13 '17 at 12:21

Community

1

answered Apr 11 '14 at 18:20

robjohn

353,833

Really make sense. – user136422 Apr 11 '14 at 18:35
Why not just say that by the rational root theorem, if there is a rational root it must be $\pm 1$? – Ovi Dec 02 '16 at 15:26
Because there are many paths that lead to the same destination. I chose a different path. – robjohn Dec 02 '16 at 15:30

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

2 Answers2