MHB Show that sin10∘ is irrational number

[kaliprasad]

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

 

[Opalg]

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

[sp]Let $x = \sin10^\circ$. The formula for $\sin3\theta$ shows that $3x - 4x^3 = \sin30^\circ = \frac12.$ So $8x^3 - 6x + 1 = 0.$ If $y=2x$ then $y^3 - 3y + 1 = 0.$ That cubic equation has no integer solutions, so by Gauss's lemma it has no rational solutions. Hence $y$, and therefore $x$, is irrational.

[/sp]

 

[Greg]

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

Spoiler

Could it be this simple?

Consider right triangle $ABC$ with $\angle{A}=80^\circ$, $\angle{B}=10^\circ$ and $\overline{AC}=1$. Then $c\sin10^\circ=1\Rightarrow\sin10^\circ=\frac1c$. As there is no Pythagorean triple with $1$ as a member, $c$ must be irrational, hence $\sin10^\circ$ is irrational.

 

[Opalg]

Spoiler

Could it be this simple?

Consider right triangle $ABC$ with $\angle{A}=80^\circ$, $\angle{B}=10^\circ$ and $\overline{AC}=1$. Then $c\sin10^\circ=1\Rightarrow\sin10^\circ=\frac1c$. As there is no Pythagorean triple with $1$ as a member, $c$ must be irrational, hence $\sin10^\circ$ is irrational.

[sp]Not quite that simple! What that argument shows is that $\sin10^\circ$ and $\cos10^\circ$ cannot both be rational. In that case, there would indeed have to be a Pythagorean triple with $1$ as a member.

The case of $\sin30^\circ$ shows that $\sin x$ can be of the form $\frac1n$ where $n$ is an integer. But of course $\cos30^\circ = \frac{\sqrt3}2$ which is not rational.

[/sp]

 

[I like Serena]

[sp]Not quite that simple! What that argument shows is that $\sin10^\circ$ and $\cos10^\circ$ cannot both be rational. In that case, there would indeed have to be a Pythagorean triple with $1$ as a member.

The case of $\sin30^\circ$ shows that $\sin x$ can be of the form $\frac1n$ where $n$ is an integer. But of course $\cos30^\circ = \frac{\sqrt3}2$ which is not rational.

[/sp]

Spoiler

By that logic that would hold for any angle, wouldn't it?
But it doesn't, since with $\alpha=\arctan(3/4)$ both $\cos\alpha$ and $\sin\alpha$ are rational.

 

[Opalg]

Spoiler

By that logic that would hold for any angle, wouldn't it?
But it doesn't, since with $\alpha=\arctan(3/4)$ both $\cos\alpha$ and $\sin\alpha$ are rational.

[sp]It holds for $\alpha=\arctan(3/4)$ because $(3,4,5)$ is a Pythagorean triple. The point is that there is no such triple having $1$ as one of its elements.

[/sp]

 

[Greg]

Spoiler

I still don't see why what I wrote is incorrect. $c$ is irrational, so must be $\sin10^\circ$.

 

[I like Serena]

[sp]It holds for $\alpha=\arctan(3/4)$ because $(3,4,5)$ is a Pythagorean triple. The point is that there is no such triple having $1$ as one of its elements.

[/sp]

Spoiler

Right.
So what that means is that there is no angle $\alpha$ that specifically has $\sin\alpha=\frac 1n$ (rational) and also a rational $\cos\alpha$.

Spoiler

I still don't see why what I wrote is incorrect. $c$ is irrational, so must be $\sin10^\circ$.

Spoiler

How do you know that $c$ is irrational?

 

[Greg]

Spoiler

Well, I can see that we have

$$\text{something}=\frac{1}{\text{something irrational}}\implies\text{ something is irrational}$$

Correct? How does the RHS being rational even enter into the conversation?

Thanks.