Show that sin10∘ is irrational number

In summary, the conversation discusses ways to show that $\sin10^\circ$ is irrational. The formula for $\sin3\theta$ is used to show that $8x^3 - 6x + 1 = 0$, where $x = \sin10^\circ$. This leads to the conclusion that $x$ is irrational. The case of $\sin30^\circ$ is also mentioned, which shows that $\sin x$ can be of the form $\frac1n$ where $n$ is an integer, but $\cos30^\circ = \frac{\sqrt3}2$ is not rational. Finally, it is mentioned that there is no Pythagorean triple with $1$ as one of its
  • #1
kaliprasad
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Show that $\sin\,10^\circ$ is irrational
 
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  • #2
kaliprasad said:
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]
 
  • #3
kaliprasad said:
Show that $\sin\,10^\circ$ is irrational

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.
 
  • #4
greg1313 said:
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]
 
  • #5
Opalg said:
[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]

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.
 
  • #6
I like Serena said:
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]
 
  • #7
I still don't see why what I wrote is incorrect. $c$ is irrational, so must be $\sin10^\circ$.
 
  • #8
Opalg said:
[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]

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

greg1313 said:
I still don't see why what I wrote is incorrect. $c$ is irrational, so must be $\sin10^\circ$.

How do you know that $c$ is irrational?
 
  • #9
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.
 

FAQ: Show that sin10∘ is irrational number

1. What is an irrational number?

An irrational number is a number that cannot be expressed as a ratio of two integers. This means that it cannot be written as a simple fraction, and its decimal representation is non-terminating and non-repeating.

2. How do you prove that sin10∘ is irrational?

To prove that sin10∘ is irrational, we can use proof by contradiction. Assume that sin10∘ is rational, meaning it can be expressed as a fraction a/b where a and b are integers. Then, using the sum and difference identities of sine, we can show that sin10∘ can be expressed as a simpler fraction. However, this contradicts our assumption that sin10∘ is irrational, therefore it must be irrational.

3. Can you provide an example of a proof by contradiction for sin10∘?

Yes, here is a proof by contradiction for sin10∘:
Assume sin10∘ is rational, so it can be expressed as a/b where a and b are integers. This means that sin10∘ = a/b.
Using the sum identity of sine, we know that sin(90∘ - 80∘) = sin90∘cos80∘ - cos90∘sin80∘.
Since sin90∘ = 1 and cos90∘ = 0, this simplifies to sin10∘ = cos80∘.
Next, using the difference identity of sine, we know that sin(90∘ - 10∘) = sin90∘cos10∘ - cos90∘sin10∘.
Since sin90∘ = 1 and cos90∘ = 0, this simplifies to sin80∘ = cos10∘.
Combining these two equations, we get cos80∘ = cos10∘, which means a = b. However, this contradicts our assumption that a and b are integers with no common factors. Therefore, our initial assumption that sin10∘ is rational must be false, and sin10∘ is irrational.

4. Are there any other methods to prove that sin10∘ is irrational?

Yes, there are a few other methods to prove that sin10∘ is irrational. One method is using the Taylor series expansion of sinx and showing that it cannot be expressed as a rational number. Another method is using the fact that sin10∘ is a root of the polynomial 8x^3 - 6x + 1, and using the rational root theorem to show that it cannot have rational roots.

5. Why is proving that sin10∘ is irrational important?

Proving that sin10∘ is irrational is important because it helps us understand the properties of irrational numbers and their relationship with trigonometric functions. It also demonstrates the power and usefulness of proof by contradiction in mathematics. Furthermore, the proof for sin10∘ can be generalized to show that sinx is irrational for any non-zero rational value of x in degrees.

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