[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.kaliprasad said:Show that $\sin\,10^\circ$ is irrational
kaliprasad said:Show that $\sin\,10^\circ$ 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.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.
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.
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[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.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.
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.
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greg1313 said:I still don't see why what I wrote is incorrect. $c$ is irrational, so must be $\sin10^\circ$.
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.
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.
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.
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.
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.