Proof of irrationality of $\arcsin(\frac{1}{4})$

Question

I was working to find a different approach to Niven's theorem from the one in my textbook taking a route via Chebyshev polynomials. It all comes to proving the irrationality of $\arcsin(\frac{1}{4})$ and I admit I am absolutely clueless how to prove this. Any help would be appreciated.

Answer 1

Before a begin I will put some observation:

1. $\arcsin(\sin (\alpha)) = \alpha$ for all $\alpha$ where the function i well defined.
2. Every transcendental number is an irrational number (I will let you research the reason).
3. I will prove that $\arcsin(\alpha)$ is transcendental for all $\alpha \neq 0$ algebraic, it is obvious that $\frac {1}{4}$ is algebraic.

Let's go:

You can use the Lindemman's Theorem:

Lindemann's Theorem: If $\alpha_1,...,\alpha_n \in \mathbb{C}$ are distinct algebraic numbers then $e^{\alpha_1}, ...,e^{\alpha_n}$ are linearment independent over $\mathbb{Q}$.

We can prove that $e^{\alpha}$ is transcendental for all alpha algebraic. That is, suppose that $e^{\alpha}$ is algebraic, so there is a polynomial $P(x) = \sum_{k=0}^n a_kx^k, \alpha_k \in \mathbb{Q}$ s.t $P(e^\alpha)=0$.

But $0 = p(e^\alpha) = \sum_{k=0}^n e^{\alpha k}$. By Lindemann Theorem we have $a_k=0, \forall k=0,...,n$. Then $p(x)=0, \forall x.$

So $e^\alpha$ is transcendental.

Now, we can prove that $\arcsin(\alpha)$ is transcendental, for all $\alpha$ algebraic. Let $\alpha$ an algebraic nonzero number, suppose that $\arcsin(\alpha)$ is algebraic so $i \cdot \arcsin(\alpha)$ is. Then $e^{i\arcsin(\alpha)}$ is transcendental (we just have proved that), but....

$e^{i\arcsin(\alpha)} = \cos(\arcsin(\alpha)) + i\sin(\arcsin(\alpha) = \cos(\arcsin(\alpha)) + i\alpha$,

Since $i \alpha$ is algebraic, we have $\cos(\arcsin(\alpha))$ is transcendental because if it was algebraic then $\cos(\arcsin(\alpha)) + i\alpha = e^{i\arcsin(\alpha)}$ would be (sum of algebraic numbers is an algebraic number).

Note that $1 = \cos^2(\arcsin(\alpha)) + \sin^2(\arcsin(\alpha))$, that is, $1 - \alpha^2 = \cos^2(\arcsin(\alpha)))$. This is a contradiction because in the right side we have an algebraic number and the left side we have a transcendental number.

Then $\arcsin(\alpha)$ is transcendental for all $\alpha \neq 0$ algebraic number.

Now take $\alpha = \frac {1}{4}$