It's know that $\sin a$ is a transcendental $t$ if $a \neq 0$ and algebraic. So

$\sin a = t$

This would imply

$\arcsin(t) = a$

And this question can be done for all inverse trigonometry.

Asked

Active

Viewed 235 times

0

It's know that $\sin a$ is a transcendental $t$ if $a \neq 0$ and algebraic. So

$\sin a = t$

This would imply

$\arcsin(t) = a$

And this question can be done for all inverse trigonometry.

Pinteco

- 2,217
- 9
- 21

4

No. The set of algebraic numbers is countable, while the set of transcendental numbers is not. Hence, there is at least one (and actually uncountable many!) transcendental numbers $\alpha$ for which $\sin^{-1} \alpha$ is also transcendental. This follows, e.g. from the continuity and monotonicity of the arcsin function.

4

$2^i$ is transcendental by the Gelfond-Schneider theorem, so $\sin(\ln(2)) = \frac{2^{i} - 1/2^{i}}{2i}$ is transcendental; $\ln(2)$ is transcendental by Lindemann's theorem. Thus $x = \sin(\ln(2))$, $\arcsin(x) = \ln(2)$ is an example where $x$ and $\arcsin(x)$ are both transcendental.

Robert Israel

- 416,382
- 24
- 302
- 603

3

You are trying to take the converse of the implication, which is not valid. It is true that if $a \neq 0$ is algebraic, then $\sin a$ is transcendental. However, there are transcendental $t$ such that $\arcsin t$ is also transcendental. As there are uncountably many transcendental numbers and only countably many algebraic numbers, most transcendental numbers (less than $1$ in absolute value) have transcendental $\arcsin$. It is hard to prove specific numbers transcendental. I would be "sure" that $\arcsin \frac e{10}$ is transcendental, but I don't have an approach to prove it.

Ross Millikan

- 362,355
- 27
- 241
- 432