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If I could guess, I would say it's not. But how to prove it?

It's know that $\sin a$ is transcendental, therefore irrational for $a$ algebraic. This is a result from Lindemann–Weierstrass theorem. Knowing this we can prove that $\arcsin a$ is also transcendental for a algebraic argument. But what about a sum, eg. $\arcsin 1/4$ + $\arcsin 1/3$.

Or $\arcsin a + \arcsin b$, for algebraic $a$ and $b$.

I found another way, but I'm not sure if it's correct.

One can show that: $\arcsin x = -i \ln(\sqrt{1-x^2}+ix)$. So,

$\arcsin a + \arcsin b = -i \ln(\sqrt{1-a^2}+ia) -i \ln(\sqrt{1-b^2}+ib) $.

$\arcsin a + \arcsin b = -i \ln[ (\sqrt{1-a^2}+ia)(\sqrt{1-b^2}+ib)] $

If $a$ and $b$ are algebraic, so it is $(\sqrt{1-a^2}+ia)(\sqrt{1-b^2}+ib)$.

Another result of Lindemann–Weierstrass theorem is that $\ln a$ is transcendental for $a$ algebraic. So for $a$ and $b$ algebraic, $\arcsin a + \arcsin b$ is transcendental.

Pinteco
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1 Answers1

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One can show that

$$\sin\left(\arcsin \frac 1 4 + \arcsin \frac 1 3\right) = \frac{1}{12} \sqrt{23 + 4\sqrt{30}}$$

which is an algebraic number. By the Lindemann-Weierstrass theorem, if $\arcsin 1/4 + \arcsin 1/3$ were rational (hence algebraic) then its sine would be transcendental, a contradiction.

In general, using the addition formula for the sine function as well as the traditional way to simplify $\cos \sin^{-1} \cdot$, you can generalize this technique to many other sums of the form $\arcsin a + \arcsin b$. You can check that

$$\sin \left(\arcsin a + \arcsin b\right) = b \sqrt{1 - a^2} + a \sqrt{1 - b^2}$$

which is algebraic whenever $a$ and $b$ are. Provided it is non-zero, we can apply the first part of this answer.