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Let $\theta= \arcsin(1/4)/\pi$. Assume $\theta$ is a rational multiple of $\pi$.

Then, there exists some $n$ such that $\sin(n\theta)=0$. This gives $\cos(n \theta)= \pm 1$.

Set $z=\cos(\theta)+i \sin(\theta)$, then $z^n= \pm 1$ and $\frac{1}{z^n}=\pm 1$.

This gives that $z$ and $\frac{1}{z}$ are algebraic integers, and hence so is $$2 i\sin(\theta)=z- \frac{1}{z}$$

Therefore, $\frac{i}{2}$ is an algebraic integer, which is a contradiction.

Let $\theta= \arcsin(1/4)$. Assume $\theta$ is a rational multiple of $\pi$.

Then, there exists some $n$ such that $\sin(n\theta)=0$. This gives $\cos(n \theta)= \pm 1$.

Set $z=\cos(\theta)+i \sin(\theta)$, then $z^n= \pm 1$ and $\frac{1}{z^n}=\pm 1$.

This gives that $z$ and $\frac{1}{z}$ are algebraic integers, and hence so is $$2 i\sin(\theta)=z- \frac{1}{z}$$

Therefore, $\frac{i}{2}$ is an algebraic integer, which is a contradiction.

Let $\theta= \arcsin(1/4)$. Assume $\theta$ is a rational multiple of $\pi$.

Then, there exists some $n$ such that $\sin(n\theta)=0$. This gives $\cos(n \theta)= \pm 1$.

Set $z=\cos(\theta)+i \sin(\theta)$, then $z^n= \pm 1$ and $\frac{1}{z^n}=\pm 1$.

This gives that $z$ and $\frac{1}{z}$ are algebraic integers, and hence so is $$2 i\sin(\theta)=z- \frac{1}{z}$$

Therefore, $\frac{i}{2}$ is an algebraic integer, which is a contradiction.

Let $\theta= \arcsin(1/4)/\pi$. Assume $\theta$ is a rational multiple of $\pi$.

Then, there exists some $n$ such that $\sin(n\theta)=0$. This gives $\cos(n \theta)= \pm 1$.

Set $z=\cos(\theta)+i \sin(\theta)$, then $z^n= \pm 1$ and $\frac{1}{z^n}=\pm 1$.

This gives that $z$ and $\frac{1}{z}$ are algebraic integers, and hence so is $$2 i\sin(\theta)=z- \frac{1}{z}$$

Therefore, $\frac{i}{2}$ is an algebraic integer, which is a contradiction.

Let $\theta= \arcsin(1/4) / \pi$. Assume it is rational.

Then, there exists some $n$ such that $\sin(n\theta)=0$. This gives $\cos(n \theta)= \pm 1$.

Set $z=\cos(\theta)+i \sin(\theta)$, then $z^n= \pm 1$ and $\frac{1}{z^n}=\pm 1$.

This gives that $z$ and $\frac{1}{z}$ are algebraic integers, and hence so is $$2 i\sin(\theta)=z- \frac{1}{z}$$

Therefore, $\frac{i}{2}$ is an algebraic integer, which is a contradiction.

Let $\theta= \arcsin(1/4)$. Assume $\theta$ is a rational multiple of $\pi$.

Then, there exists some $n$ such that $\sin(n\theta)=0$. This gives $\cos(n \theta)= \pm 1$.

Set $z=\cos(\theta)+i \sin(\theta)$, then $z^n= \pm 1$ and $\frac{1}{z^n}=\pm 1$.

This gives that $z$ and $\frac{1}{z}$ are algebraic integers, and hence so is $$2 i\sin(\theta)=z- \frac{1}{z}$$

Therefore, $\frac{i}{2}$ is an algebraic integer, which is a contradiction.