This was inspired by this post. Let,
$$q = e^{2\pi\, i/m}$$
D. Speyer's answer can be generalized as,
$$\sin\Big(\frac{\pi}{m^2}\Big) = \frac{i}{2}\Big(-q^{1/(2m)}+q^{-1/(2m)} \Big)\tag1$$
while R. Israel's Maple answer, after much simplification, apparently is,
$$\sin\Big(\frac{\pi}{m^2}\Big) = \frac{1}{2}\sqrt{2-(-q)^{1/m}q^{1/2} -(-q)^{-1/m}q^{-1/2} }\tag2$$
Q: How do we show that $(1)$ and $(2)$ are in fact equivalent?
Note that $(1)$ can be rewritten as:
$$(1) = \frac{i}{2}\left(q^{-1/2m} - q^{1/2m}\right) = \frac{1}{2}\sqrt{-\left(q^{-1/2m} - q^{1/2m}\right)^2} = \frac{1}{2}\sqrt{2 - q^{-1/m} - q^{1/m}}$$
Now $q^{-1/m} = (-q)^{-1/m}\cdot (-1)^{-1/m}$.
And $(-1)^{-1/m} = (e^{i\pi})^{-1/m} = e^{-i\pi/m} = q^{-1/2}$ so $q^{-1/m} = (-q)^{-1/m}q^{-1/2}$.
Similarly $q^{1/m} = (-q)^{1/m}q^{1/2}$.
So, picking up where we left the chain of equalities:
$$\frac{1}{2}\sqrt{2 - q^{-1/m} - q^{1/m}} = \frac{1}{2}\sqrt{2 - (-q)^{-1/m}q^{-1/2} - (-q)^{1/m}q^{1/2}} = (2)$$
