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I have an analog question of this one:

https://astronomy.stackexchange.com/questions/22388/generate-an-uniform-distribution-on-the-sky

I generate an uniform distribution on the sky, but just a patch of it, with $$\alpha_{min} \leq \alpha \leq \alpha_{max}$$ and $$\delta_{min} \leq \delta \leq \delta_{max}$$

for Right Ascention, I just have to take uniform distribution between $\alpha_{min}$and $ \alpha_{max}$.

For declination it works if I take $$\delta_{random} = \sin^{-1}\left( \mathrm{Uniform}(\sin\delta_{min},\, \sin\delta_{max}) \right)$$

What I call the declination $\delta$ is the angle related to the polar angle $\theta$, which is conventionally measured down from the pole, by this formula $\theta = \pi/2 - \delta$. Right Ascention is equivalent to angle $\phi$ on a sphere, which correspond to a longitude.

My problem is that I don't know how to demonstrate the formula for the generation of delinations $\delta_{random} = \sin^{-1}\left( \mathrm{Uniform}(\sin\delta_{min},\, \sin\delta_{max}) \right)$. Any idea how to make the proof of this?

I saw in https://astronomy.stackexchange.com/questions/22388/generate-an-uniform-distribution-on-the-sky @RobJeffries post which gives an expression of $P(\delta)$ and demonstrates that $\delta = \sin^{-1}(2P-1)$ where $P$ is a random number between 0 and 1, which is compatible to my formula with $\delta_{min} = -\frac{\pi}{2}$ and $ \delta_{max} = +\frac{\pi}{2}$, but I was not able to generalize his demonstration when for given limits.

Thank you very much in advance for your help!

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Rob Jeffries' post implies that $\sin\delta$ is uniformly distributed on $[-1,1]$. If you restrict $\delta$ to some range, you still have a uniform distribution for $\sin\delta$, and the limits are the values that $\sin\delta$ takes on at the limits of your range for delta.

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