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I would like to know how my professor had these two results:

First: Let X be a r.v uniformly distributed in [0, 2\pi], what's the density function of $Y = \sin(X)$.

He gave as an answer $$\frac{1}{\pi\sqrt{1-y^2}}$$ but I would say that the answer should be $$\frac{1}{2\pi\sqrt{1-y^2}}$$ given that the CDF is $$\frac{1}{2\pi}$$ (I used CDF and then derived it to find density function) can someone tell me where did the "2" go ?

Second, question of the same type : Let X be a r.v uniformly distributed in [0, 1], what's the density function of $$Y = a\sin(2\pi nX+\phi)$$ with $a$,$n$ and $\phi$ constants. He gave as an answer $$\frac{1}{\pi\sqrt{a^2-y^2}}$$ but I've found $$\frac{1}{2n\pi\sqrt{a^2-y^2}}$$ and I don't know where did the $2n\pi$ go.

Can someone please help me ?

the_candyman
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Ino
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    The fact that $\int_{-1}^1 \frac{dy}{\sqrt{1-y^2}}=\pi$ means that the prof's answer is correct. To see where it's going wrong, consider on what range the function $y=\sin x$ is one-to-one. – Semiclassical Mar 06 '21 at 21:43
  • Yes okay i'm wrong, but what should I do instead ? – Ino Mar 07 '21 at 12:59

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