How Does the Symmetry of Sine Influence the Distribution of Y = sin(X)?

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Suppose X ~ U[ 0, pi ]

What is the distribution of Y=sinX.

I have a solution in my notes however I don,t understand the following the second transition:
<br /> F_Y(y) = P(Y \leq y) = P(X \leq \arcsin(y)) + P(X \geq \pi - \arcsin(y)) = ...<br />

Where the P(X \geq \pi - \arcsin(y)) comes from?
 
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The sine function is not monotonic.

Y is smaller than y if either of the arguments on the RHS are fulfilled.
 
Since the sin function is symmetric about ∏/2, there are two possible ranges you need to consider for the same values(imagine drawing a straight horizontal line at any 'y' on the sin curve, there are two symmetric value intervals below it), from [0, arcsin(y)] and [∏ - arcsin(y), ∏] which can be reflected in the solution.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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