Mean & Variance of X: Solving Confusing Problem w/sin(θ)

[SomeRandomGuy]

Let X be the sin of the angle theta in radians chosen uniformly from (-pi/2,pi/2). Find the mean and variance of X. HINT: X = sin(theta). Specify the support of X and check to see if your result describes a p.d.f.

Anyone got any idea's? I managed to solve the majority of other problems and don't even know where to begin this one. Thanks for any help.

 

[HallsofIvy]

Since the sine of any angle is between -1 and 1 the "support of X" is
[-1,1]. The integral of 1 from -2\pi to 2\pi is 4\pi so the "uniform distribution" density is \frac{1}{4\pi}. The mean value of X will, of course, be
\int_{-1}^1X P(X)dX= \frac{1}{4\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}sin(\theta)cos(\theta)d\theta.

 

[SomeRandomGuy]

The integral of 1 from -2\pi to 2\pi is 4\pi so the "uniform distribution" density is \frac{1}{4\pi}.

First, i'd like to thank you for your response. I understand everything you said except what I left in quotes. Why did you integrate 1 from -2pi to 2pi? Other than that, I already had the support being [-1,1] and I know how to find mean and variance when given some P(X).