Expected value and variance of a trigonometric random variable

Question

I have $$Y=A\cos(\omega t) + c $$ where $A$ and $Y$ are random variables.

I know $E[Y]$ would look something like $$E[Y]=E[A]cos(\omega t) +c,$$

but how do I represent $V[Y]$ ?

I'm having trouble with figuring out the approach to solve for $E[Y]$ and $V[Y]$

edit - I figured it out to be $V[Y]=E[Y^2]-E^2[Y]$ and substituted the expressions to get a simplified form $V[Y]=V[A]\cos^2 \omega t$ . I don't know if this is the correct way to go about it.

Answer 1

Expectation is a linear operation, so for scalars $a,b$ and random variables $X,Y$, $E[aX + bY] = aE[X] + bE[Y]$. Taking the variance results in quadratic terms, so for example $Var(aX + bY) = a^2Var(X) + b^2Var(Y) + 2abCov(X,Y)$.