Prove mathematically that the temporal average of $\sin(at+b)$ is zero and the temporal average of $\sin^2(at+b)$ is $1/2$

Question

I started by using the Poynting formula, that is $P = 1/T \int \sin(wt)dt$. But I don't know why should we consider that $T$ is equal to $2\pi/a$ so that at the end $P$ for the $\sin(at+b)= 0$.

Answer 1

If you don't want to restrict yourself to one period ($2\pi/a$), you can let the interval go to infinity. $$\lim_{T\to \infty}\frac 1T\int_0^T\sin(at+b) dt$$ The integral is finite, $\frac 1a(\cos b-\cos(aT+b))$. When you divide by $T$ and let $T$ go to infinity, that number is zero.

You can apply the same to $\sin^2$, using $$\sin^2(at+b)=\frac 12-\frac12\cos2(at+b)$$ The integral of the $\cos$ part is once again finite, the integral of the constant term is $\frac T2$. Therefore $$\lim_{T\to \infty}\frac 1T\int_0^T\sin^2(at+b) dt=\lim_{T\to \infty}\frac1T\frac T2=\frac12$$