I have come across an expression looking like this: $$a^2-\langle a^2 \rangle =-(b^2-\langle b^2 \rangle)$$ Where the edged parentheses are for averages. a and b are some functions. Is it true then, that (either in general or under some circumstance - like a and b be periodic with the same frequency, which they in my specific case happen to be): $$a-\langle a \rangle =-(b-\langle b \rangle)$$
How to either prove or disprove it? It might be very easy, but I have managed to get myself confused here
Rearranging your condition gives $a^2 + b^2 = \langle a^2 + b^2 \rangle$, from which it follows that $a^2 + b^2$ must be constant.
So let's take, for example, $a = \cos \theta, b = \sin \theta$, and I'll assume that the averages here mean averages over a period. Here $\langle a \rangle = \langle b \rangle = 0$, so $a - \langle a \rangle = \cos \theta \neq b - \langle b \rangle = \sin \theta$.