Variance of deterministic sine wave I have read that variance of a deterministic signal is 0 because it is deterministic and does not vary over time.

In this explanation, I am considering a $\sin(t)$ wave where value at each t is a variable(it will be a random variable when we are doing measurements but I am not sure what to call it here, so just variables). We can have sampled values of $\sin(t)$ for limited number of variables but it is not related to my question.

I understand very well for a single variable that it's variance is zero. For eg: value of sine wave at $t=1$ which is given by $x=\sin(1)$ . Since $\sin(1)$ is constant every time you try to measure it, variance given by the following formula is zero as well:
$$ var(x)= E[(x-\bar x)(x-\bar x)^T] $$
I understand this is zero because $x=\bar x$ every time you calculate x.
Now, suppose you have a deterministic signal($\sin(x)$ from $0$ to $2\pi$). Now, mean of the signal is 0 and variance is just the squared sum of signal values from $0$ to $2\pi$ which is not zero. Even more, this is providing the energy of the signal.
Does that mean variance of a deterministic signal is defined and does not have to be zero. Is variance even defined for a whole signal or is this concept valid only for a single random variable.
If $x(t)$ is deterministic its variance is zero. Note that the mean of $x(t)$ (defined as an expectation) just simply equals $x(t)$:

$$\mu_x(t)=E\{x(t)\}=x(t)\tag{1}$$

where the last equality follows from the fact that $x(t)$ is deterministic. From $(1)$ we get the variance

$$E\{[x(t)-\mu_x(t)]^2\}=E\{[x(t)-x(t)]^2\}=0\tag{2}$$