Class A power amplifier: power dissipation
When the base is driven at the positive peak, the emitter-follower's emitter current is at a maximum because it must supply both the current sink current as well as the current through \$R_1\$. But this is also at the same point where the \$V_\text{CE}\$ is at a minimum. So there is a power minima here.

When the base is driven at the negative peak, the emitter-follower's emitter current is at a minimum because it now only has to supply the difference between what's required by the current sink and what is being supplied through \$R_1\$. But this is also at the same point where the \$V_\text{CE}\$ is at a maximum. Again, there is a power minima here.

So there are two power-minima in each cycle. Their minima values do not have to be equal to each other, because of the \$V_\text{BE}\$ voltage.

The two power maxima should be about the same value and will occur right when the signal voltage is near \$0\:\text{V}\$. You can work out this out simply enough. Ignoring \$V_\text{BE}\$ entirely for now (assuming the signal appears directly as the emitter voltage), the power in the BJT is roughly speaking: \$P=\left(V_+-V_\text{IN}\right)\cdot\left(I_1+\frac{V_\text{IN}}{R_1}\right)\$. Taking the derivative it works out that the maximum power occurs when \$V_\text{IN}\approx \frac{1}{2}\left(V_+-I_1\cdot R_1\right)\$. For your current setup, this occurs at about \$V_\text{IN}\approx 0\:\text{V}\$

You can test out this idea by changing your positive supply rail voltage to \$14\:\text{V}\$, for example. Then you should find that the peak power occurs at about \$V_\text{IN}\approx +2\:\text{V}\$.

That's really all there is to it.