Imagine the sin function as it dips under the $x$ axis and let that part of the curve be filled with (2-dimensional) "water" up to the axis.
We can find the amount of water that fills this area by
$$\int _{\pi }^{2\pi}\left|\sin \left(x\right)\right|dx=2$$
let the sin function be multiplied by a factor of $\left(\frac12\right)^n$ for some $n$ so that by changing $n$, the sin waves are stretched. Suppose we start out with $n=0$, fill the dip to the axis with $2$ square units of water and then change $n$. The water level will rise or fall depending on the change in $n$.
At what height or $x$-values will the water lie in terms of $n$?
This is dealing with physical measurements so always calculate the area positively.
Let me consider the case where the sine wave is amplified by factors $a_n$: this covers your case $a_n = (\frac{1}{2})^n$.
So we start with a constant volume of water, the top surface of which lies on the segment $[\pi; 2\pi]$, and we start stretching the sine wave. Then, the following "conservation of water volume" equation needs to hold $$ a_n \int_{\pi + \epsilon}^{2 \pi - \epsilon} sin(x) \mathrm{d}x = 2$$ as $\epsilon$ quantifies the horizontal changes in segment representing the top water surface. Solving for $\epsilon$ one gets $$\epsilon = arccos (\frac{1}{a_n}) $$ and the water level rised, or descended, to $sin(\pi + \epsilon)$.
Now, if the $a_n$ are increasing, the water level will continue decreasing, and the water will remain contained in the half sine wave where was originally poured in.
If on the other hand the $a_n$ are decreasing, one has to consider the condition $$ \lvert \epsilon \rvert < \frac{\pi}{2}$$ This represents the point at which the water level has reached the peaks of the neighbouring half-wave, after which the water will pour into the neighbouring valleys, and the level remain constant at the value $a_n$ (peak height of the sine wave).