whine changes slightly (while the line frequency doesn’t)
Its frequency was obviously the same
The frequency components of the whine definitely change (otherwise chances are you wouldn't have heard it). To little surprise, when you modulate the intensity of something with a temporally periodic thing (e.g. a periodic background), you modulate the tone!
Think of this: the whine comes from something (usually, transformer coils) moving due to a high-frequency (15 kHz) changing current; \$x_1(t) = \sin(2\pi f_1 t)\$.
If that current, for example, also changes at row frequency/2 (e.g., 1 kHz/2), because you're drawing the left half of the screen black and the right one white, \$x_2(t) = \operatorname{Squarewave}(2\pi f_2 t)\$ then your resulting acoustic signal is simply the product of both, \$x(t)=x_1(t)\cdot x_2(t)\$.
Now, a squarewave (like any periodic signal) is composed of a countable sum of cosines:
$$x_2(t) = \sum_{n=0}^\infty \frac{\cos((2n -1)2\pi f_2 t)}{2n -1},$$
so that the product is simply
$$x(t)=x_1(t)x_2(t) = \sum_{n=0}^\infty \frac{\cos(2\pi f_1t)\cos((2n -1) 2\pi f_2 t)}{2n -1},$$
which, thanks to trigonometry, simply is
\begin{align}
x(t)&= \sum_{n=0}^\infty \frac12\frac{\cos(2\pi f_1t+(2n -1)2\pi f_2 t)+\cos(2\pi f_1 t - (2n -1)2\pi f_2 t)}{2n -1}\\
&=\frac 12\sum_{n=0}^\infty \frac{\cos(2\pi (f_1+(2n -1)f_2) t)+\cos(2\pi (f_1-(2n -1)f_2) t)}{2n -1},\\
\end{align}
so instead of 15 kHz, you hear new frequency components at 15 kHz \$\pm\$1 kHz, 3 kHz, 5 kHz, and so on, so you've got whine frequency components at 2, 4, …, 10, 12, 14, 16, 18, 20, … kHz. The strongest parts are still the ones "close" to the original unmoodulated whine (if your modulating signal was indeed a square wave).
So: yep, there were most definitely frequencies in the whine that were caused by the modulation of the load due to image content!
