How can reflected light be visible if for every trough of a light wave, there is also a simultaneously arriving peak?

I will present first some words to explain the general idea, then I will add a more careful calculation.

Rather than "for every peak there must be a trough" it is more accurate to say "for every peak of one wave there is either a peak or a trough of some other wave". When a peak and trough cancel out, you get zero. When a peak and a peak add up, you get twice the amplitude and therefore four times the intensity (because intensity is proportional to amplitude squared). On average, for two such waves, on half of the occasions you get zero and on half the occasions you get 4 times the intensity of either wave on its own. Therefore on average you get twice the intensity of either wave on its own.

Now here is the above set out a little more fully.

Say the first wave has amplitude $A_1$, and at some given location it oscillates as $$ y_1 = A_1 cos(omega t). $$ Similarly, the second wave oscillates as $$ y_2 = A_2 cos(omega t + phi) $$ where $phi$ is the phase offset between the waves, which is here a random variable. The total oscillation is $$ y = y_1 + y_2 = A_1 cos(omega t) + A_2 cos(omega t + phi) $$ so the total amplitude is $$ I = langle y^2 rangle = langle (A_1 cos(omega t) + A_2 cos(omega t + phi))^2 rangle $$ where the angle bracket signifies an average over time and over all values of $phi$. Expanding the bracket gives $$ I = langle (A^2_1 cos^2(omega t) + A^2_2 cos^2(omega t + phi) + 2 A_1 A_2 cos(omega t) cos(omega t + phi) rangle . $$ Now the average of a sum is the sum of the averages, so we have $$ I = A^2_1 langle cos^2(omega t) rangle + A^2_2 langle cos^2(omega t + phi) rangle + 2 A_1 A_2 langle cos(omega t) cos(omega t + phi) rangle. $$ The average value of the $cos^2$ function is one half (try plotting it and this should become obvious to you). The function $cos(omega t) cos(omega t + phi)$ has an average value of zero, after you allow for all values of $phi$. If you are not sure about this, then you can try plotting graphs, or else use the formula $$ cos(theta + phi) = cos(theta) cos(phi) - sin(theta) sin(phi). $$ So the overall result is $$ I = frac{A_1^2 + A_2^2}{2}. $$

This calculation considered just two waves with a random relative phase. For many waves the argument is similar but involves a larger sum, with similar results.