Alternatively, consider the centre of mass only over the horizontal axis. Now say we only look at the 'contribution' of f + g at time t (ignoring the issues of that contribution being infinitesimal). That contribution is (f(x) + g(x)) * sin (theta) where theta is the angle of our point. Clearly this equals f(x) * sin (theta) + g(x) * sin (theta). These are the separate contributions of f(x) and g(x). The same argument holds for the centre of mass over the vertical axis (replacing sin with cos).
If we were to make the alternative explanation formal, we get back to the + being able to move outside the integral. Note that our decomposition into the horizontal and vertical part is an alternative way to de the fourier transform without complex numbers. The vertical part here is essentially the imaginary part of the fourier transform.
This is also why peaks on an fft are gaussian (finite window), and get sharper as the fft window is increased.
* for cosine, technically there is a peak at -ƒ too. this is because a real cosine signal is ambiguous whether it is "moving forward or backwards in time". Hence it has a peak at +/-ƒ. A complex exponetial (helix through time) has chirality due to the real and imag components, so it has a single peak at ƒ. And if you take a +ƒ (lefthanded) and -ƒ helix (righthanded) and add them, the complex part cancels out, leaving only a real "up and down" wave.
> what would be useful is to give an example of a set of functions which are not linearly independent.
See [1,2] for example, which (I believe) has applications in compressed sensing and dictionary learning.
i think these kinds of explanations are hilariously pointless. and i don't mean to disparage because you're just trying to answer op's question but all you've done is restated the proof in english - i.e. of course it follows from that because what you've just said is the inner product of basis functions is 0. well yes of course that's definition of orthogonal.
You can play the individual notes of a chord on one piano or several, but they still come together to produce the same chorus of frequencies.
The Fourier series representation of a waveform is itself a sum, since trigonometric functions are waveforms as well. Thus, a combination thereof should be commutative because a sum of two sums retains the properties of addition.