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I've been learning about Fourier series, and haven't found an explicit statement of this requirement for constructing any arbitrary function using just sines and cosines, so I'm asking here. Is it true that $\sin{ax},\sin{bx},\cos{cx},\cos{dx}$ are all orthogonal to each other for all distinct real $a,b,c,d$? Symbolically: $$\int \sin{ax} \sin{bx} = \int \sin{ax} \cos{cx} = \int \sin{ax} \cos{dx} = \int \sin{bx} \cos{cx}\; ...=0 $$ Is this easy to show?

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    $\begingroup$ What's the integration range? It won't work for arbitrary values. For example, look at your first integral for $a=b$. $\endgroup$
    – J.G.
    Commented Jul 3, 2020 at 17:40
  • $\begingroup$ The range is usually from $-\pi$ to $\pi$. $\endgroup$
    – markvs
    Commented Jul 3, 2020 at 17:44
  • $\begingroup$ @J.G. I suppose it's from -π to π? $\endgroup$
    – Ivan
    Commented Jul 3, 2020 at 17:48
  • $\begingroup$ @IvanZabrodin Incidentally, you can use $\pi$ to get $\pi$; you don't need to copy-paste the letter from somewhere. $\endgroup$
    – J.G.
    Commented Jul 3, 2020 at 19:58

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I'll assume the range of integration is $[-\pi,\pi]$ and that $a,b$ are nonzero integers.

$\int_{-\pi}^{\pi} \sin(a x)\cos(bx)\,dx=0$ because the integrand is odd and integrable.

If $a\neq b$, then $\int_{-\pi}^{\pi} \sin(a x)\sin(bx)\,dx=\frac{1}{2}\int_{-\pi}^{\pi} -\cos((a+b)x)+\cos((a-b)x)\,dx=0$. Similarly for the double cosine case: $\int_{-\pi}^{\pi} \cos(a x)\cos(bx)\,dx=\frac{1}{2}\int_{-\pi}^{\pi} \cos((a+b)x)+\cos((a-b)x)\,dx=0$.

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    $\begingroup$ Aren't you assuming that $a,b$ are integers? $\endgroup$
    – md2perpe
    Commented Jul 3, 2020 at 17:56

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