calculating 2 constants in a function M=\{f\in C[0,2\pi],\int_{0}^{2\pi}f(x)\sin x\,dx=\pi,\int_{0}^{2\pi}f(x)\sin2x\,dx=2\pi\} a,b\in \mathbb R,

Wendy Gutierrez

Wendy Gutierrez

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calculating 2 constants in a function

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Beginner2022-01-27Added 13 answers

The overall pattern is
where δmn is 1 when m=n and 0 otherwise.
The average value of sin2  is  12 and sines with different periods are orthogonal over a common period. You can prove the the first from sin2x=1212cos(2x) The cos(2x) integrates away. You can prove the first by using the function product identities, which gets you a sum of sine waves that integrate to zero as well. The same thing works for cosines.
This works because the sines and cosines form an orthogonal basis for the 2π periodic functions and you are effectively expanding a function in that basis.

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