Cosine transform of a Sine function

Question

We all know that cos and sin are two orthogonal functions hence their dot product is zero over a domain. Then in the cosine transform how can a sine function be represented by a cosine basis function?

PS : this question may be silly, but please be gentle enough to provide an explanation

Basically, your problem lies in the lack of precision when you say “over a domain”. See my answer. — Harald Hanche-Olsen, Dec 31 '13 at 10:01

score 0 · Accepted Answer · answered Dec 31 '13 at 10:00

0

They're orthogonal over $[-\pi,\pi]$, but not over $[0,\pi]$ – which is the domain one uses for the cosine transform. In fact, taking the cosine transform of $\sin x$ is equivalent to taking the Fourier series of $\lvert\sin x\rvert$, which is the $\pi$-periodic extension of the sine function from $[0,\pi]$ to $\mathbb{R}$.

answered Dec 31 '13 at 10:00

Harald Hanche-Olsen

31,960

i wasn't aware of the [0,π] domain of the cosine transform. your answer made it clear, thank you – Varo Dec 31 '13 at 10:11

Cosine transform of a Sine function

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