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In $\mathscr L_2$, it's easy to prove (cos$x$, sin$x$, cos$2x$, sin$2x$...cos$Nx$, sin$Nx$) is an orthogonal basis.

However, in $\mathscr L_2$, how to prove cos($kx$) is orthogonal to sin($kx$)? I am kind of lost here.

J. W. Tanner
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Busy Zhu
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1 Answers1

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In general, \begin{align*} \sin \alpha \cos \beta = \frac{1}{2}[\sin(\alpha+\beta) + \sin(\alpha - \beta)] \end{align*} So, \begin{align*} \int_{-\pi}^\pi \sin(nx) \cos(mx) dx = \frac{1}{2}\int_{-\pi}^\pi[\sin(n+m)x + \sin(n-m)x]dx = 0 \end{align*}

Tom Chen
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