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I am little bit stuck on the following problem:

Show that if a differentiable function, $f$, is orthogonal to $\cos(t)$ on $L^{2}[0,\pi]$ then $f^{\prime}$ is orthogonal to $\sin(t)$ in $L^{2}[0, \pi]$. Hint: Integrate by parts.

OK, so I figured that we must have, for a function $f$:

$$\langle f,\cos(t) \rangle = \int_{0}^{\pi} f \cos(t) dt = 0$$

If we integrate by parts, we get:

$$f \sin(t) - \int_{0}^{\pi} f^{\prime} \sin(t) dt = 0$$


$$\int_{0}^{\pi} f^{\prime} \sin(t) dt = f \sin(t)$$

But how can I prove that this latter expression equals $0$? If someone can help me along here, I would greatly appreciate it!

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up vote 5 down vote accepted

Integrate it more properly, this is an integral with endpoints: $$[f(t) \sin(t)]_{t=0}^\pi - \int_{0}^{\pi} f^{\prime} \sin(t) dt = 0$$ and that's it, because $\sin 0 = 0 =\sin\pi$.

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Oh, of course! Duh. Silly mistake on my part. Thank you very much for your answer :) – Kristian Sep 20 '12 at 15:18

Remember that $\int\limits_{0}^{\pi}f(t) \cos t dt=(f(t)\sin t )\vert_0^{2\pi} - \int\limits_{0}^{\pi} f^{\prime}(t) \sin(t) dt = 0$

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