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:
$$f \sin(t) - \int_{0}^{\pi} f^{\prime} \sin(t) dt = 0$$
Or:
$$\int_{0}^{\pi} f^{\prime} \sin(t) dt = f \sin(t)$$
But how can I prove that this latter expression equals $0$? If someone can help me along here, I would greatly appreciate it!