Is there a way to make sense of the derivative property of Fourier Transform for $f(x)=\cos(x)$?: $$\widehat{f^{''}}(\xi) = \mathcal{F}\left\{ \frac{d^2}{dx^2} f(x) \right\} = (i 2\pi \xi)^2\hat{f}(\xi)$$
I understand this property holds for integrable functions, which is an issue for $\cos$, but perhaps there is some distributional interpretation.
On the one hand, $\frac{d^2}{dx^2}\cos(x)=-\cos(x)$, so $\mathcal{F}\{-\cos(x)\}\neq (i 2\pi \xi)^2 \mathcal{F}\{\cos(x)\}$
You have to resort to distributions to get the Fourier Transform of $\cos(x)$: $$ \begin{align} \int_{-\infty}^\infty\cos(x)\,e^{-2\pi ix\xi}\,\mathrm{d}x &=\int_{-\infty}^\infty\frac{e^{ix}+e^{-ix}}2e^{-2\pi ix\xi}\,\mathrm{d}x\\ &=\int_{-\infty}^\infty\frac12\left(e^{-2\pi ix\left(\xi-\frac1{2\pi}\right)}+e^{-2\pi ix\left(\xi+\frac1{2\pi}\right)}\right)\,\mathrm{d}x\\ &=\frac12\delta\!\left(\xi-\frac1{2\pi}\right)+\frac12\delta\!\left(\xi+\frac1{2\pi}\right) \end{align} $$ Where $\delta$ is the Dirac delta function. Thus, the support of the Fourier Transform is on the set $\xi\in\left\{-\frac1{2\pi},\frac1{2\pi}\right\}$, and on that set $(i2\pi\xi)^2=-1$.