We use the definitions $$F(k)=\mathcal{F}(f(x))=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-ikx}f(x) \ dx,$$ where the inverse is defined as $$f(x)=\mathcal{F}^{-1}(F(k))=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{ikx}F(k) \ dk.$$
Consider the Fourier transform pair commonly found in tables:
Given the definition of the Fourier transform and its inverse as already defined, are these transformations symmetric? e.g. is $$\mathcal{F}\left(\frac{\sin(ax)}{x}\right)=\sqrt{\frac{\pi}{2}}H(a-k)H(a+k)$$ the same as $$\mathcal{F}^{-1}\left(\frac{\sin(ak)}{k}\right)=\sqrt{\frac{\pi}{2}}H(a-x)H(a+x)?$$
Edit:
\begin{align} \mathcal{F}^{-1}_k(F(k)\cos(ckt))&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} F(k) \cos(ckt) e^{ikx} \ dk \\ &=\frac{1}{2\sqrt{2\pi}}\int_{-\infty}^{\infty} F(k) (e^{ckti}+e^{-ckti}) e^{ikx} \ dk \\ &=\frac{1}{2\sqrt{2\pi}}\int_{-\infty}^{\infty} F(k) e^{ik(x+ct)} \ dk+\frac{1}{2\sqrt{2\pi}}\int_{-\infty}^{\infty} F(k) e^{ik(x-ct)} \ dk \\ &=\frac{1}{2}\left(f(x+ct)+f(x-ct)\right). \end{align} (recalling that $f(x)=\mathcal{F}^{-1}(F(k))$)
Almost: with your normalization, we have that $$\mathscr{F}\{f\}(x)=\mathscr{F}^{-1}\{f\}(-x)$$ i.e. $$\mathscr{F}^2\{f\}(x)=f(-x)$$ So if $$\mathscr{F}\left\{\frac{\sin(ax)}{x}\right\}(k)=\sqrt{\frac{\pi}{2}}H(a-k)H(a+k)$$ Then $$\mathscr{F}^{-1}\left\{\frac{\sin(ak)}{k}\right\}(x)=\sqrt{\frac{\pi}{2}}H(a+x)H(a-x)$$