Given any real scalar function on a vector space, $f(\mathbf{x})$, we can break it down into parts that are even and odd under parity as $$f_{\pm}(\mathbf{x}) = \frac{f(\mathbf{x}) \pm f(-\mathbf{x})}{2}$$ and reconstruct the original function from those parts as $$f(\mathbf{x}) = f_+(\mathbf{x}) + f_-(\mathbf{x}).$$
The symmetric Fourier sine and cosine transforms are defined as \begin{align} \tilde{f}_c(\mathbf{k}) & \equiv \int \operatorname{d}^dx\, f(\mathbf{x}) \frac{\cos(\mathbf{k}\cdot\mathbf{x})}{(2\pi)^{d/2}},\ \mathrm{and} \\ \tilde{f}_s(\mathbf{k}) & \equiv \int \operatorname{d}^dx\, f(\mathbf{x}) \frac{\sin(\mathbf{k}\cdot\mathbf{x})}{(2\pi)^{d/2}}, \end{align} respectively. The transforms are symmetric because they're inverted by \begin{align} f(\mathbf{x}) & = \int \frac{\operatorname{d}^dk}{(2\pi)^{d/2}} \left[\tilde{f}_c(\mathbf{k})\cos(\mathbf{k}\cdot\mathbf{x}) + \tilde{f}_s(\mathbf{k})\sin(\mathbf{k}\cdot\mathbf{x})\right]. \end{align}
Clearly, $\tilde{f}_c(\mathbf{k})$ is even under reflection of $\mathbf{k}$ because cosine is even; similarly, $\tilde{f}_s(\mathbf{k})$ is odd. Less obviously, the cosine transform only depends on the even part of $f$, and the sine transform the odd. This is because $\cos(\mathbf{k}\cdot\mathbf{x})$ is even in both $\mathbf{k}$ and $\mathbf{x}$, so the integral of $f_-(\mathbf{x})\cos(\mathbf{k}\cdot\mathbf{x})$ vanishes because the overall function is odd (similarly for $f_+$ and sine).
The even/odd function pattern would seem to suggest that I could define a combined transform function as $$\tilde{f}(\mathbf{k}) \equiv \tilde{f}_c(\mathbf{k}) + \tilde{f}_s(\mathbf{k}).$$ The question is this: in doing so would I lose any information about the original $f(\mathbf{x})$ (i.e. is it as complete as the $\left[\tilde{f}_c,\, \tilde{f}_s\right]$ pair)? What gives me pause is it seems that combining definitions gives \begin{align} \tilde{f}(\mathbf{k}) & = \int \operatorname{d}^d x \frac{f(\mathbf{x})}{(2\pi)^{d/2}} \left[\cos(\mathbf{k}\cdot\mathbf{x}) + \sin(\mathbf{k}\cdot\mathbf{x}) \right]\\ & = \int \operatorname{d}^d x \frac{f(\mathbf{x})}{(2\pi)^{d/2}} \sqrt{2} \cos\left(\mathbf{k}\cdot\mathbf{x} - \frac{\pi}{4}\right), \end{align} and given that neither the sine nor cosine transforms are complete by themselves, it seems unlikely that $\cos\left(\mathbf{k}\cdot\mathbf{x} - \frac{\pi}{4}\right)$ would be complete by itself without $\cos\left(\mathbf{k}\cdot\mathbf{x} + \frac{\pi}{4}\right)$.
Hartley transform defined (in 1D) as $$\mathscr{H}f(\omega)=\frac 1{\sqrt\pi}\int_{-\infty}^\infty f(t)\cos(\omega t-\frac\pi4)\,dt$$ is an involution, i.e. $$\mathscr{H}\left(\mathscr{H}(f)\right)=f$$ Then, you do not loose information in defining the Hartley transform function $\bar f(\mathbf{k})$ as you do. It is a real function related to the real and imaginary parts of the Fourier transform: $$\tilde f(\mathbf{k})=\tilde f_c(\mathbf{k})+\tilde f_s(\mathbf{k})=\Re\left[(1-i)\mathscr{F}(f)\right]$$