Refer to This book for this discussion.
NOTE: By Fourier Sine Transform, I am NOT referring to complex part of complex Fourier Transform (which is $0$ for even functions), but the single sided sine transform as defined below, which can be non-zero for even functions too.
Definitions from the book:
$$\mathcal{F_{c}}\{f(x)\} = \mathcal{F_{c}}(k) = \sqrt{\frac{2}{\pi}} \int\limits_{0}^{\infty} \text{f(x) cos(kx) dx}$$ $$\mathcal{F_{c}^{-1}}\{\mathcal{F_{c}}(k)\} = f(x) = \sqrt{\frac{2}{\pi}} \int\limits_{0}^{\infty} \mathcal{F_{c}}(k) \text{cos(kx) dk}$$ $$\mathcal{F_{s}}\{f(x)\} = \mathcal{F_{s}}(k) = \sqrt{\frac{2}{\pi}} \int\limits_{0}^{\infty} \text{f(x) sin(kx) dx}$$ $$\mathcal{F_{s}^{-1}}\{\mathcal{F_{s}}(k)\} = f(x) = \sqrt{\frac{2}{\pi}} \int\limits_{0}^{\infty} \mathcal{F_{s}}(k) \text{sin(kx) dk}$$
There is a derivation on page 102 of the book for the formula below. $$ \mathcal{F_{c}^{-1}} \{ \mathcal{F_{c}}(k) \mathcal{G_{c}}(k) \} = \frac{1}{\sqrt{2 \pi}} \int\limits_{0}^{\infty} f(\xi) [g(\xi+x)+g(|\xi-x|)] d\xi$$
There is a derivation on page 103 of the book for the formula below. $$ \mathcal{F_{c}^{-1}} \{ \mathcal{F_{s}}(k) \mathcal{G_{s}}(k) \} = \frac{1}{\sqrt{2 \pi}} \int\limits_{0}^{\infty} f(\xi) [g(\xi+x)+g(\xi-x)] d\xi$$
Question: The derivations seem to be correct when I follow the steps, but somehow the formula is not proving to be correct if I take concrete examples. For example, if I take $\text{f(x) = g(x) = rect(x)}$, only the formula for $\mathcal{F_{c}^{-1}} \{ \mathcal{F_{c}}(k) \mathcal{G_{c}}(k) \}$ seems to be correct, but not the formula for $\mathcal{F_{c}^{-1}} \{ \mathcal{F_{s}}(k) \mathcal{G_{s}}(k) \}$. Can someone help point to the error in the latter formula (and correct it) or prove that it actually works for $\text{f(x) = g(x) = rect(x)}$. I am not worried about the scaling factors, so go with any usual choice of definition for the rect function (boxcar).