Here we denote the Fourier transform as $$F(k)=\mathcal{F}(f(x))=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x)e^{-ikx} \ dx,$$ and the Fourier cosine transform and Fourier sine transform as \begin{align} F_c(k)=\mathcal{F}(f(x))&=\sqrt{\frac{2}{\pi}}\int_0^{\infty} f(x) \cos(kx) \ dx \\ F_s(k)=\mathcal{F}(f(x))&=\sqrt{\frac{2}{\pi}}\int_0^{\infty} f(x) \sin(kx) \ dx \end{align} respectively.
If we are trying to determine the Fourier cosine transform of an even function, $\text{e.g} \ \ \mathcal{F}_c\left(e^{-k^2t}\right),$ is this identical to the Fourier transform $\mathcal{F}\left(e^{-k^2t}\right)$? If this is true, can a similar argument be made for odd functions involving the Fourier sine transform?
Update:
Proof for Fourier cosine: \begin{align} \mathcal{F}(f(x))&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x)e^{-ikx} \ dx \\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x)\left(\cos(kx)-i\sin(kx)\right) \ dx \\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x)\cos(kx) \ dx-\frac{i}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x)\sin(kx) dx \\ &=\sqrt{\frac{2}{\pi}}\int_0^{\infty} f(x) \cos(kx) \ dx \tag{1} \\ &=\mathcal{F}_c(f(x)). \end{align} Is my proof correct? In line $(1)$, I defined $f(x)$ to be an even function. Following on from this proof, this would seem to indicate that if $f(x)$ is odd, then $\mathcal{F}(f(x))=-i\mathcal{F}_s(f(x))$. This doesn't seem consistent with my previous intuition.