Applying a Gaussian blur/kernel with a sigma of $\sigma_{gau}$ to a sine/cosine wave of frequency $f_{sin}$ will cause what percent reduction in the amplitude $p_{amp}$ (not power) of the sine wave?
Wikipedia gives this equation $f_c = \sqrt{ln(c)}*\sigma_f$ with an unclear wording on what $c$ is that I think says $\frac{1}{\sqrt{c}} = p_{amp}$ => $c = \frac{1}{{p_{amp}}^2}$. This would make the equation (Though I'm not sure its correct):
$f_{sin} = \sqrt{ln(c)}*\sigma_{gau}$
$\sqrt{ln(c)} = \frac{f_{sin}}{\sigma_{gau}}$
$c = e^{\frac{{f_{sin}}^2}{{\sigma_{gau}}^2}}$
$\frac{1}{{p_{amp}}^2} = e^{\frac{{f_{sin}}^2}{{\sigma_{gau}}^2}}$
$p_{amp} = \frac{1}{\sqrt{e^{\frac{{f_{sin}}^2}{{\sigma_{gau}}^2}}}} = e^\frac{-{f_{sin}}^2}{2{\sigma_{gau}}^2}$
And if you are a lost soul on google that equation in terms of percent reduction in power $p_{pow}$ is $p_{pow} = e^\frac{-{f_{sin}}^2}{{\sigma_{gau}}^2}$
We have for $\phi$ normalized Gaussian kernel (i.e. centered normal density with standard deviation $\sigma$) $$(e^{ix \alpha} * \phi)(x)=\frac{1}{\sigma\sqrt{2\pi }}\int_{\mathbb{R}} e^{iy \alpha}e^{-(y-x)^2/(2\sigma^2)}dy=e^{ix\alpha-\alpha^2\sigma^2/2}$$ So $$\begin{aligned}(\sin(x\alpha) * \phi)(x)&=\frac{1}{2i}\bigg(\int_{\mathbb{R}}e^{iy \alpha}e^{-(y-x)^2/(2\sigma^2)}dy-\int_{\mathbb{R}}e^{-iy \alpha}e^{-(y-x)^2/(2\sigma^2)}dy\bigg)=\\ &=e^{-\alpha^2\sigma^2/2}\sin(x\alpha)\end{aligned}$$ The net percentage change in amplitude is then $e^{-\alpha^2\sigma^2/2}-1$.