Let $\mathcal T$ lie on the unit circle, and suppose that the regular $m$-gon has vertices $a+b\omega^j$ where $\omega$ is a primitive $m$-th root of unity. Let $\zeta$ be a primitive $n$-th root of unity. Then, $a\zeta^j+b\omega^k\zeta^j$ for $0\leq k<m$ also forms a regular $m$-gon inside $\mathcal T$. Suppose $\omega^p\zeta^q=e^{\frac{2\pi}{[m,n]}i}$. Consider the following regular $m$-gons: $$a,a+b\omega,a+b\omega^2,\ldots,a+b\omega^{m-1}$$$$a\zeta^q+b\omega^p\zeta^q,a\zeta^q+b\omega^{p+1}\zeta^q,\ldots,a\zeta^q+b\omega^{p+m-1}\zeta^q.$$ These are both similar $m$-gons, so by the Mean Geometry Theorem/Gliding Principle/whatever the accepted name is, $x(a+b\omega^k)+y(a\zeta^q+b\omega^{p+k}\zeta^q)$ for some fixed $x+y=1$ forms a regular $m$-gon which is inside $\mathcal T$ for $0\leq k<m$. By adjusting $x$ and $y$, we can get any suitable rotation between $0$ and $\frac{2\pi}{[m,n]}$, which is enough. Then, the circumradius of the original $m$-gon is $|b|$, and the circumradius of the new $m$-gon is $|xb+yb\omega^p\zeta^q|$. Let $z=\omega^p\zeta^q$. Then, this is equal to $|b(1-y)+ybz|=|b+by(1-z)|=|b||1+y(1-z)|$, so since the minimum of $|1+y(1-z)|$ occurs when $1+y(1-z)$ is the foot from $0$ to the line through $1$ and $z$, which is $\frac{z+1}2$, this is also equal to $|b|\left|\frac{z+1}2\right|=|b|\cos\frac{\pi}{[m,n]}$, as desired.

Very,very nice!