Michael Maggard

2022-01-02

Better way to solve $\mathrm{cos}\left(\frac{{\gamma }^{\prime }}{2}\right)={g}_{0}$ and ${e}^{i{\beta }^{\prime }}\mathrm{sin}\left(\frac{{\gamma }^{\prime }}{2}\right)={g}_{1}$ for ${\gamma }^{\prime },{\beta }^{\prime }$?
Where ${g}_{0}$ is real, and ${g}_{1}$ is some complex number. I tried
(or) $\mathrm{sin}\left({\beta }^{\prime }\right)\mathrm{sin}\left(\frac{{\gamma }^{\prime }}{2}\right)=Im\left({g}_{1}\right)$
Then for each $\gamma \prime$, there should be two corresponding $\beta \prime$ s. However, it turned out that my solution doesn't quite work

Lynne Trussell

${e}^{i,\beta }=±\frac{{g}_{1}}{\sqrt{1-{g}_{0}^{2}}}$
and
$\beta =-i\mathrm{log}\frac{{g}_{1}}{\sqrt{1-{g}_{0}^{2}}}+k\pi$

Matthew Rodriguez

Solve one for $\mathrm{sin}\frac{{\gamma }^{\prime }}{2}$ , solve the other for $\mathrm{cos}\frac{{\gamma }^{\prime }}{2}$, and then use ${\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1$

Vasquez

Clearly $\gamma \prime ±2\mathrm{arccos}{g}_{0}\left(mod4\pi \right)$, which is consistent with , in which case ${e}^{i\beta \prime }=sgn{g}_{1}$ (except for the case ${g}_{0}=±1,{g}_{1}=0$, where $\beta \prime$ is arbitrary). This constrains $\beta \prime$ to either the even or odd multiples of $\pi$