三角函数的条件弧度恒等式

青青子衿

已知弧度$\,\phi,\,\chi,\,\psi\,$，若弧度$\,\sigma,\tau\,$满足
\begin{align*}
\dfrac{1}{\sin(\sigma)\sin(\tau)}=&\dfrac{\sin(\chi)\sin(\psi)\cos(\phi)}
{\left[\sin ^2(\phi )
-\sin^2(\chi )\right]\left[\sin^2(\phi )
-\sin ^2(\psi )\right]}
\\
&\qquad+\dfrac{\sin(\phi)\sin(\psi)\cos(\chi)}
{\left[\sin ^2(\chi)-
\sin^2(\psi)\right]\left[\sin^2(\chi)-
\sin^2(\phi)\right]}\\
&\qquad\qquad+\frac{\sin(\phi)\sin(\chi)\cos(\psi)}
{\left[\sin^2(\psi)-
\sin^2(\chi)\right]\left[\sin^2(\psi)-
\sin^2(\phi)\right]}\\

\frac{\sin(\tau)\cos(\tau)-
\sin(\sigma)\cos(\sigma)}
{\sin(\sigma)\sin(\tau)
\left[\sin^2(\sigma)-
\sin^2(\tau)\right]}=
&\dfrac{\sin(\phi)\cos(\chi)\cos(\psi)}
{\left[\sin ^2(\phi )
-\sin^2(\chi )\right]\left[\sin^2(\phi )
-\sin ^2(\psi )\right]}
\\
&\qquad+\dfrac{\sin(\chi)\cos(\phi)\cos(\psi)}
{\left[\sin ^2(\chi)-
\sin^2(\psi)\right]\left[\sin^2(\chi)-
\sin^2(\phi)\right]}\\
&\qquad\qquad+\frac{\sin(\psi)\cos(\phi)\cos(\chi)}
{\left[\sin^2(\psi)-
\sin^2(\chi)\right]\left[\sin^2(\psi)-
\sin^2(\phi)\right]}\\
\end{align*}
是否有$\,\phi+\chi+\psi-\sigma-\tau\equiv0\pmod{\pi}\,$?

青青子衿

\begin{align*}
I&=\arcsin\left(\frac{7}{25}\right)+\arcsin\left(\frac{9}{41}\right)+\arcsin\left(\frac{12}{13}\right)\\
&=\arcsin\left(\frac{756456-77\sqrt{1634309}}{1026025}\right)+\arcsin\left(\frac{756456+77\sqrt{1634309}}{1026025}\right)\\
&=\arcsin\left(\sqrt{\frac{45889657564}{83983810625}-\frac{56\sqrt{717919597834}}{177555625}}\right)+\arcsin\left(\sqrt{\frac{45889657564}{83983810625}+\frac{56\sqrt{717919597834}}{177555625}}\right)
\end{align*}
取\begin{align*}
\phi&=\arcsin\left(\frac{7}{25}\right)\\
\chi&=\arcsin\left(\frac{9}{41}\right)\\
\psi&=\arcsin\left(\frac{12}{13}\right)
\end{align*}
则\begin{align*}
\sigma&=\arcsin\left(\sqrt{\frac{45889657564}{83983810625}-\frac{56 \sqrt{717919597834}}{177555625}}\right)\\
\tau&=\arcsin\left(\sqrt{\frac{45889657564}{83983810625}+\frac{56 \sqrt{717919597834}}{177555625}}\right)
\end{align*}