다음을 풀어보세요: sinalpha=cosy

$$\sin \alpha = \cos y$$

y에 대한 해 (complex solution)

$$\left\{\begin{matrix}y=-i\ln(\frac{\sqrt{2}\left(\sqrt{2}\sin(\alpha )-\sqrt{-\cos(2\alpha )-1}\right)}{2})+2\pi n_{1}\text{, }n_{1}\in \mathrm{Z}\text{, }&\frac{\sqrt{2}\left(\sqrt{2}\sin(\alpha )-\sqrt{-\cos(2\alpha )-1}\right)}{2}\neq 0\\y=-i\ln(\frac{\sqrt{2}\left(\sqrt{2}\sin(\alpha )+\sqrt{-\cos(2\alpha )-1}\right)}{2})+2\pi n_{2}\text{, }n_{2}\in \mathrm{Z}\text{, }&\frac{\sqrt{2}\left(\sqrt{2}\sin(\alpha )+\sqrt{-\cos(2\alpha )-1}\right)}{2}\neq 0\end{matrix}\right.$$

α에 대한 해 (complex solution)

$$\left\{\begin{matrix}\alpha =-i\ln(\frac{2i\cos(y)-\sqrt{2}\sqrt{-\cos(2y)+1}}{2})+2\pi n_{1}\text{, }n_{1}\in \mathrm{Z}\text{, }&\frac{2i\cos(y)-\sqrt{2}\sqrt{-\cos(2y)+1}}{2}\neq 0\\\alpha =-i\ln(\frac{2i\cos(y)+\sqrt{2}\sqrt{-\cos(2y)+1}}{2})+2\pi n_{2}\text{, }n_{2}\in \mathrm{Z}\text{, }&\frac{2i\cos(y)+\sqrt{2}\sqrt{-\cos(2y)+1}}{2}\neq 0\end{matrix}\right.$$

y에 대한 해

$$y=-\arccos(\sin(\alpha ))+2\pi n_{1}\text{, }n_{1}\in \mathrm{Z}$$

$$y=\arccos(\sin(\alpha ))+2\pi n_{2}\text{, }n_{2}\in \mathrm{Z}$$

α에 대한 해

$$\alpha =-\arcsin(\cos(y))+2\pi n_{1}+\pi \text{, }n_{1}\in \mathrm{Z}$$

$$\alpha =\arcsin(\cos(y))+2\pi n_{2}\text{, }n_{2}\in \mathrm{Z}$$