Hallo,
hier das Ergebnis von Wolfram Alpha :
stationary points \( \quad \sin (x) \cos (y) \)
\( \sin (x) \cos (y)=0 \) at \( (x, y)=\left(2 \pi c_{2}, \frac{1}{2}\left(4 \pi c_{1}-\pi\right)\right) \) for integer \( c_{1} \) and \( c_{2} \quad \) (saddle points)
\( \sin (x) \cos (y)=0 \) at \( (x, y)=\left(2 \pi c_{2}, \frac{1}{2}\left(4 \pi c_{1}+\pi\right)\right) \) for integer \( c_{1} \) and \( c_{2} \quad \) (saddle points)
\( \sin (x) \cos (y)=0 \) at \( (x, y)=\left(2 \pi c_{2}+\pi, \frac{1}{2}\left(4 \pi c_{1}-\pi\right)\right) \) for integer \( c_{1} \) and \( c_{2} \quad \) (saddle points)
\( \sin (x) \cos (y)=0 \) at \( (x, y)=\left(2 \pi c_{2}+\pi, \frac{1}{2}\left(4 \pi c_{1}+\pi\right)\right) \) for integer \( c_{1} \) and \( c_{2} \) (saddle points)
\( \sin (x) \cos (y)=-1 \) at \( (x, y)=\left(\frac{1}{2}\left(4 \pi c_{2}-\pi\right), 2 \pi c_{1}\right) \) for integer \( c_{1} \) and \( c_{2} \quad \) (minima)
\( \sin (x) \cos (y)=1 \) at \( (x, y)=\left(\frac{1}{2}\left(4 \pi c_{2}-\pi\right), 2 \pi c_{1}+\pi\right) \) for integer \( c_{1} \) and \( c_{2} \quad \) (maxima)
\( \sin (x) \cos (y)=-1 \) at \( (x, y)=\left(\frac{1}{2}\left(4 \pi c_{2}+\pi\right), 2 \pi c_{1}+\pi\right) \) for integer \( c_{1} \) and \( c_{2} \quad \) (minima)
\( \sin (x) \cos (y)=1 \) at \( (x, y)=\left(\frac{1}{2}\left(4 \pi c_{2}+\pi\right), 2 \pi c_{1}\right) \) for integer \( c_{1} \) and \( c_{2} \quad \) (maxima)
