mwombenizhjb

2022-03-28

Determine the values of x,y such that $\mathrm{cos}x\mathrm{cos}z\mathrm{cos}y+\mathrm{sin}x\mathrm{sin}y<0$ for all z

Jambrichp2w2

Write
$x=j\pi +\xi ,\phantom{\rule{1em}{0ex}}y=k\pi +\eta ,\phantom{\rule{2em}{0ex}}|\xi |,|\eta |\le \frac{\pi }{2}$
Then the condition
$\mathrm{sin}x\mathrm{sin}y
translates into
${\left(-1\right)}^{j+k}\mathrm{sin}\xi \mathrm{sin}\eta <{\left(-1\right)}^{j+k}t\mathrm{cos}\xi \mathrm{cos}\eta \phantom{\rule{2em}{0ex}}\left(-1\le t\le 1\right)$
I shall do the case j+k even and assume for simplicity. Then we want
$\mathrm{tan}\xi \mathrm{tan}\eta
which is tantamount to $\mathrm{tan}\xi \mathrm{tan}\eta <-1$ , so that necessarily have different sign $\ne 0$. If $\xi <0<\eta$ we obtain the condition $-\mathrm{tan}|\xi |\mathrm{tan}\eta <-1$, or
$\mathrm{tan}|\xi |>\mathrm{cot}\eta =\mathrm{tan}\left(\frac{\pi }{2}-\eta \right)$
This implies $|\xi |>\frac{\pi }{2}-\eta$, and leads to a triangular shape in the second quadrant. When $\eta <0<\xi$ we obtain this shape reflected in the fourth quadrant.
The case j+k odd has to be treated in the same way.

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