Question Number 45941 by arcana last updated on 19/Oct/18
$$\mathrm{Find}\:\mathrm{0}<\theta<\mathrm{2}\pi\:\mathrm{with}\:{x},{y}\:\in\mathbb{R} \\ $$$$ \\ $$$${x}\centerdot{sin}\theta={y}\centerdot{cos}\theta \\ $$
Answered by MJS last updated on 19/Oct/18
$$\mathrm{for}\:{x}=\mathrm{0}\vee{y}=\mathrm{0}\:\mathrm{it}'\mathrm{s}\:\mathrm{trivial} \\ $$$${x}\neq\mathrm{0}\wedge{y}\neq\mathrm{0}\:\mathrm{for}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{sin}\:\theta\:=\frac{{y}}{{x}}\mathrm{cos}\:\theta \\ $$$$\mathrm{tan}\:\theta\:=\frac{{y}}{{x}} \\ $$$$\mathrm{generally}\:\theta={z}\pi+\mathrm{arctan}\:\frac{{y}}{{x}};\:{z}\in\mathbb{Z} \\ $$$$\mathrm{but}\:\mathrm{we}\:\mathrm{need}\:\theta\in\left[\mathrm{0},\:\mathrm{2}\pi\left[\right.\right. \\ $$$$−\frac{\pi}{\mathrm{2}}<\mathrm{arctan}\:\frac{{y}}{{x}}\:<\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{have} \\ $$$$\frac{{y}}{{x}}<\mathrm{0}\:\Rightarrow\:\theta=\pi+\mathrm{arctan}\:\frac{{y}}{{x}}\vee\theta=\mathrm{2}\pi+\mathrm{arctan}\:\frac{{y}}{{x}} \\ $$$$\frac{{y}}{{x}}>\mathrm{0}\:\Rightarrow\:\theta=\mathrm{arctan}\:\frac{{y}}{{x}}\vee\theta=\pi+\mathrm{arctan}\:\frac{{y}}{{x}} \\ $$