Question Number 155212 by ajfour last updated on 27/Sep/21 | ||
Commented by mr W last updated on 27/Sep/21 | ||
$$\mathrm{cos}\:\theta=\frac{{p}}{\mathrm{1}}\:\Rightarrow\mathrm{cos}\:\theta={p} \\ $$$$\frac{{c}}{\mathrm{sin}\:\theta}={p}\:\mathrm{sin}\:\theta\:\Rightarrow\mathrm{sin}^{\mathrm{2}} \:\theta=\frac{{c}}{{p}} \\ $$$$\frac{{c}}{{p}}+{p}^{\mathrm{2}} =\mathrm{1} \\ $$$${p}^{\mathrm{3}} −{p}+{c}=\mathrm{0} \\ $$$$... \\ $$ | ||