Question Number 224886 by ajfour last updated on 09/Oct/25
$${Find}\:{for}\:{acute}\:\theta,\:\mathrm{sin}\:\theta\:{and}\:\mathrm{cos}\:\theta \\ $$$${in}\:{terms}\:{of}\:\mathrm{0}<{k}<\mathrm{1}, \\ $$$${if}\:\:\:\:\frac{\mathrm{sin}\:\theta\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\mathrm{cos}\:\theta\left(\mathrm{1}−\mathrm{sin}\:\theta\right)}={k}. \\ $$
Commented by ajfour last updated on 09/Oct/25
https://youtu.be/0yTJet-ZrLw?si=eNV2L9JpjRiUepqX
Commented by ajfour last updated on 10/Oct/25
https://youtu.be/dXPGMaD4-yE?si=bisZCXDm9EDnRqxL
Answered by Ghisom_ last updated on 11/Oct/25
$$\theta=\mathrm{2arctan}\:{t} \\ $$$${t}^{\mathrm{4}} −\frac{\mathrm{2}\left({k}−\mathrm{2}\right)}{{k}}{t}^{\mathrm{3}} +\mathrm{2}{t}−\mathrm{1}=\mathrm{0} \\ $$$${t}={u}+\frac{{k}−\mathrm{2}}{\mathrm{2}{k}} \\ $$$${u}^{\mathrm{4}} −\frac{\mathrm{3}\left({k}−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{2}{k}^{\mathrm{2}} }{u}^{\mathrm{2}} +\frac{{k}^{\mathrm{3}} +\mathrm{6}{k}^{\mathrm{2}} −\mathrm{12}{k}+\mathrm{8}}{{k}^{\mathrm{3}} }{u}−\frac{\mathrm{3}{k}^{\mathrm{4}} +\mathrm{8}{k}^{\mathrm{3}} +\mathrm{72}{k}^{\mathrm{2}} −\mathrm{96}{k}+\mathrm{48}}{\mathrm{16}{k}^{\mathrm{4}} }=\mathrm{0} \\ $$$$\mathrm{for}\:\mathrm{a}\:\mathrm{useful}\:\mathrm{exact}\:\mathrm{solution}\:\mathrm{we}\:\mathrm{must}\:\mathrm{get}\:\mathrm{a} \\ $$$$“\mathrm{nice}''\:\mathrm{solution}\:\mathrm{for} \\ $$$${v}^{\mathrm{3}} +\frac{\mathrm{8}}{{k}}{v}−\frac{\mathrm{16}\left({k}−\mathrm{1}\right)}{{k}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\mathrm{since}\:\mathrm{0}<{k}<\mathrm{1}\:\mathrm{there}'\mathrm{s}\:\mathrm{only}\:\mathrm{1}\:\mathrm{real}\:\mathrm{solution} \\ $$$$\mathrm{and}\:−\infty<{v}<\mathrm{1} \\ $$$$… \\ $$