Question Number 211992 by ajfour last updated on 26/Sep/24
Commented by ajfour last updated on 26/Sep/24
$${Find}\:{r}\:{in}\:{terms}\:{of}\:{R}\:{and}\:{a}. \\ $$
Commented by ajfour last updated on 26/Sep/24
https://youtu.be/xI8mXZpX-qM?si=hInwvqv42gbBwtVx
Answered by mr W last updated on 26/Sep/24
![cos 2α=(a/R)=1−2 sin^2 α sin^2 α=((1−(a/R))/2), cos^2 α=((1+(a/R))/2) tan α=(r/(R+(√((R−r)^2 −r^2 )))) [(r/(R+(√(R(R−2r)))))]^2 =((1−(a/R))/(1+(a/R))) say k=(r/R), λ=(a/R) ⇒[(k/(1+(√(1−2k))))]^2 =((1−λ)/(1+λ)) ⇒λ(k)=(2/(1+[(k/(1+(√(1−2k))))]^2 ))−1 ⇒k(λ)=λ^(−1) (k)](https://www.tinkutara.com/question/Q212009.png)
$$\mathrm{cos}\:\mathrm{2}\alpha=\frac{{a}}{{R}}=\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\alpha \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\alpha=\frac{\mathrm{1}−\frac{{a}}{{R}}}{\mathrm{2}},\:\mathrm{cos}^{\mathrm{2}} \:\alpha=\frac{\mathrm{1}+\frac{{a}}{{R}}}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\alpha=\frac{{r}}{{R}+\sqrt{\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }} \\ $$$$\left[\frac{{r}}{{R}+\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}}\right]^{\mathrm{2}} =\frac{\mathrm{1}−\frac{{a}}{{R}}}{\mathrm{1}+\frac{{a}}{{R}}} \\ $$$${say}\:{k}=\frac{{r}}{{R}},\:\lambda=\frac{{a}}{{R}} \\ $$$$\Rightarrow\left[\frac{{k}}{\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{2}{k}}}\right]^{\mathrm{2}} =\frac{\mathrm{1}−\lambda}{\mathrm{1}+\lambda} \\ $$$$\Rightarrow\lambda\left({k}\right)=\frac{\mathrm{2}}{\mathrm{1}+\left[\frac{{k}}{\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{2}{k}}}\right]^{\mathrm{2}} }−\mathrm{1} \\ $$$$\Rightarrow{k}\left(\lambda\right)=\lambda^{−\mathrm{1}} \left({k}\right) \\ $$