Question Number 96330 by joki last updated on 31/May/20
Commented by prakash jain last updated on 31/May/20
$${area}\:{of}\:{sector}\:{AOB}=\frac{\mathrm{1}}{\mathrm{2}}\alpha{r}\mathrm{2} \\ $$$${area}\:{of}\:\bigtriangleup{AOB} \\ $$$$\:\:\:\:\:{Let}\:{X}\:\mathrm{be}\:\mathrm{mid}\:\mathrm{point}\:\mathrm{lf}\:{AB} \\ $$$$\:\:\:\:\:{AX}={BX}={r}\mathrm{sin}\frac{\alpha}{\mathrm{2}}\: \\ $$$$\:\:\:\:{AB}=\mathrm{2}{r}\mathrm{sin}\:\frac{\alpha}{\mathrm{2}} \\ $$$$\:\:\:\:\:{OX}={r}\mathrm{cos}\:\frac{\alpha}{\mathrm{2}} \\ $$$$\:\:\:\:\:{area}\:{of}\:\bigtriangleup{AOB}=\frac{\mathrm{1}}{\mathrm{2}}{AB}×{OX} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{2}{r}^{\mathrm{2}} \mathrm{sin}\:\frac{\alpha}{\mathrm{2}}\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}{r}^{\mathrm{2}} \mathrm{sin}\:\alpha \\ $$$${given}\:{that}\:{area}\:{of}\:\bigtriangleup{AOB}=\frac{\mathrm{1}}{\mathrm{3}}\left({sector}\:{AOB}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{r}^{\mathrm{2}} \mathrm{sin}\:\alpha=\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{2}}\alpha{r}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\mathrm{3sin}\:\alpha=\alpha \\ $$$$\mathrm{hence}\:\alpha\:\mathrm{satisfies}\:\mathrm{equation} \\ $$$${x}=\mathrm{3sin}\:{x} \\ $$
Commented by prakash jain last updated on 31/May/20
$$\left({iii}\right) \\ $$$${x}_{{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left({x}_{{n}} +\mathrm{3sin}\:{x}_{{n}} \right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{x}_{{n}} =\beta\:\left({assuming}\:{convergence}\right) \\ $$$$\beta=\frac{\mathrm{1}}{\mathrm{2}}\left(\beta+\mathrm{3sin}\:\beta\right) \\ $$$$\Rightarrow\beta=\mathrm{3sin}\:\beta \\ $$$$\Rightarrow{x}_{{n}} \:\mathrm{converges}\:\mathrm{to}\:\mathrm{solution}\:\mathrm{of} \\ $$$$\mathrm{equation}\:{x}=\mathrm{3sin}\:{x} \\ $$