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Question Number 96330 by joki last updated on 31/May/20

Commented by prakash jain last updated on 31/May/20

$$areaofsectorAOB=\frac{1}{2}\alpha r2$$$$areaof△AOB$$$$LetX\mathrm{be}\mathrm{mid}\mathrm{point}\mathrm{lf}AB$$$$AX=BX=r\sin \frac{\alpha }{2}$$$$AB=2r\sin \frac{\alpha }{2}$$$$OX=r\cos \frac{\alpha }{2}$$$$areaof△AOB=\frac{1}{2}AB\times OX$$$$=\frac{1}{2}2r^2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}=\frac{1}{2}{r}^2\sin \alpha$$$$giventhatareaof△AOB=\frac{1}{3}\left(sectorAOB\right)$$$$\frac{1}{2}{r}^2\sin \alpha =\frac{1}{3}\left(\frac{1}{2}\alpha r^2\right)$$$$\Rightarrow 3\sin \alpha =\alpha$$$$\mathrm{hence}\alpha \mathrm{satisfies}\mathrm{equation}$$$$x=3\sin x$$

Commented by prakash jain last updated on 31/May/20

$$\left(iii\right)$$$$x_{n+1}=\frac{1}{2}(x_n+3\sin x_n)$$$$\underset{n\to \mathrm{\infty }}{\lim }{x}_n=\beta \left(assumingconvergence\right)$$$$\beta =\frac{1}{2}(\beta +3\sin \beta )$$$$\Rightarrow \beta =3\sin \beta$$$$\Rightarrow x_n\mathrm{converges}\mathrm{to}\mathrm{solution}\mathrm{of}$$$$\mathrm{equation}x=3\sin x$$

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