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Question Number 8289 by rhm last updated on 07/Oct/16

$$whatisvalue$$$$sin36°$$$$plesegivemeanswer$$

Commented by rhm last updated on 07/Oct/16

$$siranswer$$

Commented by Rasheed Soomro last updated on 07/Oct/16

$$sin36°=\frac{\sqrt{2\sqrt{5}-10}}{4}$$

Answered by ridwan balatif last updated on 15/Oct/16

$$\mathrm{let}:{\sin 18}^{\mathrm{o}}=\sin \theta$$$${\sin 36}^{\mathrm{o}}={\cos 54}^{\mathrm{o}}$$$$\sin 2\theta =\cos 3\theta$$$$2\sin \theta \cos \theta =\cos 2\theta \cos \theta -\sin \theta \sin 2\theta$$$$2\sin \theta \cos \theta =\cos 2\theta \cos \theta -\sin \theta .\left(2\sin \theta \cos \theta \right)$$$$2\sin \theta =\cos 2\theta -{2\sin }^2\theta$$$$2\sin \theta =1-{2\sin }^2\theta -{2\sin }^2\theta$$$${4\sin }^2\theta +2\sin \theta -1=0$$$$\sin {\theta }_{1,2}=\frac{-\mathrm{b}\pm \sqrt{\mathrm{D}}}{2\mathrm{a}}$$$$\sin {\theta }_{1,2}=\frac{-2\pm \sqrt{20}}{2.4}$$$$\sin {\theta }_{1,2}=\frac{-1\pm \sqrt{5}}{4}$$$$\mathrm{we}\mathrm{take}\mathrm{positive},\mathrm{because}{\sin 18}^{\mathrm{o}}\mathrm{has}\mathrm{positive}\mathrm{value}$$$$\sin \theta =\frac{-1+\sqrt{5}}{4}$$$$\mathrm{so},{\sin 18}^{\mathrm{o}}=\frac{-1+\sqrt{5}}{4}\mathrm{and}{\cos 18}^{\mathrm{o}}=\frac{\sqrt{10+2\sqrt{5}}}{4}$$$${\sin 36}^{\mathrm{o}}={2\sin 18}^{\mathrm{o}}{\cos 18}^{\mathrm{o}}$$$${\sin 36}^{\mathrm{o}}=\left(\frac{-1+\sqrt{5}}{8}\right)\left(\sqrt{10+2\sqrt{5}}\right)$$

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