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Question Number 12085 by chux last updated on 12/Apr/17

Answered by ajfour last updated on 12/Apr/17

$$let6°=\theta$$$$\sin 24°=2\sin 12°\cos 12°$$$$let\sin 12°=xand\cos 12°=y$$$$\sin (3\theta +2\theta )=y\sin 18°+x\cos 18°$$$$\Rightarrow x\cos 18°+y\sin 18°=\sin 5\theta =\frac{1}{2}$$$$\cos (3\theta +2\theta )=y\cos 18°-x\sin 18°$$$$\Rightarrow y\cos 18°-x\sin 18°=\cos 5\theta =\frac{\sqrt{3}}{2}$$$$solvesimultaneouslyforxandy.$$$$\sin 24°=2xy$$

Answered by Joel576 last updated on 16/Apr/17

$$\mathrm{Let}\theta =18$$$$\sin 36=\cos 54$$$$\sin 2\theta =\cos 3\theta$$$$2.\sin \theta .\cos \theta ={4\cos }^3\theta -3\cos \theta$$$$2\sin \theta ={4\cos }^2\theta -3$$$$2\sin \theta =4-{4\sin }^2\theta -3$$$${4\sin }^2\theta +2\sin \theta -1=0$$$$$$$$\mathrm{Let}\sin \theta =x$$$$4x^2+2x-1=0$$$$x=\frac{-1\pm \sqrt{5}}{4}$$$$\mathrm{Because}\theta =18,\mathrm{so}\sin 18=x=\frac{\sqrt{5}-1}{4}$$$$\cos 18=\sqrt{1-{\sin }^{2}18}$$$$=\sqrt{1-{\left(\frac{\sqrt{5}-1}{4}\right)}^2}$$$$=\sqrt{\frac{16}{16}-\frac{5-2\sqrt{5}+1}{16}}$$$$=\sqrt{\frac{10+2\sqrt{5}}{16}}$$$$\cos 18=\frac{\sqrt{10+2\sqrt{5}}}{4}$$$$\cos 48=\cos (30+18)$$$$=(\cos 30.\cos 18)-(\sin 30.\sin 18)$$$$=(\frac{\sqrt{3}}{2}.\frac{\sqrt{10+2\sqrt{5}}}{4})-(\frac{1}{2}.\frac{\sqrt{5}-1}{4})$$$$=\frac{\sqrt{30+6\sqrt{5}}}{8}-\frac{\sqrt{5}-1}{8}$$$$\cos 48=\frac{\sqrt{30+6\sqrt{5}}-\sqrt{5}+1}{8}$$$$\mathrm{Let}\alpha =24$$$$\sin \alpha =\sqrt{\frac{1-\cos 2\alpha }{2}}$$$$=\sqrt{\frac{1-\cos 48}{2}}$$$$=\sqrt{\frac{\frac{8}{8}-\frac{\sqrt{30+6\sqrt{5}}-\sqrt{5}+1}{8}}{2}}$$$$=\sqrt{\frac{\left(\frac{7-\sqrt{30+6\sqrt{5}}+\sqrt{5}}{8}\right)}{2}}$$$$=\sqrt{\frac{7-\sqrt{30+6\sqrt{5}}+\sqrt{5}}{16}}$$$$\sin 24=\frac{1}{4}\left(\sqrt{7-\sqrt{30+6\sqrt{5}}+\sqrt{5}}\right)$$

Commented by mrW1 last updated on 13/Apr/17

$$pleasecheckthelast3lines.the$$$$correctresultis$$$$\sin 24°=\frac{1}{4}\sqrt{7-\sqrt{30+6\sqrt{5}}+\sqrt{5}}$$

Commented by Joel576 last updated on 16/Apr/17

$$thankyouforcorrectingmyanswer.$$$$Ihavefixedthat$$

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