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Question Number 106726 by ZiYangLee last updated on 06/Aug/20

$$\mathrm{Prove}\sin 5\theta ={16\sin }^5\theta -{20\sin }^3\theta +5\sin \theta$$$$\mathrm{Hence},\mathrm{show}\mathrm{that}\sin 6°\mathrm{is}\mathrm{an}$$$$\mathrm{irrational}\mathrm{number}.$$

Answered by Ar Brandon last updated on 06/Aug/20

$$\sin 5\theta =Im(\cos 5\theta +i\sin 5\theta )=Im{(\cos \theta +i\sin \theta )}^5$$$$=Im({\mathrm{C}}^5+5i{\mathrm{C}}^4\mathrm{S}-{10\mathrm{C}}^3{\mathrm{S}}^2-10i{\mathrm{C}}^2{\mathrm{S}}^3+{5\mathrm{CS}}^4+i{\mathrm{S}}^5)$$$$={5\mathrm{C}}^4\mathrm{S}-{10\mathrm{C}}^2{\mathrm{S}}^3+{\mathrm{S}}^5=5{(1-{\mathrm{S}}^2)}^2\mathrm{S}-10(1-{\mathrm{S}}^2){\mathrm{S}}^3+{\mathrm{S}}^5$$$$=5(1-{2\mathrm{S}}^2+{\mathrm{S}}^4)\mathrm{S}-10({\mathrm{S}}^3-{\mathrm{S}}^5)+{\mathrm{S}}^5$$$$={16\mathrm{S}}^5-{20\mathrm{S}}^3+5\mathrm{S}$$$$\Rightarrow \sin 5\theta ={16\sin }^5\theta -{20\sin }^3\theta +5\sin \theta$$

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