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Question Number 119109 by bagjagunawan last updated on 22/Oct/20

Answered by 1549442205PVT last updated on 22/Oct/20

$$\mathrm{P}=(\frac{1}{2}+\cos \frac{\pi }{20})(\frac{1}{2}+\cos \frac{3\pi }{20})(\frac{1}{2}+\cos \frac{9\pi }{20})(\frac{1}{2}+\cos \frac{27\pi }{20})$$$$=(\frac{1}{2}+\cos 9)(\frac{1}{2}+\cos 81)(\frac{1}{2}+\cos 27)(\frac{1}{2}-\cos 63)$$$$=(\frac{1}{4}+\frac{\cos 9+\cos 81}{2}+\cos 9\cos 81)$$$$\times (\frac{1}{4}+\frac{\cos 27-\cos 63}{2}-\cos 27\cos 63)$$$$=(\frac{1}{4}+\cos 45\cos 36+\frac{\cos 90+\cos 72}{2})$$$$\times (\frac{1}{4}+\sin 18\sin 45-\frac{\cos 90+\cos 36}{2})$$$$=(\frac{1}{4}+\frac{\sqrt{2}}{2}\cos 36+\frac{\sin 18}{2})(\frac{1}{4}+\frac{\sqrt{2}}{2}\sin 18-\frac{\cos 36}{2})\left(1\right)$$$$\bullet \sin 36=\cos 54\iff 2\sin 18\cos 18={4\cos }^{3}18-3\cos 18$$$$\iff 2\sin 18={4\cos }^{2}18-3=4(1-{\sin }^{2}18)-3$$$$\Rightarrow {4\sin }^{2}18+2\sin 18-1=0$$$$\sin 18=\frac{\sqrt{5}-1}{4}\Rightarrow \cos 36=1-{2\sin }^{2}18$$$$=1-\frac{3-\sqrt{5}}{4}=\frac{\sqrt{5}+1}{4}.\mathrm{Replace}\mathrm{into}\left(1\right)\mathrm{we}$$$$\mathrm{get}\mathrm{P}=(\frac{1}{4}+\frac{\sqrt{2}}{2}.\frac{\sqrt{5}+1}{4}+\frac{\sqrt{5}-1}{8})$$$$\times (\frac{1}{4}+\frac{\sqrt{2}}{2}.\frac{\sqrt{5}-1}{4}-\frac{\sqrt{5}+1}{8})$$$$=\frac{\sqrt{5}+1+\sqrt{2}(\sqrt{5}+1)}{8}\times \frac{\sqrt{2}(\sqrt{5}-1)-\sqrt{5}+1}{8}$$$$=\frac{\sqrt{10}+1+\sqrt{5}+\sqrt{2}}{8}.\frac{\sqrt{10}+1-(\sqrt{5}+\sqrt{2})}{8}$$$$=\frac{11+2\sqrt{10}-(7+2\sqrt{10})}{64}=\frac{1}{16}$$

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