Question-119109

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

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

P=((1/2)+cos(π/(20)))((1/2)+cos((3π)/(20)))((1/2)+cos((9π)/(20)))((1/2)+cos((27π)/(20))) =((1/2)+cos9)((1/2)+cos81)((1/2)+cos27)((1/2)−cos63) =((1/4)+((cos9+cos81)/2)+cos9cos81) ×((1/4)+((cos27−cos63)/2)−cos27cos63) =((1/4)+cos45cos36+((cos90+cos72)/2)) ×((1/4)+sin18sin45−((cos90+cos36)/2)) =((1/4)+((√2)/2)cos36+((sin18)/2))((1/4)+((√2)/2)sin18−((cos36)/2))(1) •sin36=cos54⇔2sin18cos18=4cos^3 18−3cos18 ⇔2sin18=4cos^2 18−3=4(1−sin^2 18)−3 ⇒4sin^2 18+2sin18−1=0 sin18=(((√5)−1)/4)⇒cos36=1−2sin^2 18 =1−((3−(√5))/4)=(((√5)+1)/4).Replace into (1)we get P=((1/4)+((√2)/2).(((√5)+1)/4)+(((√5)−1)/8)) ×((1/4)+((√2)/2).(((√5)−1)/4)−(((√5)+1)/8)) =(((√5)+1+(√2)((√5)+1))/8)×(((√2)((√5)−1)−(√5)+1)/8) =(((√(10))+1+(√5)+(√2))/8).(((√(10))+1−((√5)+(√2)))/8) =((11+2(√(10))−(7+2(√(10))))/(64))=(1/(16))

$$\mathrm{P}=\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\frac{\pi}{\mathrm{20}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\frac{\mathrm{3}\pi}{\mathrm{20}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\frac{\mathrm{9}\pi}{\mathrm{20}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\frac{\mathrm{27}\pi}{\mathrm{20}}\right) \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos9}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos81}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos27}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{cos63}\right) \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{cos9}+\mathrm{cos81}}{\mathrm{2}}+\mathrm{cos9cos81}\right) \\ $$$$×\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{cos27}−\mathrm{cos63}}{\mathrm{2}}−\mathrm{cos27cos63}\right) \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{cos45cos36}+\frac{\mathrm{cos90}+\mathrm{cos72}}{\mathrm{2}}\right) \\ $$$$×\left(\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{sin18sin45}−\frac{\mathrm{cos90}+\mathrm{cos36}}{\mathrm{2}}\right) \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{cos36}+\frac{\mathrm{sin18}}{\mathrm{2}}\right)\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{sin18}−\frac{\mathrm{cos36}}{\mathrm{2}}\right)\left(\mathrm{1}\right) \\ $$$$\bullet\mathrm{sin36}=\mathrm{cos54}\Leftrightarrow\mathrm{2sin18cos18}=\mathrm{4cos}^{\mathrm{3}} \mathrm{18}−\mathrm{3cos18} \\ $$$$\Leftrightarrow\mathrm{2sin18}=\mathrm{4cos}^{\mathrm{2}} \mathrm{18}−\mathrm{3}=\mathrm{4}\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \mathrm{18}\right)−\mathrm{3} \\ $$$$\Rightarrow\mathrm{4sin}^{\mathrm{2}} \mathrm{18}+\mathrm{2sin18}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{sin18}=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\Rightarrow\mathrm{cos36}=\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \mathrm{18} \\ $$$$=\mathrm{1}−\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{4}}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}.\mathrm{Replace}\:\mathrm{into}\:\left(\mathrm{1}\right)\mathrm{we} \\ $$$$\mathrm{get}\:\mathrm{P}=\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}.\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}+\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{8}}\right) \\ $$$$×\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}.\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}−\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{8}}\right) \\ $$$$=\frac{\sqrt{\mathrm{5}}+\mathrm{1}+\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)}{\mathrm{8}}×\frac{\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)−\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{8}} \\ $$$$=\frac{\sqrt{\mathrm{10}}+\mathrm{1}+\sqrt{\mathrm{5}}+\sqrt{\mathrm{2}}}{\mathrm{8}}.\frac{\sqrt{\mathrm{10}}+\mathrm{1}−\left(\sqrt{\mathrm{5}}+\sqrt{\mathrm{2}}\right)}{\mathrm{8}} \\ $$$$=\frac{\mathrm{11}+\mathrm{2}\sqrt{\mathrm{10}}−\left(\mathrm{7}+\mathrm{2}\sqrt{\mathrm{10}}\right)}{\mathrm{64}}=\frac{\mathrm{1}}{\mathrm{16}} \\ $$

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