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Question-146824

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Question Number 146824 by alcohol last updated on 16/Jul/21
Answered by Olaf_Thorendsen last updated on 16/Jul/21
1− w = e^((2iπ)/n) Let z_n = 1+w+w^2 +...w^(n−1) z_n = 1×((1−(e^((2iπ)/n) )^n )/(1−e^((2iπ)/n) )) = ((1−e^(2iπ) )/(1−e^((2iπ)/n) )) = 0 2− z_5 = 1+e^(i((2π)/5)) +e^((4iπ)/5) +e^((6iπ)/5) +e^((8iπ)/5) = 0 ⇒ Re(z_5 ) = 0 = 1+cos(((2π)/5))+cos(((4π)/5))+cos(((6π)/5))+cos(((8π)/5)) 3− A = cos(((2π)/5))+cos(((8π)/5)) A = 2cos(((((2π)/5)+((8π)/5))/2))cos(((((2π)/5)−((8π)/5))/2)) A = 2cos(π)cos(((3π)/5)) A = 2cos(((2π)/5)) A = 2(2cos^2 ((π/5))−1) = 4cos^2 ((π/5))−2 B = cos(((4π)/5))+cos(((6π)/5)) B = −cos((π/5))−cos((π/5)) B = −2cos((π/5)) 4− 1+A+B = 0 = 1+4cos^2 ((π/5))−2−2cos((π/5)) ⇒ 4cos^2 ((π/5))−2cos((π/5))−1 = 0 cos((π/5)) = ((2±(√(4−4×4×(−1))))/8) cos((π/5)) = ((1±(√5))/4) but cos((π/5)) > 0 ⇒ cos((π/5)) = ((1+(√5))/4) 5− cos^3 x = (((e^(ix) +e^(−ix) )/2))^3 cos^3 x = (1/8)(e^(3ix) +3e^(ix) +3e^(−ix) +e^(−3ix) ) cos^3 x = (1/4)cos(3x)+(3/4)cosx ⇒ cos(3x) = 4cos^3 x−3cosx cos(((3π)/5)) = 4cos^3 ((π/5))−3cos((π/5)) cos(((3π)/5)) = 4(((1+(√5))/4))^3 −3(((1+(√5))/4)) cos(((3π)/5)) = (1/(16))(1+3(√5)+15+5(√5))−3(((1+(√5))/4)) cos(((3π)/5)) = ((1−(√5))/4)
$$\mathrm{1}− \\ $$$${w}\:=\:{e}^{\frac{\mathrm{2}{i}\pi}{{n}}} \\ $$$$\mathrm{Let}\:{z}_{{n}} \:=\:\mathrm{1}+{w}+{w}^{\mathrm{2}} +…{w}^{{n}−\mathrm{1}} \\ $$$${z}_{{n}} \:=\:\mathrm{1}×\frac{\mathrm{1}−\left({e}^{\frac{\mathrm{2}{i}\pi}{{n}}} \right)^{{n}} }{\mathrm{1}−{e}^{\frac{\mathrm{2}{i}\pi}{{n}}} }\:=\:\frac{\mathrm{1}−{e}^{\mathrm{2}{i}\pi} }{\mathrm{1}−{e}^{\frac{\mathrm{2}{i}\pi}{{n}}} }\:=\:\mathrm{0} \\ $$$$\mathrm{2}− \\ $$$${z}_{\mathrm{5}} \:=\:\mathrm{1}+{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{5}}} +{e}^{\frac{\mathrm{4}{i}\pi}{\mathrm{5}}} +{e}^{\frac{\mathrm{6}{i}\pi}{\mathrm{5}}} +{e}^{\frac{\mathrm{8}{i}\pi}{\mathrm{5}}} \:=\:\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{Re}\left({z}_{\mathrm{5}} \right)\:=\:\mathrm{0}\:=\:\mathrm{1}+\mathrm{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)+\mathrm{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)+\mathrm{cos}\left(\frac{\mathrm{6}\pi}{\mathrm{5}}\right)+\mathrm{cos}\left(\frac{\mathrm{8}\pi}{\mathrm{5}}\right) \\ $$$$\mathrm{3}− \\ $$$$\mathrm{A}\:=\:\mathrm{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)+\mathrm{cos}\left(\frac{\mathrm{8}\pi}{\mathrm{5}}\right) \\ $$$$\mathrm{A}\:=\:\mathrm{2cos}\left(\frac{\frac{\mathrm{2}\pi}{\mathrm{5}}+\frac{\mathrm{8}\pi}{\mathrm{5}}}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{\frac{\mathrm{2}\pi}{\mathrm{5}}−\frac{\mathrm{8}\pi}{\mathrm{5}}}{\mathrm{2}}\right) \\ $$$$\mathrm{A}\:=\:\mathrm{2cos}\left(\pi\right)\mathrm{cos}\left(\frac{\mathrm{3}\pi}{\mathrm{5}}\right) \\ $$$$\mathrm{A}\:=\:\mathrm{2cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right) \\ $$$$\mathrm{A}\:=\:\mathrm{2}\left(\mathrm{2cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{5}}\right)−\mathrm{1}\right)\:=\:\mathrm{4cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{5}}\right)−\mathrm{2} \\ $$$$\mathrm{B}\:=\:\mathrm{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)+\mathrm{cos}\left(\frac{\mathrm{6}\pi}{\mathrm{5}}\right) \\ $$$$\mathrm{B}\:=\:−\mathrm{cos}\left(\frac{\pi}{\mathrm{5}}\right)−\mathrm{cos}\left(\frac{\pi}{\mathrm{5}}\right) \\ $$$$\mathrm{B}\:=\:−\mathrm{2cos}\left(\frac{\pi}{\mathrm{5}}\right) \\ $$$$\mathrm{4}− \\ $$$$\mathrm{1}+\mathrm{A}+\mathrm{B}\:=\:\mathrm{0}\:=\:\mathrm{1}+\mathrm{4cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{5}}\right)−\mathrm{2}−\mathrm{2cos}\left(\frac{\pi}{\mathrm{5}}\right) \\ $$$$\Rightarrow\:\mathrm{4cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{5}}\right)−\mathrm{2cos}\left(\frac{\pi}{\mathrm{5}}\right)−\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{cos}\left(\frac{\pi}{\mathrm{5}}\right)\:=\:\frac{\mathrm{2}\pm\sqrt{\mathrm{4}−\mathrm{4}×\mathrm{4}×\left(−\mathrm{1}\right)}}{\mathrm{8}} \\ $$$$\mathrm{cos}\left(\frac{\pi}{\mathrm{5}}\right)\:=\:\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\mathrm{but}\:\mathrm{cos}\left(\frac{\pi}{\mathrm{5}}\right)\:>\:\mathrm{0}\:\Rightarrow\:\mathrm{cos}\left(\frac{\pi}{\mathrm{5}}\right)\:=\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\: \\ $$$$\mathrm{5}− \\ $$$$\mathrm{cos}^{\mathrm{3}} {x}\:=\:\left(\frac{{e}^{{ix}} +{e}^{−{ix}} }{\mathrm{2}}\right)^{\mathrm{3}} \\ $$$$\mathrm{cos}^{\mathrm{3}} {x}\:=\:\frac{\mathrm{1}}{\mathrm{8}}\left({e}^{\mathrm{3}{ix}} +\mathrm{3}{e}^{{ix}} +\mathrm{3}{e}^{−{ix}} +{e}^{−\mathrm{3}{ix}} \right) \\ $$$$\mathrm{cos}^{\mathrm{3}} {x}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\left(\mathrm{3}{x}\right)+\frac{\mathrm{3}}{\mathrm{4}}\mathrm{cos}{x} \\ $$$$\Rightarrow\:\mathrm{cos}\left(\mathrm{3}{x}\right)\:=\:\mathrm{4cos}^{\mathrm{3}} {x}−\mathrm{3cos}{x} \\ $$$$\mathrm{cos}\left(\frac{\mathrm{3}\pi}{\mathrm{5}}\right)\:=\:\mathrm{4cos}^{\mathrm{3}} \left(\frac{\pi}{\mathrm{5}}\right)−\mathrm{3cos}\left(\frac{\pi}{\mathrm{5}}\right) \\ $$$$\mathrm{cos}\left(\frac{\mathrm{3}\pi}{\mathrm{5}}\right)\:=\:\mathrm{4}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\right)^{\mathrm{3}} −\mathrm{3}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\right) \\ $$$$\mathrm{cos}\left(\frac{\mathrm{3}\pi}{\mathrm{5}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{1}+\mathrm{3}\sqrt{\mathrm{5}}+\mathrm{15}+\mathrm{5}\sqrt{\mathrm{5}}\right)−\mathrm{3}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\right) \\ $$$$\mathrm{cos}\left(\frac{\mathrm{3}\pi}{\mathrm{5}}\right)\:=\:\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$

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