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Question Number 125686 by mathocean1 last updated on 12/Dec/20

show that cos((4π)/5)+cos((2π)/5)+1=0

showthatcos4π5+cos2π5+1=0

Commented by bramlexs22 last updated on 13/Dec/20

cos ((4π)/5) = cos (π−(π/5))=−cos (π/5) cos ((4π)/5) = −cos 36° cos ((2π)/5) = cos 72° (•) cos ((4π)/5)+cos ((2π)/5)= −cos 36°+cos 72° = −2sin 54° sin 18° = −2cos 36° sin 18° = −2(1−2sin^2 18°)sin 18° = −2(1−2((((√5)−1)/4))^2 )((((√5)−1)/4)) =−2(1−2(((3−(√5))/8)))((((√5)−1)/4)) =−2(1−(((3−(√5))/4)))((((√5)−1)/4)) =−2((((√5)+1)/4))((((√5)−1)/4))= −(1/2) cos ((4π)/5)+cos ((2π)/5)+1=−(1/2)+1=(1/2)

cos4π5=cos(π−π5)=−cosπ5cos4π5=−cos36°cos2π5=cos72°(∙)cos4π5+cos2π5=−cos36°+cos72°=−2sin54°sin18°=−2cos36°sin18°=−2(1−2sin218°)sin18°=−2(1−2(5−14)2)(5−14)=−2(1−2(3−58))(5−14)=−2(1−(3−54))(5−14)=−2(5+14)(5−14)=−12cos4π5+cos2π5+1=−12+1=12

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