Question Number 104676 by qwertyu last updated on 23/Jul/20

Commented byDwaipayan Shikari last updated on 23/Jul/20

cos36°+cos108°=cos36°−cos72°  =(((√5)+1)/4)−(((√5)−1)/4)=(1/2)

Commented byOlafThorendsen last updated on 23/Jul/20

36° and 72° are not  remarkable angles sir.  You should prove that  cos36° = (((√5)+1)/4) and cos72° = (((√5)−1)/4)

Commented byDwaipayan Shikari last updated on 23/Jul/20

θ=18°  sin5θ=1  2θ+3θ=(π/2)  sin2θ=cos3θ  2sinθcosθ=4cos^3 θ−3cosθ  2sinθ=4cos^2 θ−3  2sinθ=1−4sin^2 θ  4sin^2 θ+2sinθ−1=0  sinθ=((−2+(√(4+16)))/(2.4))=(((√5)−1)/4)  sin18°=(((√5)−1)/4)=cos72°  cos72°=cos^2 36°−sin^2 36°=2cos^2 36°−1  cos36°=(((√5)+1)/4)

Commented byOlafThorendsen last updated on 23/Jul/20

Waouh. Great method sir!

Answered by bemath last updated on 23/Jul/20

cos a + cos  b = 2cos (((a+b)/2)) cos(((a−b)/2))  cos 36°+cos 108°=   2cos 72° cos 36° = 2sin 18°cos 36°  = 2sin 18° (1−2sin ^2 18°)  = 2((((√5)−1)/4))(1−2((((√5)−1)/4))^2 )  = (1/2)((√5)−1)(1−2(((6−2(√5))/(16))))  =(1/2)((√5)−1)(1−(((3−(√5))/4)))  =(1/8)((√5)−1)((√5)+1)  = (1/8)×4= (1/2) ★

Commented byOlafThorendsen last updated on 23/Jul/20

you should explain why  sin18° is equal to (((√5)−1)/4) sir.

Commented by6478 last updated on 23/Jul/20

cos a + cos  b = 2cos (((a+b)/2)) cos(((a−b)/2))  cos 36°+cos 108°=   2cos 72° cos 36° = 2sin 18°cos 36°  = 2sin 18° (1−2sin ^2 18°)  = 2((((√5)−1)/4))(1−2((((√5)−1)/4))^2 )  = (1/2)((√5)−1)(1−2(((6−2(√5))/(16))))  =(1/2)((√5)−1)(1−(((3−(√5))/4)))  =(1/8)((√5)−1)((√5)+1)  = (1/8)×4= (1/2) ★

Commented bybemath last updated on 23/Jul/20

Answered by OlafThorendsen last updated on 23/Jul/20

a = cos36°  b = cos108°  a+b = cos36°+cos108° = cos(π/5)+cos((3π)/5)  a+b = cos(π/5)−cos((2π)/5)  a+b = cos(π/5)−(2cos^2 (π/5)−1)  a+b = a−2a^2 +1 (1)  a+b = cos(π/5)+cos((3π)/5)  a+b = 2cos((((3π)/5)+(π/5))/2)cos((((3π)/5)−(π/5))/2)  a+b = 2cos((2π)/5)cos(π/5)  a+b = 2(2a^2 −1)a (2)  (1) and (2) :  2(2a^2 −1)a = a−2a^2 +1  4a^3 +2a^2 −3a−1 = 0  (a+1)(4a^2 −2a−1) = 0  a = cos(π/5) = −1 : impossible  4a^2 −2a−1 = 0  a = ((1±(√5))/4)  a = cos(π/5) = ((1−(√5))/4)<0 : impossible  Finally a = cos(π/5) = ((1+(√5))/4)  Then b = cos((3π)/5) = −cos((2π)/5) = −(2a^2 −1)  b = 1−2(((1+(√5))/4))^2  = ((1−(√5))/4)  a+b = ((1+(√5))/4)+((1−(√5))/4) = (1/2)

Answered by 1549442205PVT last updated on 23/Jul/20

36°+54°=90°⇒sin36°=cos54°  ⇒2sin18°cos18°=cos3.18°=4cos^3 18°−3cos18°  ⇒2sin18°=4cos^2 18°−3  ⇒2sin18°=4(1−sin^2 18°)−3  ⇔4sin^2 18°+2sin18°−1=0  ⇔(2sin18°+(1/2))^2 −(5/4)=0⇔2sin18°+(1/2)=((√5)/2)  ⇒sin18°=(((√5)−1)/4)⇒cos72°=sin18°=(((√5)−1)/4)  cos 72°=2co^2 36°−1=(((√5)−1)/4)  ⇒cos^2 36°=((3+(√5))/8)=((6+2(√5))/(16))⇒cos36°=(((√5)+1)/4)  cos108°=cos(90°+18°)=−sin18°=((1−(√5))/4)  Therefore,cos36°+cos108°=  ((1+(√5))/4)+((1−(√5))/4)=(1/2)