MQ69 --- Trigonometry

Evaluate Sin18°
Sol
w=72度
5w=360度
3w+2w=360度
3w=360度－2w
Sin3w=Sin(360度－2w)
Sin3w=－Sin2w
Ssinw－4Sin^3 w=－2SinwCosw
Sinw>0
3－4Sin^2 w=－2Cosw
3－4+Ccos^2 w=－2Cosw
4Cos^2 w+2Cosw－1=0
Cosw=(－2+√20)/8 (負不合)
Cosw=(√5－1)/4
Cos72度=(√5－1)/4
Sin18度=(√5－1)/4

cis5θ = (cisθ)^5
cis5θ = (cosθ)^5 + 5i(sinθ)(cosθ)^4 - 10(sin²θ)(cos³θ) - 10i(sin³θ)(cos²θ) + 5cosθ(sinθ)^4 + i(sinθ)^5
comparing real parts,

cos5θ = 5cosθ(sinθ)^4 - 10cos³θsin²θ + (cosθ)^5
cos5θ = cosθ[5(1-cos²θ)² - 10sin²θ(1-cos²θ)) + (cosθ)^4]
cos5θ = cosθ[16(sinθ)^4-20sin²θ+5]
put θ=18°
cos 90° = cos18°[16(cos18°)^4 - 20cos²18 +5]

since cos°18 > 0
so
16(cos18°)^4 - 20cos²18° +5 = 0
solving gives
cos²18° = (5±sqrt5)/8

note: 1 > cos²18° > cos²30° = 0.75
so cos²18° = (5-sqrt5)/8 rejected (~0.34549150)
so cos²18° = (5+sqrt5)/8 (~0.90450850)
sin²18° = 1 - (5+sqrt5)/8 = (3-sqrt5)/8
so sin18° = sqrt[(3-sqrt5)/8]