a-maths~~~~~~~

a)L.H.S.= (2cos2θ+2cosθ+1)sin(θ/2)
=2cos2θsin(θ/2)+2cosθsin(θ/2)+sin(θ/2)
=sin(2θ+θ/2)-sin(2θ-θ/2)+sin(θ+θ/2)-sin(θ-θ/2)+sin(θ/2)
=sin(5θ/2)-sin(3θ/2)+sin(3θ/2)-sin(θ/2)+sin(θ/2)
=sin(5θ/2)
=R.H.S.
cos2θ=2cos2 θ – 1, 所以
[2(2cos2 θ – 1)+2cosθ+1]sin(θ/2)=sin(5θ/2)
(4cos²θ+2cosθ-1)sin(θ/2)=sin(5θ/2)
bi)(4cos²θ+2cosθ-1)sinθ/2=sin(5θ/2)
4cos²θ+2cosθ-1= sin(5θ/2)/ sin(θ/2)
代x = cosθ,
sin(5θ/2)/ sin(θ/2)=0
sin(5θ/2) = 0
5θ/2 = π, 2π
θ= 2π/5, 4π/5
所以cos2π/5及cos4π/5為方程4x²+2x-1=0的根。
ii)cos²(2π/5)+cos²(4π/5)
=[cos(2π/5)+cos(4π/5)]2-2[cos(2π/5)cos²(4π/5)]
=(-2/4)2-2(-1/4)
=3/4

(2cos2θ+2cosθ+1)sinθ/2=sin5θ/2

先把sinθ/2 乘進去
變成 2cos2θsinθ/2+2cosθsinθ/2+sinθ/2 = sin5θ/2

(2cos2θsinθ/2)+(2cosθsinθ/2)+(sinθ/2 - sin5θ/2) =0
跟據公式
變成
(sin5θ/2 - sin3θ/2) + (sin3θ/2 - sinθ/2) + (2cos3θ/2sin2θ/2)=0
中間兩個sin3θ/2 cancell 左, 餘下
sin5θ/2 - sinθ/2 + (2cos3θ/2sin2θ/2)=0
-(sinθ/2 - sin5θ/2) + (2cos3θ/2sin2θ/2) =0
-(2cos3θ/2sin2θ/2) + (2cos3θ/2sin2θ/2) =0
0=0