a-maths!!

(a) prove (2cos2θ+2cosθ+1)sinθ/2=sin(5θ/2)。
(2cos2θ+2cosθ+1)sinθ/2 = 2cos2θsinθ/2+2cosθsinθ/2+sinθ/2
= sin(5θ/2)-sin(3θ/2)+sin(3θ/2)-sinθ/2+sinθ/2
= sin(5θ/2)
Because cos(2θ)=2cos²θ - 1
(2cos2θ+2cosθ+1)sinθ/2= (2(2cos²θ - 1)+ 2cosθ+1) sinθ/2
=(4cos²θ+2cosθ-1)sinθ/2=sin(5θ/2)

(b)(i) 4x²+2x-1=0
Let x = cosθ , 4cos²θ+2cosθ-1= 0
By the result of (a),
sin(5θ/2) = 0,
because of two roots, 5θ/2 = 2π, or 5θ/2 = 4π
θ = 2π/5, θ = 4π/5

(ii) by result of (i), 4cos²(2π/5)+2cos(2π/5)-1= 0 ------------(1)
4cos²(4π/5)+2cos(4π/5)-1= 0 ------------(2)
Eq(1) + Eq(2): 4cos²(2π/5)+4cos²(4π/5)+2cos(4π/5)+2cos(2π/5) = 2
Because cos(2θ)=2cos²θ - 1
4cos²(2π/5)+4cos²(4π/5)+4cos²(2π/5)+2cos(2π/5)-2 = 2
4cos²(2π/5)+4cos²(4π/5)+(4cos²(2π/5)+2cos(2π/5)-1)-1 = 2
By previous result, 4cos²(2π/5)+2cos(2π/5)-1 = 0
4cos²(2π/5)+4cos²(4π/5)-1 = 2
cos²(2π/5)+cos²(4π/5) = 3/4