what is the answer to (π/π/2)cos(π/π/2Ψ)cos^2(Ψ)-2sin(π/π/2Ψ)cos(Ψ)sin(Ψ)

If  Π/Π/2  =  (Π/Π)/2  = 1/2

If  Π/Π/2  =  Π/(Π/2)   = 2

I'm going to assume the second answer.

So, I'm going to make Π/Π/2Ψ  =  Π/(Π/2Ψ)  = 2Ψ.

Problem:  2cos(2Ψ)cos²(Ψ) - 2sin(2Ψ)cos(Ψ)(sin(Ψ) 

             =  2[ cos(2Ψ)cos²(Ψ) - sin(2Ψ)cos(Ψ)(sin(Ψ) ]

Since sin(2X) = 2sin(X)cos(X):

             =  2[ cos(2Ψ)cos²(Ψ) - sin(Ψ)cos(Ψ)cos(Ψ)(sin(Ψ) ]

             =  2[ cos(2Ψ)cos²(Ψ) - sin²(Ψ)cos²(Ψ) ]

             =  2cos²(Ψ)[ cos(2Ψ)  - sin²(Ψ) ]

             =   2cos²(Ψ)[ 1 - 2sin²(Ψ) - sin²(Ψ) ]

             =  2cos²(Ψ)[ 1 - 3sin²(Ψ) ] 

From here ... ?