{topic -- trigonometric identities }
{ difficulty -- easy } -- 5 points:
full solutions next week (or when someone gets it).

week #1:
prove:

LHS ⇒ RHS:
Identities:
[1] cos(2A) = 2cos²(A) - 1 = 1 - 2sin²(A)
[2] sin(2A) = 2sin(A)cos(A)
[3] sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
[4] cos(A + B) = cos(A)cos(B) - sin(A)sin(B)

cos(x) - cos(x + 2Θ)
= cos(x) - (cos(x)cos(2Θ) - sin(x)sin(2Θ))    [4]
= cos(x) - cos(x)(1 - 2sin²(Θ)) + sin(x)(2sin(Θ)cos(Θ))    [1] [2]
= cos(x) - cos(x) + 2sin²(Θ)cos(x) + 2sin(Θ)sin(x)cos(Θ)
= 2sin²(Θ)cos(x) + 2sin(Θ)sin(x)cos(Θ)
= 2sin(Θ)(sin(Θ)cos(x) + sin(x)cos(Θ))
= 2sin(Θ)sin(x + Θ)

2xy/9 is the answer to this problem