(a) Explain why AP = 2cosθ.
(b) Find a similar expression for BP in terms of θ.
(c) Explain why the coordinates of pt P is (cos2θ, sin2θ).
(d) Prove that triangle APQ is similar to triangle ABP.
(e) Hence deduce that
(i) sin2θ = 2sin θcosθ
(ii) cos2θ = 2cos^2 θ - 1
(iii) tan2θ = (2tanθ)/(1-tan^ 2 θ)
Could someone please help me on this problem?? I've worked out parts a and b but added them in just in case it is required for the other parts.