(1) in Rt△ABC, AB=acosθ, AC=asinθ, S 1 = 12ab? Ac = AC= 12a2sinθcosθ(3 θ (3 points)

Let the side length of a square be x, then BP = xsinθ, AP = xcosθ,

By BP+AP=AB, xsinθ+xcosθ = acosθ, so x = asinθ cosθ 1+sinθ cosθ.

So S2 = X2 = (asinθ cosθ1+sinθ cosθ) 2 (6 points).

(2)S 1S2= 12？ (1+sinθ cosθ) 2sinθ cosθ = (1+12sinθ) 2sinθ =1sin2θ+1,(8 points)

Let t=sin2θ, because 0 < θ < π 2,

So 0 < 2θ < π, then t = sin2θ ∈ (0, 1)( 10 point).

So s1S2 =1t+14t+1= g (t), g' (t) =? 1t2+ 14