a-maths!!!!!!

a)SR = a sinθ
OQ = a sinθ/ tan 2α
OR = a cosθ
QR = PS = a cosθ- a sinθ/ tan 2α
= ( a sin 2αcosθ- a sinθcos 2α) / sin 2α
= a sin ( 2α- θ) / sin 2α
b)K = SR x PS
= a sinθ{ a sin ( 2α- θ) / sin 2α}
= a2 sinθsin ( 2α- θ) / sin 2α
= (a2 / 2 sin2α){ 2 sinθsin ( 2α- θ) }
= ( a2 / 2 sin 2α){ cos ( 2θ- 2α) – cos 2α}
= ( a2 / 2 sin 2α){ cos ( 2α- 2θ) – cos 2α}
K的極值(當cos ( 2α- 2θ) = 1):
K = ( a2 / 2 sin 2α)( 1 – cos 2α)
= ( a2 / 4 sinαcosα)( 2 sin2α)
= a2 tanα/ 2

(a) In triangle SOR, sin α = SR/a
SR = a sin α

∠OPQ = 180 - 90 -2α = 90-2α
∠OPS = 90+90-2α = 180-2α
In triangle OPS, a/sin (180-2α) = PS/sin(2α-θ)
a/sin2α=PS/sin(2α-θ)
Hence PS = asin(2α-θ)/sin2α