Proving identity

Let O be the center, r be the radius, so the area of the semi circle is r^2 π/2.
As angle PAB = a rad, so the angle POB = 2a rad, and POB = (π - 2a) rad.
area of the sector POA = r^2 (π - 2a) / 2
area of the triangle POB = (r^2 sin 2a) / 2
As area of the sector POA + area of the triangle POB = r^2 (π/4), therefore
r^2 (π - 2a) / 2 + (r^2 sin 2a) / 2 = r^2 (π/4)
==> π - 2a + sin 2a = π/2 . . . (multiply both sides by 2/r^2 )
==> 2a - sin 2a = π/2

Sorry, I think you are asking to prove that 2a - sin 2a = pi/2

ps. area of triangle = (1/2) a b sin C

2012-10-25 16:54:36 補充：
May be the question is angle PBA = a rad, then angle POA = 2a rad, then
r^2 (2a) / 2 + r^2 sin (π - 2a) / 2 = r^2 (π/4)
==> 2a + sin 2a = π/2

ps. sin (π - 2a) = sin 2a