When t=π/ 12, the auxiliary angle is -π/6, which is the corresponding complex number.

p 1 = cosπ/6-isπ/6，

When t=π/4, the auxiliary angle is π/6, which corresponds to a complex number.

P2 = cosπ/6+isπ/6，

A(0，π)

The area of the graphics area swept by the vector AP.

It is Δ Δ AP1p 2 area+bow area.

√ΔAP1p 2 area =Δop1p 2 area.

∴ is Δ Δ op1p 2 area+bow area.

= area of sector OP 1P2

= 1×π/3=π/3

Not π/6