点P在直径AB=1的半圆上移动,过P作圆的切线PT且PT=1,角PAB=a,问a为何值时四边形ABPT面积最大?我得3pi/16

Answer 1

不对
答案是 pi/8
S ABPT = S APT + S APB
=1/2*cosa*1*sin(pi/2-a)+1/2*cosa*sina
=(1+cos2a+sin2a)/4
=(根号2sin(2a+pi/4)+1)/4
2a+pi/4=pi/2时取最大值
a=pi/8