题目内容：

有关三角函数的最值问题
在△ABC中,A、B、C分别为三角形内角,a、b、c为其所对边,已知2√2*(sin^2A-sin^2C)=(a-b)sinB,△ABC外接圆半径为√2.
1、求周长范围 2、求a²+b²的范围

优质解答

(1)根据正弦定理 ∵△ABC外接圆半径为√2.
∴a/sinA=b/sinB=c/sinC=2√2.
又2√2*(sin^2A-sin^2C)=(a-b)sinB
即2√2*(a²/8-c²/8)=(a-b)√2b/4
即a²-c²=ab-b²
b²+a²-c²=ab
cosC=1/2
∠C=60°
a+b+c
=2√2(sinA+sinB+sinC)
=2√2(sinA+sin(A+60°)+√3/2)
=2√2(sinA+sinA/2+√3cosA/2+√3/2)
=2√2(3*sinA/2+√3cosA/2+√3/2)
=2√2[√3(√3sinA/2+cosA/2)+√3/2]
=2√2(√3*sin(A+30°)+√3/2)
∵A属于(0,2π/3)
A+π/6属于(π/6,5π/6)
sin(A+π/6)属于(1/2,1]
∴2√2(√3*sin(A+30°)+√3/2)属于(2√6,3√6].
(2)a²+b²
=2√2((sinA)²+sin(A+π/6)²)
=2√2((sinA)²+(sinA/2+√3cosA/2)²)
=2√2(5sinA/4+√3sinA*cosA/2+3(1-(sinA)²)/4)
=2√2((sinA)²/2+√3sin2A/4)
=2√2(1-cos2A+√3sin2A)/4
=√2(1+2sin(2A-π/6))/2
可得sin(2A-π/6)属于(-1/2,1]
a²+b²属于(0,3√2/2]