三角形△ABC中，B=60°，求sinA*sinA+sinC*sinC的取值范围

三角形△ABC中，B=60°，求sinA*sinA+sinC*sinC的取值范围

S
= (sinA)^2 + (sinC)^2
= (sin(2π/3-C) )^2 + (sinC)^2
= ( (√3/2)cosC + (1/2)sinC )^2 +(sinC)^2
= (3/4)(cosC)^2 + (√3/2)sinC.cosC + (5/4)(sinC)^2
=(3/8)cos2C + 3/8 +(√3/4)sin2C + (5/8) - (5/8)cos2C
= (√3/4)sin2C - (1/4)cos2C + 1
=(1/2)sin(2C -π/6) + 1
(1/2)sin(-π/6) + 1< S < (1/2)sin(π/2) + 1
3/4 < S < 3/2
3/4 < (sinA)^2 + (sinC)^2 <3/2

C=180-B-A=120-A
所以sin²A+sin²C
=sin²A+sin²(120-A)
=sin²A+(sin120cosA+cos120sinA)²
=sin²A+3/4*cos²A-√3/2*sinAcosA+1/4*sin²A
=5/4*sin²A+3/4*cos²A-√3/2*sinAcosA
=5/4*(1-cos2A)/2+3/4*(1+cos2A)/2-√3/2*sin2A/2
=1-1/2(1/2*cos2A+√3/2*sin2A)
=1-1/2(sin30cos2A+cos30sin2A)
=1-1/2*sin(30+2A)

030<30+2A<270
所以-1所以1/2≤1-1/2*sin(30+2A)<3/2

由楼上得：sinA+sinC=2sinB
sinA+sinC=2sin[(A+C)/2]cos[(A-C)/2](和差化积)=2sin[(π-B)/2]cos[(A-C)/2]=4sin(B/2)cos(B/2)
∴2sin(B/2)=cos[(A-C)/2]∈(0,1] B≤60° (等号成立时是等边△)

正余弦定理做