问题补充：
已知三角形ABC为锐角三角形,若C=π/3,求函数y=cos²A+sin²B的值域
答案：
C=60 A+B=120
cos^2A=1-sin^2A
y=sin^2B-sin^2A+1
=(sinB+sinA)(sinB-sinA)+1
=2sin(A+B)/2*cos(A-B)/2*2cos(A+B)/2*sin(A-B)/2+1
=sin(A+B)sin(A-B)+1
=√3/2*sin(2A-120)+1 0
======以下答案可供参考======
供参考答案1:
解:因为三角形ABC为锐角三角形,所以当A=B=派/3时,函数y=cos^2A＋sin^2B最小,且最小为1,又因为cos^2A＋sin^2B供参考答案2:
已知三角形ABC为锐角三角形,若C=π/3,求函数y=cos²A+sin²B的值域
解析：∵⊿ABC为锐角三角形，C=π/3
∴A+B=2π/3==>B=2π/3-A
设f(x)=(cos(x))^2+(sin(2π/3-x))^2
=1+1/2cos(2x)+1/2cos(π/3-2x)
=1+1/2cos(2x)+1/2(1/2cos(2x)+√3/2sin(2x))
=1+3/4cos(2x)+√3/4sin(2x))
=1+√3/2sin(2x+π/3)
∴f(x)= 1+√3/2sin(2x+π/3) (x∈(0,2π/3)
∴函数y=cos²A+sin²B的值域为(1-√3/2,1+√3/2]
供参考答案3:
∵⊿ABC为锐角三角形，C=π/3
∴A+B=2π/3==>B=2π/3-A
A、B∈（0，π/2）设f(x)=(cos(x))^2+(sin(2π/3-x))^2
=1+1/2cos(2x)+1/2cos(π/3-2x)
=1+1/2cos(2x)+1/2(1/2cos(2x)+√3/2sin(2x))
=1+3/4cos(2x)+√3/4sin(2x))
=1+√3/2sin(2x+π/3)
∴f(x)= 1+√3/2sin(2x+π/3) (x∈(0,π/2)
2x+π/3 ∈(π/3,4π/3)
∴y∈(7/4,√3/2 +1)
