Question

锐角三角形ABC已知∠A.B.C的对边分别为abc,且2sin^2A-cos2A=2

锐角三角形ABC已知∠A.B.C的对边分别为abc,且2sin^2A-cos2A=2,则函数y=2sin^2B+sin(2B+π/6)的值域为多少?还要过程，谢谢。

Answer 1

∵2sin^2A-cos2A=2
∴(1-cos2A)-cos2A=2
∴cos2A=-1/2
∵A是锐角
∴2A=2π/3,A=π/3
∴C=π-A-B=2π/3-B
∵C是锐角 ∴2π/3-B<π/2
∴π/6<B<π/2
y=2sin^2B+sin(2B+π/6)
=1-cos2B+sin2Bcosπ/6+cos2Bsinπ/6
=1+√3/2sin2B-1/2cos2B
=1+sin(2B-π/6)
∵π/6<B<π/2
∴π/3<2B<π
∴π/6<2B-π/6<5π/6
∴1/2<sin(2B-π/6)≤1
∴3/2<1+sin(2B-π/6)≤2
∴所求函数值域为(3/2,2]