Question

已知函数f(x)=2sin²(x+π/4)-根号3cos2x,x∈[π/4,π/2].设x=α时f(x)取到最大值.

已知函数f(x)=2sin²(x+π/4)-根号3cos2x,x∈[π/4,π/2].设x=α时f(x)取到最大值.
(1)求f(x)的最大值及α的值;
(2)在△ABC中,角A,B,C所对的边分别为a,b,c,A=α-π/12,且sinBsinC=sin²A,求b-c的值.

Answer 1

（1）f(x)=2sin²(x+π/4)-√3cos2x
=(sinx+cosx)²-√3cos2x
=1+sin2x-√3cos2x
=2(cosπ/3sin2x-sinπ/3cos2x)+1
=2sin(2x-π/3)+1
∴f(x)的最大值=3
此时2x-π/3=2kπ+π/2
即α=kπ+5π/12
（2）sinBsinC=sin²A
A=5π/12-π/12=π/3
sinBsinC=sin²A
bc=a²=b²+c²-2bccosπ/3
∴b²-2bc+c²=0→(b-c)²=0
∴b-c=0