已知函数fx=cos^2（x-pai／6）-sin^2（x）

(1)
f(x)= [cos(x-π/6)]^2 - (sinx)^2
f(π/12) = (cos(π/12))^2 - (sin(π/12))^2
=cos(π/6)
=√3/2
(2)
f(x) =[cos(x-π/6)]^2 - (sinx)^2
= [(√3/2)cosx + (1/2)sinx]^2 - (sinx)^2
= (1/4)( 3(cosx)^2 + 2√3sinxcosx -3(sinx)^2 )
= (1/4) ( 3cos2x + √3sin2x )
= (√3/2)((√3/2)cos2x+ (1/2)sin2x)
= (√3/2)sin(2x+π/3)
max f(x) = √3/2

第一题应该是pi/12吧？