f(x)=cos²(x-π／6)-sin²x ；1、求f(π/12)的值；2、若對于任意的x∈[0,π/2],都有f(x)≤c,求實數(shù)c的取值范圍
1、f(π/12)=cos²(π/12-π/6)-sin²(π/12)=cos²(-π/12)-sin²(π/12)=cos²(π/12)-sin²(π/12)
=cos(π/6)=(√3)/2.
2、f(x)=[(√3/2)cosx+(1/2)sinx]²-sin²x=(3/4)cos²x+(√3/2)sinxcosx+(1/4)sin²x-sin²x
=(3/4)(cos²x-sin²x)+(√3/4)sin2x=(3/4)cos2x+(√3/4)sin2x=(√3/4)[(√3)cos2x+sin2x]
=(√3/2)[(√3/2)cos2x+(1/2)sin2x]=(√3/2)[cos2xcos(π/6)+sin2xsin(π/6)]
=(√3/2)cos(2x-π/6)≦(√3)/2；故c≧(√3)/2,這就是c的取值范圍.