已知函数f（x）=cos^2(x-π/6）-sin^2x

（|） 求f（π/12）的值
（||）求函数f(X)在{0，π/2}上的最大值

求求你
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f(x)=cos²(x-π/6)-sin²x

f(π/12)
=cos²(π/12-π/6)-sin²(π/12)
=cos²(-π/告汪12)-sin²(π/12)
=cos²(π/12)-sin²(π/袜圆仔12) 两倍角公式
=cos[2(π/12)]
=cosπ/6
=√3/2

f(x)=cos^2(x-π/6)-sin^2x
=（2cos^2(x-π/6)-1+1）/2+(1-2sin^2x-1)/2
=[cos(2x-π/3)+cos2x]/2
=(cos2xcosπ/3+sin2xsinπ/3+cos2x)/2
=根号3（2分之根号3倍的cos2x+2分之sin2x)/2
=根号3/2sin(2x+π/3)
0<=x<=Pai/2,则有Pai/3<=2x+Pai/3<=4Pai/3
故有-根号3/2<=sin(2x+Pai/3)<=1
故最大值是:根号腔携3/2*1=根号3/2.

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