已知函数f（x）=cos^2(x-π/6）-sin^2x

题目

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

答案

f(x)=cos²(x-π/6)-sin²x
f(π/12)
=cos²(π/12-π/6)-sin²(π/12)
=cos²(-π/12)-sin²(π/12)
=cos²(π/12)-sin²(π/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