在区间（-π/6,π/6）内,求（1+根号3）cosx+（1-根号3）sinx的最大值及取得这个最大值的x
问题描述：
在区间（-π/6,π/6）内,求（1+根号3）cosx+（1-根号3）sinx的最大值及取得这个最大值的x
答
(1+√3)cosx +(1-√3)sinx
= cosx + sinx + √3(cosx-sinx)
= √2(sinπ/4cosx+cosπ/4sinx) + √6(cosxcosπ/4-sinxsinπ/4)
= √2sin(x+π/4) + √6cos(x+π/4)
= √8{1/2sin(x+π/4)+√3/2cos(x+π/4)}
= √8{sin(x+π/4)cosπ/3+cos(x+π/4)sinπ/3}
= √8sin(x+π/4+π/3}
= √8sin(x+7π/12)
x∈（-π/6,π/6）
x+7π/12∈（5π/12,3π/4)
当x+7π/12=π/2时,(1+√3)cosx +(1-√3)sinx取最大值√8
此时x = -π/12