求函数f(x)=sin(x+π/6)^2+cos(x+π/3)^2的最大值和最小值请写的详细点，谢谢！！！！！
热心网友
原式=sin(x+π/6)^2+cos[(x+π/6)+π/6]^2再求解
热心网友
原式＝[1-cos(2x+π/3)]/2+[1+cos(2x+2π/3)]/2=1+[cos(2x+2π/3)-cos(2x+π/3)]/2=1+(cos2xcos2π/3-sin2xsin2π/3-cos2xcosπ/3+sin2xsinπ/3)/2=1+(-0.5cos2x-0.5cos2x)=1-cos2x所以当x=2kπ+π/2(k包含与z),最大值２ x=2kπ,最小植０
热心网友
f(x)=sin(x + π/6)^2 + cos(x + π/3)^2=sin(x + π/6)^2 + 1 - sin(x + π/3)^2=1 + [sinx·cos(π/6)+cosx·sin(π/6)]^2-[sinx·cos(π/3)+cosx·sin(π/3)]^2=1 + [(√3/2)sinx + (1/2)cosx + (1/2)sinx + (√3/2)cosx]·[(√3/2)sinx + (1/2)cosx - (1/2)sinx - (√3/2)cosx]=1 + [(√3+1)(√3-1)/4](sinx + cosx)(sinx - cosx)=1 + (1/2)(six^2 - 1 + sinx^2)=1/2 + sinx^2当x = 2kπ(k = 0，±1，±2，±3……)时，f(x)有最小值f(x)min = 1/2当x = 2kπ ± π/2(k = 0，±1，±2，±3……)时，f(x)有最大值f(x)max = 1/2 + 1 = 3/2。