Question

已知函数f(x)=sin(2x+π/6)-cos(2x+π/3)+2(cosx)^2 ⑴求f(π/12)的值 ⑵求f(x)的最大值及相应x的值

Answer 1

f(π/12)=根号3+1

追问

有木有详细过程？

Answer 2

化简f(x),得
f(x)=sin2xcosπ/6+cos2xsinπ/6-cos2xcosπ/3+sin2xsinπ/3+2cos²x-1+1
=√3/2sin2x+1/2cos2x-1/2cos2x+√3/2sin2x+cos2x+1
=√3sin2x+cos2x+1
=2(√3/2 sin2x+1/2cos2x)+1
=2(sinπ/3sin2x+cosπ/3cos2x)+1
=2cos(2x-π/3)+1
∴f(π/12)=2cos(π/6-π/3)+1=2cosπ/6+1=√3+1
∵-1≤cos(2x-π/3)≤1
当2x-π/3=2kπ时，cos(2x-π/3)=1
此时，2x=2kπ+π/3 x=kπ+π/6 (k是整数）
∴当x=kπ+π/6时，f(x)有最大值3