Question

已知函数f(x)=(sinx+cosx)^2-2cos^2x(1)求f(x)最小正周期。(2)比较f(-π/12)和f(π/6)大小

已知函数f(x)=(sinx+cosx)^2-2cos^2x(1)求f(x)最小正周期。(2)比较f(-π/12)和f(π/6)大小

Answer 1

f(x)=(sinx+cosx)^2-2cos^2x=1--2cos^2x+2sinxcosx=sin2x-cosx=√2sin(2x-π/4)
f(x)最小正周期T=π
f(-π/12)=-√2sin(5π/12)<0
f(π/6)=)=√2sin(π/12)>0
所以f(-π/12)<(π/6)

Answer 2

f(x)=(sinx＋cosx)²－2cos2x=1＋2sinxcosx－cos2x=1＋sin2x－cos2x=1＋√2sin(2x－π/4)，周期为π，第二问就代入计算就可以了。

Answer 3

(sinx+cosx)^2=1+sin2x 周期为T=π
2cos^2x=1+cos4x周期为T=π/2
所以f(x)的最小正周期为π

f(-π/12)=sin-π/6-cosπ/3=-1<0
f(π/6)=sinπ/3-cos2π/3=sinπ/3+cosπ/3>0
所以f(-π/12)<f(π/6)

Answer 4

DSAD