Question

已知函数f（x）=4cossin（x+π/6）-1

1）求f（x）的最小正周期
2）求f（x）在区间上【-π/6,π/4】的最大值和最小值

Answer 1

f(x)=4cosx(sinxcosπ/6+cosxsinπ/6)+1
=2√3sinxcosx+2cos²x-1+2
=√虚枝搭3sin2x+cos2x+2
=2(sin2x*√3/2+cos2x*1/2)+2
=2(sin2xcosπ/6+cos2xsinπ/搭卖6)+2
=2sin(2x+π/差拿6)+2
所以T=2π/2=π

-π/6<=2x+π/6<=2π/3
所以-1/2<=sin(2x+π/6)<=1
所以最大值=2×1+2=4
最小值=2×(-1/2)+2=1

Answer 2

(x)=4cosx(sinxcosπ/6+cosxsinπ/6)+1
=2√3sinxcosx+2cos²旅运x-1+2
=√3sin2x+cos2x+2
=2(sin2x*√3/拆慧梁2+cos2x*1/2)+2
=2(sin2xcosπ/6+cos2xsinπ/6)+2
=2sin(2x+π/6)+2
T=2π/2=π

-π/6<=2x+π/6<=2π/3
-1/碧没2<=sin(2x+π/6)<=1
2×1+2=4
2×(-1/2)+2=1

Answer 3

http://dayidata2.jiandan100.cn/M00/03/06/dU_r7lA98N_l8SqqAAC8zW6aDsU731.jpg