Question

数学题，高一部分，在线等，要求质量和过程拜托了！钱随便开

1.求函数y=2cos(x+π/4)cos(x-π/4)+根号三×xin2x的值域和最小周期？
2.设函数f(x)=根号2/2×cos(2x+π/4)+sin²x，求函数f（x）的最小正周期?

Answer 1

1.
y=2cos(x+π/4)cos(x-π/4)+√3sin2x
=cos[(x+π/4)+(x-π/4)]+cos[(x+π/4)-(x-π/4)]+√3sin2x【这里用了和差化积公式，不懂请追问】
=cos2x+√3sin2x
=2sin(2x+π/6)
所以值域是[-2,2]
最小正周期是T=2π/2=π
2.
f(x)=(√2/2)×cos(2x+π/4)+sin²x
=(√2/2)×[cos2x*(√2/2)-sin2x*(√2/2)]+(1-cos2x)/2
=(1/2)*cos2x-(1/2)*sin2x+(1-cos2x)/2
=1/2-(1/2)*sin2x
=(1-sin2x)/2
所以最小正周期是T=2π/2=π

Answer 2

1、y=2(cosx·√2/2-sinx·√2/2)(cosx·√2/2+sinx·√2/2)+√3sin2x
=2(cos²x/2-sin²/2)+√3sin2x
=cos²x-sin²x+√3sin2x
=cos2x+√3sin2x
=2sin(2x+π/6).
值域为[-2,2]，周期为π.

2、f(x)=√2/2(cos2x·√2/2-sin2x·√2/2)+sin²x
=cos2x/2-sin2x/2+(1-cos2x)/2
=-1/2sin2x+1/2.
T=π.

Answer 3

te/2×cos(2x+π/4)+sin²x，求函数f（x）的最小正周期?