1.求函数y=7-4sinxcosx+4cosx平方-4cosx四次方,的最大值和最小值2.已知函数f(x)=sin方wx+根号3sinwx(wx+π/2)的最小正周期为π.求函数f(x)在区间〔0,2π/3〕上的取值范围3.已知tan(π/4+α)=2,tanβ=1/2.(1)求tanα的值（2）求sin(α+β)-2sinαcosβ/2sinαsinβ+cos(α+β)

(1)y=7-4sinxcosx+4cosx^2-4cosx^4
=7-4sinxcosx+4（1-cosx^2）cosx^2
=7-4sinxcosx+4sinx^2*cosx^2
=7-4sinxcosx+4（sinx*cosx）^2
=4（sinx*cosx）^2-4sinxcosx+1+6
=（2sinx*cosx-1）^2+6
所以：6

1.
y=7-4sinxcosx+4cos^2x-4cos^4x
=7-2*(2sinxcosx)+4cos^2x(1-cos^2x)
=7-2sin2x+4cos^2x sin^2x
=7-2sin2x+(2sinxcosx)^2
=6+1-2sin2x+sin^2(2x)
=6+(1-sin2x)^2
-1 ≤ sin2x ≤ 1
0 ≤ 1-sin2x ≤ 2
6 ≤ 6+(1-sin2x)^2 ≤ 10
最大值10，最小值6
2.
f(x)=sin^2(wx)+根号3sinwx sin(wx+π/2)
=1/2[1-cos(2wx)] + 根号3sin(wx) cos[π/2-(wx+π/2)]
=1/2-1/2 cos(2wx) + 根号3sin(wx) cos[-wx+]
=1/2-1/2 cos(2wx) + 根号3sin(wx) cos(wx)
=1/2-1/2 cos(2wx) + 根号3 /2 sin(2wx)
=根号3 /2 sin(2wx) -1/2 cos(2wx) +1/2
= sin(2wx)cosπ/6 -cos(2wx) sinπ/6 +1/2
= sin(2wx-π/6) +1/2
∵最小正周期为π
∴2π/(2w)=π，w=1
∴f(x) = sin(2x-π/6) +1/2
∵x∈〔0,2π/3〕
∴2x-π/6 ∈〔-π/6，7π/6〕
2x-π/6 = -π/6，或7π/6时，f(x）取最小值：-1/2+1/2=0
2x-π/6 = π/2时，f(x）取最大值：1+1/2=3/2
3.
tan(π/4+α)=2
(tanπ/4+tanα) / (1-tanπ/4 tanα) = 2
(1+tanα) / (1- tanα) = 2
1+tanα = 2- 2tanα
tanα=1/3
tanβ=1/2
[ sin(α+β)-2sinαcosβ ] / [ 2sinαsinβ+cos(α+β)]
=[ sinαcosβ+cosαsinβ-2sinαcosβ ] / [ 2sinαsinβ+cosαcosβ-sinαsinβ]
=[ cosαsinβ-sinαcosβ ] / [ sinαsinβ+cosαcosβ]
【分子分母同除以cosαcosβ：】
=(tanβ-tanα) / (tanαtanβ+1）
=(1/2-1/3)/(1/2*1/3+1)
=(1/6) / (1/7)
=1/7

1)y=7-4sinxcosx+4cosx^2-4cosx^4
=7-4sinxcosx+4（1-cosx^2）cosx^2
=7-4sinxcosx+4sinx^2*cosx^2
=7-4sinxcosx+4（sinx*cosx）^2
=4（sinx*cosx）^2-4sinxcosx+1+6
=（2sinx*cosx-1）^2+6
=(sin2x-1)^2+6 因为（sin2x-1）^2∈[0,4],所以这个函数值域为[6,10]
2)这个估计你打的时候出了问题,第二个可能是sinwxsin(wx+π/2),而不是sinwx(wx+π/2),改后
f（x）=（1-cos2wx）/2+根号3 sinwx coswx=0.5-cos2wx/2+根3sin2wx/2
=cosπ/6sin2wx-sinπ/6cos2wx+0.5
=sin(2wx-π/6)+0.5,因为它的周期是π,所以w=1,
所以f（x）=sin(2x-π/6)+0.5
x∈〔0,2π/3〕 则 2x∈〔0,4π/3〕 则 2x-π/6∈〔-π/6,7π/6〕
则sin(2x-π/6)∈[-0.5,1] 则f(x)∈[0,1.5] 这个是第二题的答案
3）tan(π/4+α)=（tanπ/4+tanα）/(1-tanπ/4tanα)=（1+tanα）/(1-tanα)=2
所以 tanα+1=2-2tanα,解得tanα=1/3 这个是第一小题答案
sin(α+β)-2sinαcosβ/2sinαsinβ+cos(α+β)
=(sinαcosβ+cosαsinβ-2sinαcosβ)/(2sinαsinβ+cosαcosβ-sinαsinβ)
这个式子分子分母同除cosαcosβ得
原式=（tanα+tanβ-2tanα）/(2tanαtanβ+1-tanαtanβ）
=（tanβ-tanα）/(tanαtanβ+1）
因为tanα1/3,tanβ=1/2 所以原式=(1/6)/(7/6)=1/7
好累,纯粹手打,