作业帮三角形ABC,角BAC=45°,高是AD,BD=2,DC=3,求面积2.向量a=(2cosx/2,tan(x/2+π/4),向量b=(根号2sin(x/2+π/4),tan(x/2-π/4))设f（x）=a.求f（x）在[0，π]上单调区间3.三角形ABC,A+C=2B,1/cosA+1/cosC=负（根号2）/cosB,A>C,求A.B.
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三角形ABC,角BAC=45°,高是AD,BD=2,DC=3,求面积2.向量a=(2cosx/2,tan(x/2+π/4),向量b=(根号2sin(x/2+π/4),tan(x/2-π/4))设f（x）=a.求f（x）在[0，π]上单调区间3.三角形ABC,A+C=2B,1/cosA+1/cosC=负（根号2）/cosB,A>C,求A.B.
三角形ABC,角BAC=45°,高是AD,BD=2,DC=3,求面积
2.向量a=(2cosx/2,tan(x/2+π/4),向量b=(根号2sin(x/2+π/4),tan(x/2-π/4))
设f（x）=a.求f（x）在[0，π]上单调区间
3.三角形ABC,A+C=2B,1/cosA+1/cosC=负（根号2）/cosB,A>C,求A.B.C.
三角形ABC,角BAC=45°,高是AD,BD=2,DC=3,求面积2.向量a=(2cosx/2,tan(x/2+π/4),向量b=(根号2sin(x/2+π/4),tan(x/2-π/4))设f（x）=a.求f（x）在[0，π]上单调区间3.三角形ABC,A+C=2B,1/cosA+1/cosC=负（根号2）/cosB,A>C,求A.B.
BD = 2 ,DC=3
=> BC = 5
let AD = h,∠ABC = θ
then ∠BCA = 135°-θ
tanθ = h/2 (1)
tan(135°-θ) = h/3
(tan135° + tanθ)/( 1- tan135°tanθ) = h/3
(tanθ-1)/(1+tanθ) = h/3
3tanθ - 3 = h+ htanθ
tanθ(3-h) = 3+h
tanθ = (3+h)/(3-h) (2)
equaling (1) and (2)
(3+h)/(3-h) = h/2
6+2h=3h-h^2
h^2-h+6 =0
(h-3)(h+2) =0
h = 3
or h = -2 ( rejected,h>0)
面积
= (1/2) .BC.h
= (1/2).5.3
= 4
(2)
f(x) = a.b
= (2cosx/2)√2sin(x/2+π/4) + tan(x/2+π/4)tan(x/2-π/4)
= 2cosx/2(sinx/2 + cosx/2) - 1
= sinx + (2(cosx/2)^2-1)
= sinx + cosx
= √2( sin(π/4 + x) )
单调区间
增加[0,π/4]
减小[π/4,π]