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.](/uploads/image/z/6386546-2-6.jpg?t=%E4%B8%89%E8%A7%92%E5%BD%A2ABC%2C%E8%A7%92BAC%3D45%C2%B0%2C%E9%AB%98%E6%98%AFAD%2CBD%3D2%2CDC%3D3%2C%E6%B1%82%E9%9D%A2%E7%A7%AF2.%E5%90%91%E9%87%8Fa%3D%282cosx%2F2%2Ctan%28x%2F2%2B%CF%80%2F4%29%2C%E5%90%91%E9%87%8Fb%3D%28%E6%A0%B9%E5%8F%B72sin%28x%2F2%2B%CF%80%2F4%29%2Ctan%28x%2F2-%CF%80%2F4%29%29%E8%AE%BEf%EF%BC%88x%EF%BC%89%3Da.%E6%B1%82f%EF%BC%88x%EF%BC%89%E5%9C%A8%5B0%EF%BC%8C%CF%80%5D%E4%B8%8A%E5%8D%95%E8%B0%83%E5%8C%BA%E9%97%B43.%E4%B8%89%E8%A7%92%E5%BD%A2ABC%2CA%2BC%3D2B%2C1%2FcosA%2B1%2FcosC%3D%E8%B4%9F%EF%BC%88%E6%A0%B9%E5%8F%B72%EF%BC%89%2FcosB%2CA%3EC%2C%E6%B1%82A.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.
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]