Question

高一数学 第二问

Answer 1

追答

Answer 2

2x^2-(√5+1)x+m=0
(1)
sum of roots
sinθ+cosθ=(√5+1)/2
1+2sinθcosθ = (6+2√5)/4
2sinθcosθ= (1+√5)/2

product of roots
sinθcosθ = m/2
m= 2sinθcosθ
=(1+√5)/2

(2)
tanθ.sinθ/(tanθ-1) +cosθ/(1-tanθ)
=(tanθ.sinθ-cosθ)/(tanθ-1)
=[(sinθ)^2-(cosθ)^2]/ (sinθ - cosθ)
= sinθ+cosθ
=(1+√5)/2

Answer 3

tana=sina/cosa
分式上下同乘cosa
原式=sina^2/(sina-cosa)+cosa^2/(cosa-sina)
=(sina^2-cosa^2)/(sina-cosa)
=sina+cosa

Answer 4

我用x来举例
通分为[(tanx)^2*cosx-cosx]/(tanx-1)=cosx(tanx-1)(tanx+1)/(tanx-1)
=cosx(tanx+1)=sinx+cosx=(根号3+1）/2

Answer 5

先化简：分子分母同乘cos，得sin2/（sin-cos）-cos2/（sin-cos）
=（sin2-cos2）/（sin-cos）=sin+cos得解

Answer 6

喀！……