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设x1=cosa,x2=cosb
x1+x2=cosa+cosb=√2sin50°
(x1+x2)^2=2(sin50°)^2=(cosa)^2+2cosa*cosb+(cosb)^2
x1*x2=cosa*cosb=(sin50°)^2-1/2
(x1+x2)^2-2x1*x2=1
=(cosa)^2+2cosa*cosb+(cosb)^2-2cosa*cosb
=(cosa)^2+(cosb)^2
∴(cosa)^2=(sinb)^2
sina=cosb
a+b=π/2

x1+x2=cosa+cosb=√2sin50°
=sina+cosa
=√2sin(a+45°)
=√2sin50°
a=5°
b=85°
b-2a=75°
tan(b-2a)=2+√3

∵0°<α<β<90° ∴ cosα > cosβ
方程x²-(√2×sin50°)x+sin²50°-½=0的两个实根为:
x = (sin50° ± cos50°)/ √2
即:cosα = (sin50° + cos50°)/ √2 = sin(50°+45°)=cos 5° ∴ α = 5°
cosβ=(sin50...

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∵0°<α<β<90° ∴ cosα > cosβ
方程x²-(√2×sin50°)x+sin²50°-½=0的两个实根为:
x = (sin50° ± cos50°)/ √2
即:cosα = (sin50° + cos50°)/ √2 = sin(50°+45°)=cos 5° ∴ α = 5°
cosβ=(sin50° - cos50°)/ √2 = sin(50°-45°)= cos85° ∴ β = 85°
tan(β - 2α) = tan75° = tan(45°+30°)= (tan45° + tan30°) / (1-tan45°tan30°) = 2+√3

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