证明
已知x、y∈(0,π/2),tanx=3tany,求证:x-y≤π/6。
tanx=3tany>tany ∴x>y→x-y∈(0,π/2). 而(√3tany-1)^2≥0 →2tany/(1+3(tany)^2)≤√3/3. 又,tan(x-y) =(tanx-tany)/(1+tanxtany) =2tany/(1+3(tany)^2). ∴tan(x-y)≤tan(π/6). 因此,x-y≤π/6。
x、y∈(0,π/2),tanx=3tany, ∴x-y∈(0,π/2), tan(x-y)=(tanx-tany)/(1+tanxtany)=2tany/[1+3(tany)^2]<=1/√3, ∴x-y<=π/6.
答:x∈(0,π/2)时(第一象限), 设直角坐标系中原点为O 设单位圆与横轴正向交于A 角x的终边与单位圆交于P,与过A点且垂直于横轴的直线交于Q, 过P作横轴的...详情>>
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