www.fltk.net > 已知锐角α,β满足tAn(α%β)=sin2β,求证2tAn2β=tAnα+tAnβ

已知锐角α,β满足tAn(α%β)=sin2β,求证2tAn2β=tAnα+tAnβ

过程如下

α为锐角,且有2tan(π-α)-3cos(π2+β)+5=0,可得-2tanα+3sinβ+5=0

由cos2α+cos2β+cos2γ=1联想到锐角α、β、γ是长方体的对角线与过一个顶点的三条棱所成

tanα=1/7 sinβ=√10/10 cosβ=√[1-(sinβ)^2]=3√10/10

1-coa2a=sinacosa,2sin²a=sinacosa,则tana=si

1 cosβ =√5/5, sinα = √10/10 sin²α+cos&

tanα=7 cosβ=2√5/5 tanβ = √5/(2√5) =1/2 tan2β

由3(sinα)^2+2(sinβ)^2=1得3(sinα)^2+1-cos2β=1 所以3(s

∵cos a=3/5 ∴sin a=√(1-cos^2 a)=4/5 ∴tan a=sin a

^2是平方 由于α、β∈(0,π/2),所以α+β∈(0,π),sin(α+β)>0 且α

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