α、β、γ是锐角,1=sin^2α+sin^2β+sin^2γ=1-cos(α+β)cos(α-β)+sin^2γ,得cos(α+β)cos(α-β)=sin^2γ,所以cos(α+β)＞0,α+β＜π/2,同理β+γ＜π/2,α+γ＜π/2,于是α+β+γ＜0.75π ①.假设α+β+γ≤π/2,则cos(α+β)cos(α-β)=sin^2γ≤sin^2(π/2-α-β)=cos^2(α+β),有cos(α-β)≤cos(α+β),sin(α)sin(β)≤0,矛盾,所以π/2＜α+β+γ ②,由①、②可知命题得证.