由 sinαcosβ=[sin(α+β)+sin(α-β)]/2可知
sin(a+β)cosa-1/2[sin(2a+β)-cosβ]
=1/2[sin(2a+β)+sinβ]-1/2[sin(2a+β)-cosβ]
=1/2[cosβ+sinβ]=1/2
所以cosβ+sinβ=1
又 (sinβ)2+(cosβ)2=1 且0