已知sin(a+β)cosa-1/2[sin(2a+β)-cosβ]=1/2,0
问题描述：
已知sin(a+β)cosa-1/2[sin(2a+β)-cosβ]=1/2,0
答
sin(a+b)cosa-1/2sin[(a+b)+a]+1/2cosb=1/2
sin(a+b)cosa-1/2[sin(a+b)cosa+cos(a+b)sina]+1/2cosb=1/2
sin(a+b)cosa-cos(a+b)sina+cosb=1
sin[(a+b)-a]+cosb=1
sinb+cosb=1
根据0
答
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