(sina/2+cosa/2)^2+2sin^2(π/4-a/2)=
问题描述：
(sina/2+cosa/2)^2+2sin^2(π/4-a/2)=
答
(sina/2+cosa/2)^2+2sin^2(π/4-a/2)
=sin²a/2+cos²a/2+2sina/2cosa/2+2{1-cos[2(π/4-a/2)]}/2
=1+sina+1-cos(π/2-a)
=2+sina-sina
=2
答
(sina/2+cosa/2)²+2sin²(π/4-a/2)=sin²a/2+cos²a/2+2sina/2cosa/2+2sin²(π/4-a/2)=1+sina+1-cos[2(π/4-a/2)]=2+sina-cos(π/2-a)=2+sina-sina=2