求cosα+sinα，式子见上传的图

cos2α=-(2^0.5/2)sin(α-π/4)
(cosα)^2-(sinα)^2=-(2^0.5/2)(sinαcosπ/4-cosαsinπ/4)
(cosα-sinα)(cosα+sinα)=-(2^0.5/2)(sinα×(2^0.5/2)-cosα(2^0.5/2))
(cosα-sinα)(cosα+sinα)=(1/2)(cosα-sinα)
∵sin(α-π/4)≠0
∴α-π/4kπ
α≠kπ+π/4
cosα≠sinα
cosα+sinα=1/2

利用公式:cos 2a=cos^2-sin^2=(cosa-sina)×(cosa+sina)和sin(a-??/4)=ㄏ2/2×sina-ㄏ2/2×cosa(由于找不到符号,所以用"??"表示3.14,"ㄏ"表示根号),所以cos2a÷sin(a-??/4)=[(cosa-sina)×(cosa+sina)]÷ㄏ2/2×(sina-cosa)=-ㄏ2/2,再约分化简为:cosa+sina=-1/2