蓝色夏天雨
幼苗
共回答了17个问题采纳率:88.2% 举报
(1+sinA+cosA)[sin(1/2)A-cos(1/2)A]/√(2+2cosA)
=[(sin1/2A)^2+(cos1/2A)^2+2sin(1/2)Acos(1/2)A+(cos1/2A)^2-(sin1/2A)^2[sin(1/2)A-cos(1/2)A]/√(2+2cosA)
={[sin(1/2)A+cos(1/2)A]^2+[(cos1/2A)^2-(sin1/2A)^2]}[sin(1/2)A-cos(1/2)A]/√(2+2cosA)
={[sin(1/2)A+cos(1/2)A]^2+[sin(1/2)A+cos(1/2)A][cos(1/2)A-sin(1/2)A]}[sin(1/2)A-cos(1/2)A]/√(2+2cosA)(其实就是提取公因式)
={[sin(1/2)A+cos(1/2)A][sin(1/2)A+cos(1/2)A+cos(1/2)A-sin(1/2)A]}[sin(1/2)A-cos(1/2)A]/√(2+2cosA)
=[sin(1/2)A+cos(1/2)A][2cos(1/2)A][sin(1/2)A-cos(1/2)A]/√(2+2cosA)
=-[(cos1/2A)^2-(sin1/2A)^2]][2cos(1/2)A]/√(2+2cosA)
=[-2cos(1/2)A*cosA]/√2[(sin1/2A)^2+(cos1/2A)^2+(cos1/2A)^2-(sin1/2A)^2]
=[-2cos(1/2)A*cosA]/√2[2(cos1/2A)^2]
=[-2cos(1/2)A*cosA]/[2cos(1/2)A]
=-cosA
累死我了.解起来容易.打起来真麻烦.如果看不懂就问我哦~~
1年前
3