高中問題 求解
回答 (2)
2012-10-19 4:16 am
✔ 最佳答案
sinA - cosA=1/根號2(sinA - cosA)^2=1/2
(sinA)^2+(cosA)^2-2*sinA*cosA=1/2
1/2=2*sinA*cosA
求sinA+cosA=?
(sinA+cosA)^2=(sinA)^2+(cosA)^2+2*sinA*cosA
=1+1/2=3/2
sinA+cosA=(根號6)/2
2012-10-19 4:35 am
設p=SinA+CosA>0
(SinA－CosA)^2+(SinA+CosA)^2=(1/√2)^2+p^2
2=(1/2)+p^2
p^2=3/2=6/4
p=√6/2
(SinA－CosA)^2+(SinA+CosA)^2=(1/√2)^2+p^2
2=(1/2)+p^2
p^2=3/2=6/4
p=√6/2
收錄日期: 2021-04-30 16:57:50
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20121018000010KK05835