Trigonometry
回答 (2)
2012-05-09 2:54 am
✔ 最佳答案
left=sin(-θ)cos(90+θ)+cos^2 (180-θ)left=(-sinθ)(-sinθ)+(-cosθ)^2
left=sin^2θ+cos^2θ
left=1
2012-05-07 3:31 am
sin(-θ)cos(90+θ)+cos^2 (180-θ) =1
L.H.S.=sin(-θ)cos(90+θ)+cos^2 (180-θ)
=(-sinθ)(-sinθ)+(-cos θ)^2
** sin(-θ)=-sin θ ; cos (90+θ)=-sinθ ; cos(180-θ)=-cosθ
=(sinθ)^2+(cosθ)^2
=1
=R.H.S
so it is identity.
L.H.S.=sin(-θ)cos(90+θ)+cos^2 (180-θ)
=(-sinθ)(-sinθ)+(-cos θ)^2
** sin(-θ)=-sin θ ; cos (90+θ)=-sinθ ; cos(180-θ)=-cosθ
=(sinθ)^2+(cosθ)^2
=1
=R.H.S
so it is identity.
參考: me
收錄日期: 2021-04-29 21:45:44
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120506000051KK00609