A. Maths
回答 (5)
2008-01-14 3:59 am
✔ 最佳答案
L.H.S = sin²(θ －270ﾟ)＋cos²(90ﾟ + θ) + tan²(θ －360ﾟ) =[-sin(270ﾟ-θ)]2+[-sinθ]2+[-tan(360ﾟ-θ)]2
=[-sin(180ﾟ+90ﾟ-θ)]2+sin2θ+[-tan(180ﾟ+180ﾟ-θ)]2
=[sin(90ﾟ-θ)]2+sin2θ+[tan(180ﾟ-θ)]2
=[cosθ]2+sin2θ+[-tanθ]2
=cos2θ+sin2θ+tan2θ
=1+tan2θ
=sec2θ
=R.H.S
圖片參考：http://hk.yimg.com/i/icon/16/3.gif
2008-01-15 1:57 am
ans:R.S.H
2008-01-14 2:43 am
我計左幾次都only計倒sec²θ 咋... show俾你睇下丫
sin²(θ －270ﾟ)＋cos²(90ﾟ + θ) + tan²(θ －360ﾟ)
＝[sin－(270ﾟ －θ)]² + [cos(90ﾟ + θ)]² + [tan－(360°－θ)]²
＝[－sin(270ﾟ －θ)]² + (－sinθ)² + [－tan(360°－θ)]²
＝cos²θ +sin²θ +tan²θ
＝1+tan²θ
＝1+(sec²θ－1)
＝sec²θ
你check 下丫=]
sin²(θ －270ﾟ)＋cos²(90ﾟ + θ) + tan²(θ －360ﾟ)
＝[sin－(270ﾟ －θ)]² + [cos(90ﾟ + θ)]² + [tan－(360°－θ)]²
＝[－sin(270ﾟ －θ)]² + (－sinθ)² + [－tan(360°－θ)]²
＝cos²θ +sin²θ +tan²θ
＝1+tan²θ
＝1+(sec²θ－1)
＝sec²θ
你check 下丫=]
2008-01-14 2:33 am
L.H.S = sin²(θ －270ﾟ)＋cos²(90ﾟ + θ) + tan²(θ －360ﾟ)
=[-sin(270ﾟ-θ)]2+[-sinθ]2+[-tan(360ﾟ-θ)]2
=[-sin(180ﾟ+90ﾟ-θ)]2+sin2θ+[-tan(180ﾟ+180ﾟ-θ)]2
=[sin(90ﾟ-θ)]2+sin2θ+[tan(180ﾟ-θ)]2
=[cosθ]2+sin2θ+[-tanθ]2
=cos2θ+sin2θ+tan2θ
=1+tan2θ
=sec2θ
=R.H.S
2008-01-13 18:35:25 補充：
should R.H.S. be sec²θ ?
2008-01-24 13:07:36 補充：
copy me answer!
=[-sin(270ﾟ-θ)]2+[-sinθ]2+[-tan(360ﾟ-θ)]2
=[-sin(180ﾟ+90ﾟ-θ)]2+sin2θ+[-tan(180ﾟ+180ﾟ-θ)]2
=[sin(90ﾟ-θ)]2+sin2θ+[tan(180ﾟ-θ)]2
=[cosθ]2+sin2θ+[-tanθ]2
=cos2θ+sin2θ+tan2θ
=1+tan2θ
=sec2θ
=R.H.S
2008-01-13 18:35:25 補充：
should R.H.S. be sec²θ ?
2008-01-24 13:07:36 補充：
copy me answer!
2008-01-14 2:10 am
∵sin²(θ －270ﾟ)＋cos²(90ﾟ + θ) + tan²(θ －360ﾟ) = secθ
∴所以這是對的
∴所以這是對的
收錄日期: 2021-04-23 21:51:44
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https://hk.answers.yahoo.com/question/index?qid=20080113000051KK03093