証明以下之三角恆等式﹕
tan2∅ -sin2∅ ≡ tan2∅ -sin2∅
簡化以下問題﹕
(1+cosθ)/sin^2θ -1/(1-cosθ )
求中三級三角比問題解答
2014-09-19 5:18 pm
回答 (2)
2014-09-29 4:57 am
✔ 最佳答案
∅ = empty setφ = phi
They are different symbols.
2014-09-28 20:57:44 補充：
Please read:
圖片參考：https://s.yimg.com/rk/HA00430218/o/1265126181.png
2014-09-20 1:52 am
(a+b)(a-b)=a^2 - b^2
sin^2 θ + cos^2 θ=1
(1+cos θ)/(sin^2 θ) - 1/(1-cos θ)
=[ (1+cos θ)(1-cos θ) - (sin^2 θ) ]/ [ (sin^2 θ)(1-cos θ) ]
=( 1 - cos^2 θ - sin^2 θ ) / [ (sin^2 θ)(1-cos θ) ]
=( 1 - 1 ) / [ (sin^2 θ)(1-cos θ) ]
=0/ [ (sin^2 θ)(1-cos θ) ]
=0
sin^2 θ + cos^2 θ=1
(1+cos θ)/(sin^2 θ) - 1/(1-cos θ)
=[ (1+cos θ)(1-cos θ) - (sin^2 θ) ]/ [ (sin^2 θ)(1-cos θ) ]
=( 1 - cos^2 θ - sin^2 θ ) / [ (sin^2 θ)(1-cos θ) ]
=( 1 - 1 ) / [ (sin^2 θ)(1-cos θ) ]
=0/ [ (sin^2 θ)(1-cos θ) ]
=0
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140919000051KK00018