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Prove that 1-2 sinθ cosθ =（cosθ- sinθ）^2
prove數學問題
2009-05-14 2:16 am
回答 (4)
2009-05-14 5:36 am
✔ 最佳答案
SolFor L.H.S.,
note that 2sinθcosθ=sin2θ
therefore, 1-2sinθcosθ=1-sin2θ
For R.H.S.,
(cosθ-sinθ)^2=cos^2θ-2cosθsinθ+sin^2θ
note that 2sinθcosθ=sin2θ and sin^2θ+cos^2θ = 1
=sin^2θ+cos^2θ - 2 sin θ cos θ = 1 - sin 2θ
=L.H. S.
Therefore,
1-2 sinθ cosθ =（cosθ- sinθ）^2
2009-05-16 10:11 pm
R.H.S.=（cosθ- sinθ）^2
=cos^2θ-2sinθcosθ+sin^2θ
=1-2sinθcosθ
=L.H.S.
=cos^2θ-2sinθcosθ+sin^2θ
=1-2sinθcosθ
=L.H.S.
2009-05-14 2:28 am
R.H.S= (cosθ- sinθ)^2
=(cosθ)^2 +(sinθ)^2 -2sinθcosθ
=1- 2sinθcosθ
=L.H.S
=(cosθ)^2 +(sinθ)^2 -2sinθcosθ
=1- 2sinθcosθ
=L.H.S
參考: 自己
2009-05-14 2:26 am
1-2 sinθ cosθ =（cosθ- sinθ）^2
sol
1-2sinθ cosθ
=cos^2θ+sin^2θ-2sinθ cosθ
=cos^2θ-2sinθ cosθ+sin^2θ
=(cosθ- sinθ)^2
=(cosθ- sinθ)^2
=cos^2θ-2sinθ cosθ+sin^2θ
=cos^2θ+sin^2θ-2sinθ cosθ
=
sol
1-2sinθ cosθ
=cos^2θ+sin^2θ-2sinθ cosθ
=cos^2θ-2sinθ cosθ+sin^2θ
=(cosθ- sinθ)^2
=(cosθ- sinθ)^2
=cos^2θ-2sinθ cosθ+sin^2θ
=cos^2θ+sin^2θ-2sinθ cosθ
=
收錄日期: 2021-04-15 14:44:44
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090513000051KK01146