請回答問題，並寫出步驟。Tank you very much！
1) Prove that ( sinθ﹣cosθ)^2 = 1 ﹣2 sinθcosθ
2) Prove that sinθ + ( cosθ) / ( tanθ) = 1 / sinθ
3) Prove that sinθ/ cosθ + cosθ/ sinθ = 1 / ( sinθcosθ)
4) Prove that 1 / (1 + cosθ) + 1/ ( 1 ﹣cosθ) = 2 / sin^2θ
利用畢氏定理求三角比的值
a) If cosθ= 2 / ( √13 ) ，evaluate sinθ and tanθ without using calculator .
(F.2)三角恆等式我的證明
2009-05-09 8:46 pm
回答 (1)
2009-05-09 9:23 pm
✔ 最佳答案
1) L.H.S.=( sinθ﹣cosθ)2
=(sin2θ+cos2θ-2sinθcosθ)
=1-2sinθcosθ
=R.H.S.
2)
L.H.S.=sinθ + ( cosθ) / ( tanθ)
=sinθ+(cos2θ/sinθ)
=(sin2θ+cos2θ)/sinθ
= 1 / sinθ
=R.H.S.
3)
L.H.S.=sinθ/ cosθ + cosθ/ sinθ
=(sin2θ+cos2θ)/(cosθsinθ)
=1/(sinθcosθ)
=R.H.S.
4)
L.H.S.=1 / (1 + cosθ) + 1/ ( 1 ﹣cosθ)
=2/[(1+cosθ)(1-cosθ)]
=2/(1-cos2θ)
=2/sin2θ
=R.H.S.
利用畢氏定理求三角比的值
a)
sinθ=3/√13 or sinθ=-3/√13
tanθ=3/2 or tanθ=-3/2
收錄日期: 2021-04-23 20:41:44
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090509000051KK00716