Jade 發問於 科學及數學數學 · 2 月前

how to show x = 1 / (1+2costheta)?

OAB is isosceles, OA=OB, AB=1

how to show x = 1 / (1+2costheta)?

Please help.

更新:

Hi, Mr. 老怪物

how do you get = (1/2)/(cos^2(θ/2)+cos^2(θ/2)tan(θ/2)/tan(θ))?

更新 2:

Hi, Mr. 老怪物, finally I got it. Thanks a bunch!

Attachment image

2 個解答

評分
  • ?
    Lv 7
    2 月前
    最愛解答

    2021/01/13: 

    昨天看到你在回答區問的, 我就補充了.

    (1/2)tan(θ)(1+tan^2(θ/2))/(tan(θ)+tan(θ/2))

          = (1/2)tan(θ)sec^2(θ/2)/(tan(θ)+tan(θ/2)) 

          = (1/2)(1/cos^2(θ/2))/(1+tan(θ/2)/tan(θ)) 

          = (1/2)/(cos^2(θ/2)+cos^2(θ/2)tan(θ/2)/tan(θ))

    請再看看還有哪裡寫得不夠清楚.

    =============================

    作 線段OC⊥AB邊 於 C.

    則 ∠COX = θ/2,

    ∴ CX/OX = sin(θ/2)

       OC/OX = cos(θ/2)

       (1/2)/OA = sin(3θ/2)

       OC/OA = cos(3θ/2)

    ∴ OC = cot(3θ/2)/2

       OX = cot(3θ/2)/(2cos(θ/2))

       CX = cot(3θ/2)tan(θ/2)/2

       AX = 1/2 - cot(3θ/2)tan(θ/2)/2

          = (1/2)(1-tan(θ/2)/tan(3θ/2))

          = (1/2)[1-tan(θ/2)(1-tan(θ)tan(θ/2))/(tan(θ)+tan(θ/2))]

          = (1/2)tan(θ)(1+tan^2(θ/2))/(tan(θ)+tan(θ/2))

          = (1/2)tan(θ)sec^2(θ/2)/(tan(θ)+tan(θ/2))

          = (1/2)(1/cos^2(θ/2))/(1+tan(θ/2)/tan(θ))

          = (1/2)/(cos^2(θ/2)+cos^2(θ/2)tan(θ/2)/tan(θ))

          = (1/2)/(cos^2(θ/2)+sin(θ/2)cos(θ/2)/tan(θ))

          = (1/2)/[(1+cos(θ))/2 +(1/2)sin(θ)/tan(θ)]

          = 1/(1+2cos(θ))

  • 2 月前

    Hi, Mr. 老怪物

    how do you get    = (1/2)/(cos^2(θ/2)+cos^2(θ/2)tan(θ/2)/tan(θ))?

還有問題嗎?立即提問即可得到解答。