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-無言- asked in 科學數學 · 6 years ago

三角函數計算

Suppose C:x^2+(y-1)^2=r^2 and y=sinx have only one intersection and the x-coordinate of this intersection point is α,

then [sinα+sin3α-4(cosα)^2]/(α*cosα) = _______.

Ans: -4

Update:

關於微分的那一部份

如果不用微分處理

是否有其他解法?

先謝謝螞蟻雄兵知識長~

2 Answers

Rating
  • 6 years ago
    Favorite Answer

    SupposeC:x^2+(y-1)^2=r^2 and y=Sinx haveonly one intersection and the x-coordinate

    of this intersection point is “α”,then (Sinα+Sin3α-4Cos^2 α)/(α*Cosα)=?

    Sol

    x^2+(y-1)^2=r^2與y=Sinx只有一交點

    (0,1)到y=Sinx距離=r

    x^2+(y-1)^2=r^2,y=Sinx

    x^2+(Sinx-1)^2=r^2

    x^2+(y-1)^2=r^2

    2x+2(y-1)y’=0

    y’=x/(1-y)

    y=Sinx

    y’=Cosx

    x/(1-y)=Cosx

    1-y=x/Cosx

    y=1-x/Cosx=Sinx

    1-α/Cosα=Sinα

    1-Sinα=α/Cosα…………………

    Sinα+Sin3α-4Cos^2 α

    = Sinα+3Sinα-4Sin^3α-4Cos^2 α

    =4Sinα(1-Sin^2α)-4Cos^2 α

    =4SinαCos^2 α-4Cos^2 α

    =4Cos^2 α(Sinα-1)

    =4Cos^2 α(Sinα-1)

    =-4αCosα

    So

    (Sinα+Sin3α-4Cos^2 α)/(α*Cosα)=-4

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  • 6 years ago

    我也認為 螞蟻雄兵 ( 知識長 ) 這個微分計法很強勁!

    ╭∧---∧╮

    │ .✪‿✪ │

    ╰/) ⋈ (\\╯

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