三角函數計算

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

我也認為 螞蟻雄兵 ( 知識長 ) 這個微分計法很強勁！

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