Add.Maths question

EASY
sinA-cosA=x...(1)
cos2A=y^2...(2)
由(2)
cos2A=y^2
(cos^2 A-sin^2 A)=y^2
(cos A+sin A)(cos A-sin A)=y^2
x(cos A+sin A)=y^2
x^2(cos A+sin A)^2=y^2
x^2(1+2sinAcosA)=y^2...(*)
現在考慮sinA-cosA=x
(sinA-cosA)^2=x^2
1-2sinAcosA=x^2
1-x^2=2sinAcosA
(*)變成 x^2(1+1-x^2)=y^2
x^2(2-x^2)=y^2
x^4-2x^2+y^2=0


2008-12-21 20:33:25 補充：
少少錯
cos2A=y^2

(cos^2 A-sin^2 A)=y^2

(cos A+sin A)(cos A-sin A)=y^2

-x(cos A+sin A)=y^2

x^2(cos A+sin A)^2=y^4

x^2(1+2sinAcosA)=y^4...(*)

現在考慮sinA-cosA=x

(sinA-cosA)^2=x^2

1-2sinAcosA=x^2

1-x^2=2sinAcosA

(*)變成 x^2(1+1-x^2)=y^4

x^2(2-x^2)=y^4

x^4-2x^2+y^4=0

sinA-cosA=x-------(1)
cos2A=y^2---------(2)

將(1)兩邊2次方
sin^2A + cos^2A - 2sinAcosA = x^2
1 - sin2A = x^2---(3) (sin^2A+cos^2A=1)(2sinAcosA=sin2A)

從2可找出sin2A 畫圖 用畢氏定理求直角三角形既邊
即底 = y^2
斜邊 = 1
現在要找出高(設b)
1^2 = (y^2)^2 + b^2
1 = y^4 + b^2
b = (開方 1 - y^4)
sin2A = 高/斜邊 = (開方 1 - y^4)/1 = (開方 1 - y^4)
將sin2A = (開方 1 - y^4)代入(3)
1 - (開方 1 - y^4) = x^2

2008-12-21 20:42:02 補充：
想答案好睇d可以寫成
1 - x^2 = (開方 1 - y^4)
兩邊2次方
1 - 2x^2 + x^4 = 1 - y^4
x^4 - 2x^2 + y^4 = 0