兩題數學問題(跟圓有關的)(20點!!)

圖片參考：http://imgcld.yimg.com/8/n/AD01636710/o/101111260804313869498640.jpg


圖片參考：http://imgcld.yimg.com/8/n/AD01636710/o/101111260804313869498651.jpg


圖片參考：http://imgcld.yimg.com/8/n/AD01636710/o/101111260804313869498662.jpg

希望對你有幫助...

1. 設S=﹛﹙x，y﹚︱(x+y)^2+4(x+y)+3≦0﹜，
T=﹝(x，y)︱x^2+y^2+2x+2y+1≦0﹞，則S∩T之區域面積=?
Sol
(x+y)^2+4(x+y)+3<=0
(x+y)^2+4(x+y)+4<=1
(x+y+2)^2－1^2<=0
(x+y+2－1)(x+y+2+1)<=0
(x+y+1)(x+y+3)<=0
x+y+1<=0，x+y+3>=0
S=｛(x，y)｜x+y+1<=0，x+y+3>=0｝
x^2+y^2+2x+2y+1<=0
(x^2+2x+1)+(y^2+2y+1)<=1
(x+1)^2+(y+1)^2<=1
T=｛(x，y)︱(x+1)^2+(y+1)^2<=1｝
(－1，－1)至x+y+1=0距離=｜－1－1+1｜/√2=1/√2<1
(－1，－1)至x+y+3=0距離=｜－1－1+3｜/√2=1/√2<1
S∩T之區域面積
=π－2(π/4－1/2)
=π－π/2+1
=1+π/2

2. 0°<θ<90°，求2/Cos^2 θ+8/Sin^2 θ的最小值?
Sol
0°<θ<90°
0°<2θ<180°
0<Sin2θ<1
1/Sin2θ>1
2/Cos^2 θ>0，8/Sin^2 θ<0
(2/Cos^2 θ+8/Sin^2 θ)/2
=(2Sec^2 θ+8Csc^2 θ)/2
>=√(2Sec^2 θ*8Csc^2 θ)
=4Secθ*Cscθ
=8/Sin2θ
>=8
2/Cos^2θ+8/Sin^2 θ>=16