3(sinα)^2+2(sinβ)^2=2sinα则(sinα)^2+(sinβ)^2取值范围^
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/25 01:46:46
3(sinα)^2+2(sinβ)^2=2sinα则(sinα)^2+(sinβ)^2取值范围^3(sinα)^2+2(sinβ)^2=2sinα则(sinα)^2+(sinβ)^2取值范围^3(si
3(sinα)^2+2(sinβ)^2=2sinα则(sinα)^2+(sinβ)^2取值范围^
3(sinα)^2+2(sinβ)^2=2sinα则(sinα)^2+(sinβ)^2取值范围^
3(sinα)^2+2(sinβ)^2=2sinα则(sinα)^2+(sinβ)^2取值范围^
因为,3(sinα)^2+2(sinβ)^2=2sinα
所以,3(sinα)^2<=2sinα
所以,sinα(3sinα-2)<=0
则,0<=sinα<=2/3
(sinα)^2+(sinβ)^2=1/2[2(sinα)^2+2(sinβ)^2]
=(1/2)[3(sinα)^2+2(sinβ)^2-(sinα)^2]
=(1/2)[2sinα-(sinα)^2]
=(1/2)[1-(1-sinα)^2]
0<=sinα<=2/3
-2/3<=-sinα<=0
1/3<=1-sinα<=1
1/9<=(1-sinα)^2<=1
-1<=-(1-sinα)^2<=-1/9
0<=1-(1-sinα)^2<=8/9
0<=(1/2)[1-(1-sinα)^2]0<=4/9
则(sinα)^2+(sinβ)^2取值范围[0,4/9]
化简sin(α+β)+sin(α-β)+2sinαsin(3π/2-β)=
求证:sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα
求证sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα
求证sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα
求证sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα
证明:sin(2α+β)/sinα - 2cos(α+β)=sinβ/sinα
sin(α+β)sin(α-β)=sin^2α-sin^2β的推导过程
若sin^2β-sin^2α=m,则sin(α+β)sin(α-β)
已知sin(2α+ β)=5sinβ,求证:2sin(α+ β)=3sinα
已知sin(α+β)*sin(α-β)=-1/3,求sin^2α-sin^2β的值
求证:sin²α+sin²β+2sinαsinβcos(α+β)=sin²(α+β)
sinα+sinβ=1/3 求sinα-(cosβ)^2最大值
已知3sin²α+2sin²β=2sinα,求cos²α+cos²β的取值范围已知3sin²α+2sin²β=2sinα则有2sin²β=2sinα-3sin²α即sin²β=sinα-1.5sin²α所以cos²β=1-sin²β=1-(sinα-1.5sin²α)=1-
sin(α+β)=2/3,sin(α-β)=3/5,sinα+sinβ=1/2,求cos(α+β)/2*sin(α-β)/2
若 3sin^2α +2sin^2β =2sinα 求y=sin^2α+sin^2β的最大值
已知:3sin^2α+2sin^2β=2sinα,求sin^2α+sin^2β的值
已知3sin^2α—2sinα+2sin^2β=0 求sin^2α+sin^2β的取值范围
已知3sin^2α+2sin^2β=2sinα,求sin^2α+sin^2β的值