3sinα^2+2sinβ^2=1,3sin2α-2sin2β=0,αβ均为锐角,求α+2β的值

化简sin(α+β)+sin(α-β)+2sinαsin(3π/2-β)= 已知sin(α+β)*sin(α-β)=-1/3,求sin^2α-sin^2β的值 sinα+sinβ=1/3 求sinα-(cosβ）^2最大值 s = 2*sin(x)-sin(2*x)+2/3*sin(3*x)-1/2*sin(4*x)+2/5*sin(5*x)用matlab画图,求教啊 求和S=sin^2 1度+sin^2 2度+sin^2 3度+...+sin^2 88度+sin^2 89度 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 sinα+sinβ=2sin(α+β)/2cos(α-β)/2; (1) sinα-sinβ=2cos(α+β)/2sin(α-β)/2; (2) cosα+cosβ=2sinα+sinβ=2sin(α+β)/2cos(α-β)/2; (1)sinα-sinβ=2cos(α+β)/2sin(α-β)/2; (2)cosα+cosβ=2cos(α+β)/2cos(α-β)/2; (3)cos-cosβ=-2sin(α+β)/2s 已知sin（2α+ β）=5sinβ,求证：2sin（α+ β）=3sinα sinα=-2cosα,求sin^2α-3sinαcosα+1 S(1-sin^2x)d(sinx)为什么=sinx-1/3sin^3x? sin(α+β)=1/2 ,sin(α-β)=1/3 .2sinαcosβ等于多少 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 已知3(sinα)^2-2sinα+2(sinβ)^2=0,试求(sinα)^2+(sinβ)^2的取值范围 3(sinα)^2+2(sinβ)^2=2sinα则(sinα)^2+(sinβ)^2取值范围^