3(sinα)^2 + 2(sinβ)^2= 2sinα試求(sinα)^2+(sinβ)^2的取值范圍.

熱心網友

根據 3(sinα)^2 + 2(sinβ)^2= 2sinα == (sinβ)^2 = sinα-3/2*(sinα)^2因為(sinβ)^2≥0，所以sinα-3/2*(sinα)^2≥0 ==0≤sinα≤2/3Y=(sinα)^2+(sinβ)^2 = (sinα)^2 + sinα-3/2*(sinα)^2=-1/2*(sinα)^2 + sinα=-1/2*(sinα-1)^2 + 1/2 ①顯然當sinα=2/3時，Y|max = 8/18當sinα=0時，Y|min = 0所以 0≤(sinα)^2+(sinβ)^2≤8/18注：一定要注意sinα的取值范圍，否則很可能在①式中取sinα=1，從而出錯。