Question

3sin^2x+-2sin^2y=2sinx sin^2x+-sin^2y=?

Answer 1

3sin^2x±2sin^2y=2sinx

sin^2x±sin^2y = (3sin^2x±2sin^2y-sin^2x)/2

= (2sinx-sin^2x)/2

= { 1 - (sinx-1)^2 }/2

-1≤sinx≤1

-2≤sinx-1≤0

0≤(sinx-1)^2≤4

-4≤-(sinx-1)^2≤0

-3≤ 1 - (sinx-1)^2 ≤ 1

-3/2 ≤ {1 - (sinx-1)^2 }/2 ≤ 1/2

即：sin^2x±sin^2y∈【-3/2,1/2】