Y 2 = 3 * (sin α) 2/(2-cos α) 2, 2 stands for square.

=3*( 1-(cosα)^2)/(2-cosα)^2

Let cosα=x and y 2 = (3-3x 2))/(2-x) 2.

=-3- 12/(x-2)-(3/(x-2))^2

Let t=3/(x-2), then t belongs to -3,-1, y 2 =-T2-4t-3,

Y 2 belongs to 0, 1.

Y belongs to-1, 1.

2, I can't be sure, because ab is two internal angles of an acute triangle, the relationship between sina and cosa can't be determined, sinb and cosb can't be determined, and the ranking from big to small can't be determined.

Of course, it is also possible that you didn't write the question correctly!

As far as f(x+ 1)=-f(x) is known,

Then f(x+2)=-f(x+ 1)=f(x)

That is, the period of f(x) is 2, and f(x) is an even function and a subtraction function on -3 and -2.

Then f(x) is a decreasing function at-1, and a increasing function at 0, 1.

Meanwhile, f(x+ 1)=-f(x), and f(-0.5)=-f(0.5).

And f(x) is an even function, then f(-0.5)=f(0.5),

Then f(0.5)=0,

To sum up, where f(x) is 0, 1 and f(0.5)=0, it is increasing function.

Because the relationship between Sina and cosa cannot be determined, the size of the rest cannot be determined!