Let f (x) = (sin ^ 4 x-cos ^ 4 X-5) / (cos2x + 2); (a) prove that f (x) = 3 / (2 sin ^ 2 x-3) - 1 (b) find the range of F (x)

Let f (x) = (sin ^ 4 x-cos ^ 4 X-5) / (cos2x + 2); (a) prove that f (x) = 3 / (2 sin ^ 2 x-3) - 1 (b) find the range of F (x)

f(x)=(sin^4 x-cos^4 x-5)/(cos2x+2)
=(sin^2 x-cos^2 x-5)/(cos2x+2)
=-(cos2x+5)/(cos2x+2)
=-1-3/(cos2x+2)
=-1-3/(1-2sin^2 x+2)
=3/(2sin^2x-3)-1
It's over
b)、f(x)==3/(2sin^2x-3)-1,sin^2x∈【0,1】
So 2Sin ^ 2x-3 ∈ [- 3, - 1]
So f (x) max = - 2, f (x) min = - 4
That is, the range of F (x) is [- 4, - 2]