1.求y=cos²x+sinx+1(|x|≤1)的值域 2.求函数y=sin²x-5/2sinx+5

(cosx-sinx+1)/(cosx+sinx+1)
令x=2t
=(cost^2-sint^2-2sintcost+sint^2+cost^2)/(cost^2-sint^2+2sintcost+sint^2+cost^2)
=2cost(1-sint)/2cost(1+sint)
=(1-sint)/(1+sint)
万能公式,得证

y=cos²x+sinx+1=1-sinx2+sinx+1
=-(sinx2-sinx)+2
=-(sinx2-sinx+1/4)+2+1/4
=-(sinx-1/2)2+9/4
-11
所以至于为0

你想说的公式是asinx+bcosx=√(a²+b²)sin(x+φ) 其中sinφ=b/a,
cosx+sinx+1
=√2sin(x+π/4)+1
-1≤sin(x+π/4)≤1
1-√2≤√2sin(x+π/4)+1≤√2+1
函数的值域为[1-√2,√2+1]
2kπ-π/2≤x+π/4≤2kπ+π/2 (k∈Z)时,即2kπ-3π/4≤x≤2kπ+π/4 (k∈Z),函数单调递增.
2kπ+π/2≤x+π/4≤2kπ+3π/2 (k∈Z)时,即2kπ+π/4≤x≤2kπ+5π/4 (k∈Z),函数单调递减.