求函数y=-cos^2x-sinx+1 x∈[π/4,3π/4]
问题描述：
求函数y=-cos^2x-sinx+1 x∈[π/4,3π/4]
如题
答
求函数y=-cos^2x-sinx+1 x∈[π/4,3π/4]的值域.
y=sin^2x-sinx
=(sinx-1/2)^2-1/4
x∈[π/4,3π/4],
√2/2≤sinx≤1,
当sinx=√2/2,ymin=(1-√2)/2.
当sinx=1时,ymax=0.
∴值域[=(1-√2)/2,0].