Question

函数f(x)=(sinx-1)/√[(sinx-1)^2+(cosx-1)^2](0≤x≤2π)的值域

Answer 1

f(x)=(sinx-1)/√((sinx-1)^2+(cosx-1)^2)
=-(1-sinx)/√((1-sinx)^2+(1-cosx)^2)
=-1/√[1+(1-cosx)^2/(1-sinx)^2]

(1-cosx)^2/(1-sinx)^2=(2cos²(x/2)) ^2/(sinx/2-cosx/2) ^ 4
=4(cos(x/2)) ^4/(sinx/2-cosx/2) ^ 4 ……分子分母同除以cos(x/2))^4可得下式
=4/(tan(x/2)-1)^4

所以f(x)=-1/√[1+4/(tan(x/2)-1)^4]
(tan(x/2)-1)^4≥0，所以f(x)min=-1

因为Sinx≤1，所以(sinx-1)/√((sinx-1)^2+(cosx-1)^2) ≤0，
f(x)max=0

故值域为[-1,0]