Question

求函数y=cos+2/sin-1的值域

Answer 1

y=(cosx+2)/(sinx-1)

ysinx-y=cosx+2

ysinx-cosx=y+2

√(y²+1)sin(x-t)=y+2,t=arctan(1/y)

sin(x-t)=(y+2)/√(y²+1)

故|y+2|/√(y²+1)≤1

y²+4y+4≤y²+1

y≤-3/4

此即为值域.