Question

已知X=(0,∏）,求f(x)=(1+cos2x+8sinx^2)/sin2x的最小值

Answer 1

f(x)=(1+cos2x+8(sinx)^2)/(sin2x)

=(1+(1-2(sinx)^2)+8(sinx)^2)/(2sinx*cosx)

=(1+3(sinx)^2)/(sinx*cosx)

=(4(sinx)^2+(cosx)^2)/(sinx*cosx)

=4sinx/cosx + cosx/sinx

设t=sinx/cosx --> cosx/sinx=1/t

f(x)=4t+1/t

当00 -->t=sinx/cosx>0

--> f(x)=4t+1/t ≥ 2√(4t*1/t)=4

即f(x)的最小值是4(当π/2