第1个回答 2011-02-16
f(x)=cosx*√[(1+sinx)/(1-sinx)]+sinx*√[(1+cosx)/(1-cosx)]
=cosx*√[(1+sinx)(1+sinx)/(1-sinx)(1+sinx)]+sinx*√[(1+cosx)(1+cosx)/(1-cosx)(1+cosx)]
=cosx * (1+sinx)/|cosx| +sinx*(1+cosx)/|sinx|
1)x∈(-π/2,0),在第三象限,则cosx为负,sinx也为负
则f(x)= - 1-sinx-1-cosx = 2-sinx-cosx
f(-π/4)=2 -√2/2-√2/2=2-√2
2)x∈(π/2,π),在第二象限,则cosx为负,sinx也为正
则f(x)= -1-sinx+1+cosx = cosx -sinx = -√2 sin(x-π/4)
x∈(π/2,π),则x-π/4 ∈(π/4,3π/4)
sin(x-π/4)∈(√2/2,1)
则-√2 sin(x-π/4)∈(-√2, -1〕
即当x∈(π/2,π)时,求函数f(x)的值域为(-√2, -1〕
第2个回答 2011-02-16
f(x)=cosx*√[(1+sinx)/(1-sinx)]+sinx*√[(1+cosx)/(1-cosx)]
=cosx*√[(1+sinx)(1+sinx)/(1-sinx)(1+sinx)]+sinx*√[(1+cosx)(1+cosx)/(1-cosx)(1+cosx)]
=cosx * (1+sinx)/|cosx| +sinx*(1+cosx)/|sinx|
1).x∈(-π/2,0),在第四象限,则cosx为正,sinx为负
则f(x)=cosx * (1+sinx)/cosx - sinx*(1+cosx)/sinx=1+sin(x)-1-cos(x)=sin(x)-cos(x)
当x=-π/4时,f(-π/4)=-√2
2)x∈(π/2,π),在第二象限,则cosx为负,sinx也为正
则f(x)= -1-sinx+1+cosx = cosx -sinx=√2sin(x-π/4)
则f(x)的值域是[-√2,√2]