急~~~~~一道数学题：已知函数f(t)=√[(1-t)/(1+t)],g(x)=cosx*f(sinx)+sinx*f(cosx),x属于（π，17π/12）

f(t) = [(1-t)/(1+t)]^(1/2),
定义域，
(1 - t)/(1 + t) >= 0, t 不等于 -1.
(1-t)(1+t) >= 0,

-1 < t <= 1.

x属于（π，17π/12）时，
-1 < sinx < 0, -1 < cosx < 0.

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

= {(1-sinx)^2/[1-(sinx)^2]}^(1/2)

= {[(1-sinx)/cosx]^2}^(1/2)

= -(1-sinx)/cosx

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

= {(1-cosx)^2/[1-(cosx)^2]}^(1/2)

= {[(1-cosx)/sinx]^2}^(1/2)

= -(1-cosx)/sinx

g(x) = cosx*f(sinx)+sinx*f(cosx)

= -(1-sinx) -(1-cosx)

= -2 + sinx + cosx

= 2^(1/2)[sinxcos(π/4) + cosxsin(π/4)] - 2

= 2^(1/2)sin(x + π/4) - 2

又，
x属于（π，17π/12）时，
x + π/4 属于（5π/4，5π/3）.

5π/4 = π + π/4,
5π/3 = π + π/2 + π/6,

-2^(-1/2) = sin(5π/4) > sin(x + π/4) > sin(3π/2) = -1

-1 > 2^(1/2)sin(x + π/4) > -2^(1/2)

-3 > 2^(1/2)sin(