已知tan(π/4-a)=1/3,a属于(0,π/4).(1)求(a)=(sin2a-2cos^2a)/(1+tana)的值.

tan(π/4-a)=1/3
-tan(a-π/4)=1/3
tan(a-π/4)=-1/3
(tana-tanπ/4)/(1+tanπ/4tana)=-1/3
(tana-1)/(1+tana)=-1/3
3(tana-1)=-1-tana
3tana-3=1+tana
4tana=2
tana=1/2
(sin2a-2cos^2a)/(1+tana)
=(sin2a-2cos^2a)/(1+1/2)
=(2sinacosa-2cos^2a)/(3/2)
=(2sinacosa-2cos^2a)/[3/2*(sin^2a+cos^2a)]分子分母同时除以cos^2a
=(2sinacosa/cos^2a-2cos^2a/cos^2a)/[3/2*(sin^2a/cos^2a+cos^2a/cos^2a)]
=(2sinacosa/cos^2a-2cos^2a/cos^2a)/[3/2*(sin^2a/cos^2a+cos^2a/cos^2a)]
=(2tana-2)/[3/2*(tan^2a+1)]
=(2*1/2-2)/{3/2*[(1/2)^2+1]}
=-1/{3/2*5/4}
=-1/{15/8}
=-8/15
a∈(0,π/4)
a-π/4∈(-π/4,0)
tan(a-π/4)=-1/3,
sin(a-π/4)=-√10/10
cos(a-π/4)=3√10/10
b∈(0,π/2)
b+3π/4∈(3π/4,5π/4)
sin(3π/4+b)=-√5/5
b+3π/4∈(π,5π/4)
cos(3π/4+b)=-2√5/5
a+b∈(0,3π/4)
sin[(3π/4+b)+(a-π/4)]
=sin(3π/4+b+a-π/4)
=sin(π/2+a+b)
=cos(a+b)
sin[(3π/4+b)-(a-π/4)]
=sin(3π/4+b)cos(a-π/4)-cos(3π/4+b)sin(a-π/4)
=-√5/5*3√10/10-(-2√5/5)*(-√10/10)
=-15√2/50-10√2/50
=-1/2
cos(a+b)=-1/2
a+b=2π/3