半角公式题目

解:
1.由于:
-3pi/2<x<-pi
则:X为第二象限角
则:sinx>0,cosx<0
又cosx=-4/5
则:sinx=根号[1-cos^2(x)]=3/5

[sin(pi/2-2x)-2sin^2(pi/4-x)]/[1+tan(pi/4-x)]

={cos2x-2*[1-cos(pi/2-2x)]/2}/[1+tan(pi/4-x)]

={cos2x-1+sin2x}/[1+tan(pi/4-x)]

={cos2x+sin2x-1}/{1+[tan(pi/4)-tanx]/[1+tan(pi/4)*tanx]}

={sin2x+cos2x-1}/[1+(1-tanx)/(1+tanx)]

=[sin2x+cos2x-1]/[2/(1+tanx)]

=(sinx+cosx)[2sinxcosx+2cos^2(x)-2]/2cosx

=-21/100

2.
cos(a+b)cos(pi/4-b)-sin(a+b)sin(pi/4-b)=-4/5
则:cos[(a+b)+(pi/4-b)]=-4/5
cos(a+pi/4)=-4/5
由于：a属于(pi/4,3pi/4)
则：(a+pi/4)属于(pi/2,pi)
则：sin(a+pi/4)>0
则：sin(a+pi/4)
=根号[1-cos^2(a+pi/4)]
=3/5
则：tan(a+pi/4)
=sin(a+pi/4)/cos(a+pi/4)
=-3/4
则：
[tana+tan(pi/4)]/[1-tan(pi/4)tana]=-3/4
(tana+1)/(1-tana)=-3/4
则：tana=-7=sina/cosa
cosa<0
又(sina)^2+(cosa)^2=1
则：sina=7