已知cos(a+b)cos(π/4 -b)-sin(a+b)sin(π/4 -b)=-4/5,且π/4<a<3π/4求sin2a 与sina的值

cos(a+b)cos(π/4 -b)-sin(a+b)sin(π/4 -b)
=cos[(a+b)+(π/4-b)]
=cos(a+π/4)=-4/5
cosacosπ/4-sinasinπ/4=-4/5
cosa-sina=-4√2/5
两边平方
cos²a+cos²a-2sinacosa=32/25
1-2sinacosa=32/25
sin2a=2sinacosa=-7/25

sina-cosa=4√2/5
-sinacosa=7/50
即sina+(-cosa)=4√2/5
sina*(-cosa)=7/50
所以sina和-cosa是方程x²-4√2x/5+7/50=0的根
(x-√2/5)(x-7√2/5)=0
x=√2/5,x=7√2/5
π/4<a<3π/4
所以sinπ/4<sina<sinπ/2
√2/2<sina<1
所以sina=7√2/5

cos(a+b)cos(π/4 -b)-sin(a+b)sin(π/4 -b)
=cos(a+b+π/4 -b)
=cos(a+π/4 )
=cosacosπ/4-sinasinπ/4
=√2/2*(cosa-sina)
=-4/5
sina-cosa=4√2/5
(sina-cosa)^2=1-2sinacosa=1-sin2a=(4√2/5)^2=32/25
sin2a=1-32/25=-7/25

设sina=x
则：√(1-x^2)=x-4√2/5
1-x^2=x^2+32/25-8√2x/5
50x^2-40√2x+7=0
解出x 便是sina