若log2/1(sin a)+log2/1(sin b)=2,(27^cos a)^cos b=1/9,cos(2a+2b)=?

题目

若log2/1(sin a)+log2/1(sin b)=2,(27^cos a)^cos b=1/9,cos(2a+2b)=?

答案

log2/1(sin a)+log2/1(sin b)=2
sina*sinb=-1
(27^cos a)^cos b=1/9
3^(3cosacosb)=3^(-2)
cosacosb=-2/3
cos(a+b)=cosacosb-sinasinb=-2/3+1=1/3
cos(2a+2b)=cos2(a+b)=2[cos(a+b)]^2-1
=2*(1/3)^2-1=-7/9