Question

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

Answer 1

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