已知cos(π/6-a)=1/3,则cos(5π/6+a)sin(2π/3-a=?

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已知cos(π/6-a)=1/3,则cos(5π/6+a)sin(2π/3-a=?已知cos(π/6-a)=1/3,则cos(5π/6+a)sin(2π/3-a=?已知cos(π/6-a)=1/3,则

已知cos(π/6-a)=1/3,则cos(5π/6+a)sin(2π/3-a=?
已知cos(π/6-a)=1/3,则cos(5π/6+a)sin(2π/3-a=?

已知cos(π/6-a)=1/3,则cos(5π/6+a)sin(2π/3-a=?
cos ( 5π/6+a ) = cos ( π - (5π/6+a) ) = cos ( π/6-a ) = 1/3
sin (2π/3-a) = cos (π /2 - (2π/3-a) ) = cos (π/6-a) =1/3
所以 原式= 1/9