sin(π/6-a)=cos(π/2-π/6+a)
=cos(π/3+a)=1/3
sin(π/3+a)=√(1-1/9 )=+(-)2√2/3
cos(2π/3+a)=cos[π/3+(π/3+a)]
=cosπ/3cos(π/3+a)-sinπ/3sin(π/3+a)
=1/2cos(π/3+a)-√3[sin(π/3+a)]/2
=1/2*1/3-√3/2*[+(-)2√2/3]
=1/6-(+)√6/3
sin(π/6-a)=cos(π/2-π/6+a)
=cos(π/3+a)=1/3
sin(π/3+a)=√(1-1/9 )=+(-)2√2/3
cos(2π/3+a)=cos[π/3+(π/3+a)]
=cosπ/3cos(π/3+a)-sinπ/3sin(π/3+a)
=1/2cos(π/3+a)-√3[sin(π/3+a)]/2
=1/2*1/3-√3/2*[+(-)2√2/3]
=1/6-(+)√6/3