cos(π/3＋a)=cos[π/2－(π/6－a)]=sin(π/6－a)=－1/3
注：利用的是cos(π/2－a)=sina
cos(π/3+a)=__-1/3___
cos(π/3+a)=cos(-π/3-a)=
=sin(-π/3-a+π/2)
=-sin(π/6-a)
=-1/3
sin(π/6-a)= -1/3
则cos(π/3+a)=sin[π/2-(π/3+a)]=sin(π/6-a)=-1/3
cos(2π/3+2a)
=cos[2(π/3+a)]
=2cos²(π/3+a)-1
=2sin²[π/2-(π/3+a)]-1
=2sin²(π/6-a)-1
cos(π/3+a)= sin[π/2 - (π/3 +a)] =sin(π/6-a) =-1/3
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