1、cos(2π/3－θ)＝sin[π/2－(2π/3－θ)]＝sin(﹣π/6＋θ)＝﹣sin(π/6－θ)＝﹣m
2、∵半径为1,弧长8π/3 ∴转过的角度为：8π/3
∴cos(﹣8π/3)＝cos(﹣2π/3)＝cos(2π/3)＝﹣cos(π/3)＝﹣1/2
∴sin(﹣8π/3)＝sin(﹣2π/3)＝﹣sin(2π/3)＝﹣sin(π/3)＝﹣√3/2
∴Q点的坐标是(﹣1/2,﹣√3/2 )
1、已知sin(π/6-θ)=m,则cos(2π/3-θ)=
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