1、已知sin(π/6-θ)=m,则cos(2π/3-θ)=2、点P从点(1,0)出发,沿单位圆X平方+Y平方=1按顺时针方向运动了8π/3弧长到达Q点,则Q点的坐标是()

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1、已知sin(π/6-θ)=m,则cos(2π/3-θ)=2、点P从点(1,0)出发,沿单位圆X平方+Y平方=1按顺时针方向运动了8π/3弧长到达Q点,则Q点的坐标是()
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1、已知sin(π/6-θ)=m,则cos(2π/3-θ)=2、点P从点(1,0)出发,沿单位圆X平方+Y平方=1按顺时针方向运动了8π/3弧长到达Q点,则Q点的坐标是()
1、已知sin(π/6-θ)=m,则cos(2π/3-θ)=
2、点P从点(1,0)出发,沿单位圆X平方+Y平方=1按顺时针方向运动了8π/3弧长到达Q点,则Q点的坐标是()

1、已知sin(π/6-θ)=m,则cos(2π/3-θ)=2、点P从点(1,0)出发,沿单位圆X平方+Y平方=1按顺时针方向运动了8π/3弧长到达Q点,则Q点的坐标是()
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 )

sin(π/6-θ)=m,
cos(2π/3-θ)
=cos(π/2+π/6-θ)
=cos[π/2+(π/6-θ)]
=-sin(π/6-θ)
=-m
按顺时针方向运动了8π/3
即Q点在第一象限,
x=cosπ/3=1/2
y=sinπ/3=√3/2
则Q点的坐标是(1/2,√3/2)