若sin(π/6-α)=1/3 则cos(2π/3+α)=?
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若sin(π/6-α)=1/3则cos(2π/3+α)=?若sin(π/6-α)=1/3则cos(2π/3+α)=?若sin(π/6-α)=1/3则cos(2π/3+α)=?sin(π/6-α)=1/
若sin(π/6-α)=1/3 则cos(2π/3+α)=?
若sin(π/6-α)=1/3 则cos(2π/3+α)=?
若sin(π/6-α)=1/3 则cos(2π/3+α)=?
sin(π/6-α)=1/3
则：cos²(π/6-α)=1-sin²(π/6-α)=8/9所以,cos(π/6-α)=-2√2/3或2√2/3cos(2π/3+α)=cos[5π/6-(π/6-α)] =(-√3/2)cos(π/6-α)+(1/2)sin(π/6-α)当cos(π/6-α)=-2√2/3时,上式=(1+2√6)/6；
当cos(π/6-α)=2√2/3时,上式=(1-2√6)/6
所以,cos(2π/3+α)=(1+2√6)/6或cos(2π/3+α)=(1-2√6)/6
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