已知sin(π/3 +α)=-1/3,则cos(7π/6 -α)=?

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已知sin(π/3+α)=-1/3,则cos(7π/6-α)=?已知sin(π/3+α)=-1/3,则cos(7π/6-α)=?已知sin(π/3+α)=-1/3,则cos(7π/6-α)=?cos(

已知sin(π/3 +α)=-1/3,则cos(7π/6 -α)=?
已知sin(π/3 +α)=-1/3,则cos(7π/6 -α)=?

已知sin(π/3 +α)=-1/3,则cos(7π/6 -α)=?
cos(7π/6 -α)
=cos(π+π/6 -α)
=-cos(π/6 -α)
=-cos(π/2-π/3-α)
=-sin(π/3+α)
=-(-1/3)
=1/3