已知cos(α-π/3)=2/3,求cos(2π/3+α)+tan^2(4π/3-α)的值
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已知cos(α-π/3)=2/3,求cos(2π/3+α)+tan^2(4π/3-α)的值
![已知cos(α-π/3)=2/3,求cos(2π/3+α)+tan^2(4π/3-α)的值](/uploads/image/z/19865500-52-0.jpg?t=%E5%B7%B2%E7%9F%A5cos%28%CE%B1-%CF%80%2F3%29%3D2%2F3%2C%E6%B1%82cos%282%CF%80%2F3%2B%CE%B1%29%2Btan%5E2%284%CF%80%2F3-%CE%B1%29%E7%9A%84%E5%80%BC)
已知cos(α-π/3)=2/3,求cos(2π/3+α)+tan^2(4π/3-α)=cos(π-π/3+α)+tan^2(π+π/3-α)
=-cos(α-π/3)+sin^2(α-π/3)/cos^2(α-π/3)=-2/3+5/4=7/12
=-cos(α-π/3)+sin^2(α-π/3)/cos^2(α-π/3)=-2/3+5/4=7/12
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