化简:[ cos(α-π/2)/sin(5π/2+α) ] ×sin(α-2π) × cos(2π-α)
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化简:[ cos(α-π/2)/sin(5π/2+α) ] ×sin(α-2π) × cos(2π-α)
RT.
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![化简:[ cos(α-π/2)/sin(5π/2+α) ] ×sin(α-2π) × cos(2π-α)](/uploads/image/z/6407760-48-0.jpg?t=%E5%8C%96%E7%AE%80%EF%BC%9A%5B+cos%28%CE%B1-%CF%80%2F2%29%2Fsin%285%CF%80%2F2%2B%CE%B1%29+%5D+%C3%97sin%28%CE%B1-2%CF%80%29+%C3%97+cos%EF%BC%882%CF%80-%CE%B1%EF%BC%89)
原式=sinα/cosα *sinα*cosα
=sin^2α
=sin^2α
α∈(0,π/2 ),比较 sin(cosα) 与cos(sinα)大小
化简:[ cos(α-π/2)/sin(5π/2+α) ] ×sin(α-2π) × cos(2π-α)
已知(3sinα+cosα)/(3cosα-sinα)=2,则2-3sin(α-3π)sin(1.5π-α)-[cos(
化简[sin(π/2-α)cos(π/2-α)]/cos(π+α)-[sin(π-α)cos(π/2-α)]/sin(π
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已知α∈(π,2π) sinα+cosα=1/5 求sinα*cosα sinα-cosα
若sinα+cosαsinα−cosα=2,则sin(α-5π)•sin(3π2-α)等于( )
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比较大小sin(cosα)与cos(sinα)(0<α<π/2)
sin(π/2+α)·cos(π/2-α)/cos(π+α)+sin(π-α)·cos(π/2+α)/sin(π+α)=