化简sin(-α)cos(2π+α)/cos(α-π) 要步骤,谢谢
已知0<α<π,sinα+cosα=1\5,求sinα-cosα的值 我很急,写一下步骤,谢谢
(1+sinα+cosα)*[sin(α/2)-cos(α/2)]/√(2+2cosα)化简 (3/2*π
化简{sin(π+α)^3cos(-α)cos(π+α)}/{tan(π+α)^3cos(-π-α)^2}
化简sin(α-5π)/cos(3π-α)×cos(π/2-α)/sin(α-3π)×cos(8π-α)/sin(-α-
化简[sin(π/2-α)cos(π/2-α)]/cos(π+α)-[sin(π-α)cos(π/2-α)]/sin(π
化简:sinα-√3 cosα+2cos(α-π/3)
1.化简sin²α+cosαcos(π/3+α)-sin²(π/6-α)
比较大小sin(cosα)与cos(sinα)(0<α<π/2)
α∈(0,π/2 ),比较 sin(cosα) 与cos(sinα)大小
化简:sin^2αcos^2β-cos^2αsin^2β+cos^2α-cos^2β具体步骤,
已知3sinα+cosα=0,求sin²α+2sinα*cosα-3cos²α的值,要求有步骤
化简:[ cos(α-π/2)/sin(5π/2+α) ] ×sin(α-2π) × cos(2π-α)