化简cos(α-π/2)/sin(α+5π/2)*sin(α-2π)*cos(2π-α),因为我刚学,所以要具体点的过程
化简cos(α-π/2)/sin(α+5π/2)*sin(α-2π)*cos(2π-α),因为我刚学,所以要具体点的过程
cos(π/2+α)=-sinα转换过程
化简sin(α-5π)/cos(3π-α)×cos(π/2-α)/sin(α-3π)×cos(8π-α)/sin(-α-
(1+sinα+cosα)*[sin(α/2)-cos(α/2)]/√(2+2cosα)化简 (3/2*π
化简:[ cos(α-π/2)/sin(5π/2+α) ] ×sin(α-2π) × cos(2π-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
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
已知tan(π-α)=2,求sinα-2sinαcosα-cosα/4cosα-3sinα的值
已知cos(π/2+α)=sin(α-π/2) 求sin(π-α)-cos(π+α)/cos(5π/2 -α)+2sin
已知 sinα+2cos(5π/2+α)/cos(π-α)-sin(π/2-α)=-1/4 求(sinα+cosα)平方