..sin^2(π+α)*cos(π+α)*cos(-α-2π)/tan(π+α)*sin^3(π/2+α)*sin(-α-2π)
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bmLD i)95<-e_\g2 nN
α∈(0,π/2 ),比较 sin(cosα) 与cos(sinα)大小
sin(α-π)=2cos(α-2π)求cos平方α-sin平方αcosα
设α∈(0,π/2),则cosα,sin(cosα),cos(sinα)的大小关系为什么?
设角a=-35π/6,则2sin(π+α)cos(π-α)-cos(π+α)/ 1+sin^2α+sin(π-α)-cos^2(π+α)sina=sin(-35π/6)=sin(-6π+π/6)=1/2 cosα=根号3/22sin(π+α)cos(π-α)-cos(π+α)/ [1+sin^2α+sin(π-α)-cos^2(π+α)]=2(-sinα)(-cosα)+cosα/[1+sin²
若cosα>sinα,(-π/2
已知6sin²α+sinαcosα-2cos²α=0,α∈(π/2,π),求值1).(sinα-3cosα)/(sinα-cosα);2).sinαcosα-sin²α;3).sin²α-3cosαsinα-2
化简[sin(π/2-α)cos(π/2-α)]/cos(π+α)-[sin(π-α)cos(π/2-α)]/sin(π+α)
sin(α-β)sin(β-r)-cos(α-β)cos(r-β) 1.sin(α-β)sin(β-r)-cos(α-β)cos(r-β)2.( tan4分之5π+tan12分之5π)/(1-tan12分之5π)3.[ sin(α+β)-2sinαcosβ]/2sinαsinβ+cos(α+β)
【三角函数】已知sin(α-π)=2cos(2π-α),求[sin(π-α)+5cos(2π-α)]/[3cos(π-α)-sin(-α)]
已知α∈(π,2π) sinα+cosα=1/5 求sinα*cosα sinα-cosα
若sin(α-π)=2cos(2π-α),求(sinα+5cosα)/(sinα-3cosα)的值,
已知tan(3π+α)=2,求:1、(sinα+cosα)²;2、sinα-cosα/2sinα+cosα
已知tan(π-α)=2,求sin²α-2sinαcosα-cos²α/4cos²α-3sin²α+1
已知tan(π-α)=2,求sinα-2sinαcosα-cosα/4cosα-3sinα的值
若(sinα+cosα)/(sinα-cosα)=2 则sin(α-5π)·sin(3π/2-α)=?
若sinα+cosα/sinα-cosα=2,则sin(α-5π)*sin(3π/2-α)等于?
化简:2sin(π+α)cos(π-α)
sin(π+α)+cos(5π-α))/(sin(α-6π)+2cos(2π+α)=(-sinα-cosα)/(sinα+2cosα)为什么相等