sin^2(π+α)-cos(π+α)cos(-α)+1= 2

若cosα>sinα,(-π/2 利用和差角公式化简 (2)sin(π/3+α)+sin(π/3-α)(2)sin(π/3+α)+sin(π/3-α)(3)cos(π/4+α)-cos(π/4-α)(4)cos(60°+α)+cos(60°-α)(5)sin(α-β)cosβ+cos(α-β)sinβ(6)cos(α+β)cosβ+sin(α+β)sinβ α∈（0,π／2 ）,比较 sin(cosα) 与cos(sinα)大小 比较大小sin(cosα)与cos(sinα)（0＜α＜π/2) 化简:sin(2π-α)cos(π+α)cos(11π/2-α)/cos(π-α)sin(3π-α)sin(-π-α)sin(9π/2+α) cos^2(π/4+α)=(cos(π/4)cosα-sin(π/4)sinα)² cos^2(π/4+α)是怎么等于(cos(π/4)cosα-sin(π/4)sinα)²的? 若α∈（-π/2+2kπ,2kπ）（k∈Z),则sinα,cosα,tanα的大小关系是A.tanα＞sinα＞cosαB.tanα＞cosα＞sinαC.tanα＜sinα＜cosαD.tanα＜cosα＜sinα [sin(5π-α)cos(-π-α)]/[cos(α-π)cos(π/2+α)]化简 化简sin(-π/2-α)sin(πα)cos(-α-π)/cos(π-α)sin(3π α) 求值【sin(2π-α)sin(π+α)cos(-π-α)】/【sin(3π-α)-cos(π-α)】 sin(α-π)=2cos（α-2π）求cos平方α-sin平方αcosα (1+sinα+cosα)*[sin(α/2)-cos(α/2)]/√(2+2cosα)化简 （3/2*π cosαcos[(2k+1)π]-sinαsin[(2k+1)π]为什么等于-cosα 设α∈（0,π/2）,则cosα,sin(cosα),cos(sinα)的大小关系为什么? 化简（sin α/2 + cos α/2）² + 2sin²（π/4-α/2）A 2+sin α B 2C 2+sin α-cosαD 2+sinα+cosα 我怎么算都是 1+sinα+cosα! 已知α,β,c∈（0,π/2）,sinα+sinβ=sinc,cosβ+cosc=cosα,求β-α的值 sin(π-α)=-2sin(π/2+α),则sinαcosα等于 sin(π-α)=-2sin(π/2+α),求sinα·cosα