若sin(π-α)+cos(2π-α)＝1 ,则sin4α+cos4α+sin(π+α)·cos(4π-α)的值为_________.★ 试题详解：∵ 已知化为sinα+cosα＝1,则 sinα·cosα＝0 原式＝sin4α+cos4α-sinαcosα＝(sin2α+cos2α)2-2sin2αcos2α-sinαcosα＝1为什

若cosα>sinα,(-π/2 若sin(α-π)=2cos(2π-α),求（sinα+5cosα）/（sinα-3cosα）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? 若(sinα+cosα)/(sinα-cosα)=2,求sin(α-5π)sin(3π/2-α)的值 若cosα/sin(α-π/4)=-根号2/2,则cosα+sinα的值为、 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) 若sin(α-π）＝2cos（2π-α),求[sin(π－α）＋5cos（2π－α）]／[3cos（π－α）－sin（－α）]的值 若sin(π-α)+cos(2π-α)＝1 ,则sin4α+cos4α+sin(π+α)·cos(4π-α)的值为_________.★ 试题详解：∵ 已知化为sinα+cosα＝1,则 sinα·cosα＝0 原式＝sin4α+cos4α-sinαcosα＝(sin2α+cos2α)2-2sin2αcos2α-sinαcosα＝1为什 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? [sin(5π-α)cos(-π-α)]/[cos(α-π)cos(π/2+α)]化简 若sinα=1/2,则sin(π/4-α)cos(π/4-α)= 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