已知2(1 sinβ∧2)=3cosα∧2，3(cosα sinα)∧2＝1 2(sinβ cosβ)∧2求cos(2α 2β)

显示全部楼层 · 2013-6-23 20:09:39

2(1+sin2β)=3cos2α，3(cosα+sinα)2＝1+2(sinβ+cosβ)2；由第二式：3+3sin2α=3+2sin2β，化简得 sin2α=(2/3)sin2β；由第一式：2+(1-cos2β)=(3/2)(1+cos2α)，化简得 cos2α=1-(2/3)cos2β；∴ sin22α+cos22α=(4/9)sin22β+[1-(2/3)cos2β]2；即 1=1+1-(4/3)cos2β，∴ cos2β=3/4；cos(2α+2β)=cos2αcos2β-sin2αsin2β=[1-(2/3)cos2β]*cos2β-(2/3...