√1/2+1/2√1/2+1/2cos2α﹙∏＜α＜3∏/2﹚化简

求证：cos2αcos2β=1/2{cos2（α+β）+cos2(α-β)} 化简√(2+cos2+sin^2 1) sin²α*sin²β+cos²α*cos²β-1/2cos2αcos2β=（1-cos2α）/2*（1-cos2β）/2+（1+cos2α）/2*（1+cos2β）/2-1/2cos2α*cos2β=1/4（1+cos2α*cos2β-cos2α-cos2β）+1/4（1+cos2α*cos2β+cos2α+cos2β）-1/2cos2α*cos2β=1/4+1/ cos2次方1°加cos2次方2°加cos2次方3°.cos2次方88°＋cos2次方89° 化简√ (2+ cos2-sin平方1)的结果是 1+cos2θ+2sin²θ=2 1-cos2θ/1+cos2θ=tan²θ 化简2sin2a/1+cos2a*cos2/cos2a cos(1+2)等于cos1+cos2吗? 根号下1减sin(π+2)cos(π+2)化简是多少啊A sin2-cos2 B cos2-sin2 C +-(sin2-cos2) D sin2+cos2 (sin2α-cos2α+1)/(1+tanα)=2sin2αcos2α 为什么 (2sin2α/1+cos2α)*(cosα)^2/cos2α=? (2sin2α/1+cos2α)*(cosα)^2/cos2α=多少 (2sin2α/1+cos2α)*(cos^2/cos2α)=? 化简 [2sin2α/(1+cos2α)]*[cos^2α/cos2α] ((1+cos2α)/(3sin2α))*((2sin^2(α))/cos2α)的化简 ((1+cos2α)/(3sin2α))*((2sin^2(α))/cos2α)的化简 急 ((1+cos2α)/(3sin2α))*((2sin^2(α))/cos2α)的化简 求证1+2cos2α-cos2α=2 好多年没碰了