√1/2+1/2√1/2+1/2cos2α﹙∏<α<3∏/2﹚化简
求证:cos2αcos2β=1/2{cos2(α+β)+cos2(α-β)}
(sin2α-cos2α+1)/(1+tanα)=2sin2αcos2α 为什么
(2sin2α/1+cos2α)*(cosα)^2/cos2α=?
化简(sinα-cosα)^2-1/-cos2α
化简:sin^2αsin^2β+cos^2αcos^2β-1/2cos2αcos2β
化简(sinα)^2*(sinβ)^2+(cosα)^2(cosβ)^2-1/2cos2αcos2β
化简sinα^2sinβ^2+cos^2cosβ^2-1/2cos2αcos2β
化简:sin²αsin²β+cos²αcos²β-1/2cos2αcos2β
化简:sin²αsin²β+cos²αcos²β—1/2cos2αcos2β
化简 sin²αsin²β+cos²αcos²β-1/2(cos2αcos2β)
化解sin^2αsin^2β+cos^2αcos^2β-1/2cos2αcos2β
已知tan(α+β)/2=1/2,求cos2αcos2β-{cos(α-β)}^2的值