化简sin^2asin^2b+cos^2acos^2b-1/2cos2acos2brt

化简sin^2acos^b-cos^2asin^2b+cos^a-cos^2b 在三角形ABC中,cos A cos B+cos Asin B+sin Asin B=2,则三角形ABC是 化简sin^2asin^2b+cos^2acos^2b-1/2cos2acos2brt 化简sin²Asin²B+cos²Acos²B-1/2cos2Acos2B 求证：sin^2acos^2b-cos^2asin^2b=cos^2b-cos^2a 求证：sin^2A+sin^2B-sin^2Asin^2B+cos^2Acos^2B=1 sin acos b=1/2,则cos asin b的取值范围 已知4sin asin b=根号2,4cos a cos b=根号6,则cos 2a+cos 2b的值是? 化简：sin^2a+sin^2β-sin^2asin^2β+cos^2acos^2β 在△ABC中 若sin^2Acos^2B-cos^2Asin^在△ABC中 若sin^2Acos^2B-cos^2Asin^2B=sin^2c求△ABC的形状、 化简cos^2x-sin^2x+2倍根3化简成Asin（）+~的形式 求证：sin^2A+sin^2B-sin^2Asin^2B+cos^2Acos^2B=1题中的^2为平方. 1-1/4sin^2 2A-sin^2B-cos^4Asin(A+B)=3/5sin(A-B)=-4/5 在△ABC中 若sin^2Acos^2B-cos^2Asin^2B=sin^2c,判断△ABC的形状.但不知错哪了,sin^2Acos^2B-cos^2Asin^2B=sin^2c=sin^2(π-(A+B))=sin^2(A+B)=(sinAcosB+sinBcosA)^2=sin^2Acos^2B+2sinAcosBsinBcosA+sin^2Bcos^2A约分得,2cos^2Asin^2B+2sinAcosBsi 化简sin^2asin^2b+cos^2acos^2b-1/2(cos2acos2b)=感觉这题有点怪啊 y=sin平方x-1/2怎么化成y=Asin（ax+b)或者cos的 化简cos(a+b)cos(a-b)+sin^2b 将f(x)=2sin^4 x+2cos^4 x+cos^2 x-3化简化成y=Asin(wx+b)+c形式