=Sin2α＋sin2β-4Sinαcosαsinβ cosβ+(cos^α-Sin^α)× (cos^β-Sin^β)
=Sin2α+sin2β+(cosαcosβ-SinαSinβ)^-(cosαSinβ+Sinαcosβ)^
=Sin2α+sin2β+cos^(α+β)-Sin^(α+β)
=1+Sin2α+sin2β-2Sin^(α+β)
=1+2(Sinαcosα+Sinβcosβ-Sinαcosα-Sinβcosβ)
=1
化简sin2α+sin2β-sin2αsin2β+cos2αcos2β
=Sin2α+sin2β+cos^(α+β)-Sin^(α+β)=1+Sin2α+sin2β-2Sin^(α+β)=1+2(Sinαcosα+Sinβcosβ-Sinαcosα-Sinβcosβ)=1
sin2^α+sin2^β-sin^2αsin2^β+cos^2αcos2^β 化简
答案是1sin�0�5α+sin�0�5β-sin�0�5αsin�0�5β+cos�0�5αcos�0�5β=sin�0�5α(1-sin�0�5β)+cos�0�5αc...
化简sin^2α+sin^2β-sin^2αsin^2β+cos^2cos^2β
化简sin²α+sin²β-sin²αsin²β+cos²αcos²β 解：原式=sin²α+sin²β-sin²αsin²β+(1-sin²α)(1-sin²β)=sin²α+sin²β-sin²αsin²β+(1-sin²α-sin²β+sin&#...
化简sin平方α+sin平方β-sin平方αsin平方β+cos平方αcos平方β
sin^2α+sin^2β-sin^2αsin^2β+cos^2αcos^2β =sin^2α+sin^2β-(1-cos^2α)(1-cos^2β)+cos^2αcos^2β =sin^2α+sin^2β-1+cos^2α+cos^2β-cos^2αcos^2β+cos^2αcos^2β =sin^2α+sin^2β+cos^2α+cos^2β-1 =1+1-1 =1 ...
化简:sin^2α+sin^2β+cos^2αcos^2β=
原式=1-cos^2α+sin^2β+cos^2αcos^2β =cos^2α(cos^2β-1)+sin^2β+1 =-cos^2αsin^2β+sin^2β+1 =sin^2β(1-cos^2α)+1 =sin^2αsin^2β+1
化简:sin²αsin²β+cos²αcos²β-½cos2αcos2β
sin²αsin²β+cos²αcos²β-(1\/2)cos2αcos2β =sin²αsin²β+cos²αcos²β-(1\/2)(1-2sin²α)(1-2sin²β)=sin²αsin²β+cos²αcos²β-(1\/2)(1-2sin²α-2sin²β+4sin&...
证明sin^2α+sin^2β-sin^2α×sin^2β+cos^2α×cos^2β=1
(sinα)^2+(sinβ)^2-(sinαsinβ)^2+(cosαcosβ)^2 =(1-cos2α)\/2+(1-cos2β)\/2+[cosαcosβ+sinαsinβ][cosαcosβ-sinαsinβ =1-1\/2(cos2α+cos2β)+cos(α-β)cos(α+β) =1-1\/2*2cos(α-β)cos(α+β)+cos(α-β)cos(α+β) =1 求采纳 ...
证明:sinα^2+sinβ^2-sinα^2sinβ^2+cosα^2cosβ^2=1
sinα^2+sinβ^2-sinα^2sinβ^2+cosα^2cosβ^2=sinα^2(1-sinβ^2)+cosα^2cosβ^2+sinβ^2=sinα^2cosβ^2+cosα^2cosβ^2+sinβ^2=(sinα^2+cosα^2)cosβ^2+sinβ^2=cosβ^2+sinβ^2=1
急求化简(sinα)^2+(sinβ)^2-(sinα)^2(sinβ)^2+(cosα)^2(cosβ...
(sinα)^2+(sinβ)^2-(sinα)^2(sinβ)^2+(cosα)^2(cosβ)^2 =(sinα)^2(1-(sinβ)^2)+(cosα)^2(cosβ)^2+(sinβ)^2 =(sinα)^2(cosβ)^2+(cosα)^2(cosβ)^2+(sinβ)^2 =((sinα)^2+(cosα)^2)(cosβ)^2+(sinβ)^2 =(cosβ)^2+(sinβ)^2 ...
化解sin^2αsin^2β+cos^2αcos^2β-1\/2cos2αcos2β
是不是：sin^2αsin^2β+cos^2αcos^2β-1\/（2cos2αcos2β）1\/（2cos2αcos2β）还是1\/2cos2α*cos2β?
