化简 sin²αsin²β+cos²αcos²β-1/2(cos2αcos2β)

问题描述:

化简 sin²αsin²β+cos²αcos²β-1/2(cos2αcos2β)

我做了一下,1/2是对滴~~
zxqsyr 想复杂了。没必要吧~。你随便找一个1/2的看看过程,看不懂hi我,很乐意为您解答疑惑。(if i can^)写一遍解题过程太累了。三角函数的题要自己玩味滴,一开始做是有恐惧感,多做些题就好了

sin²αsin²β+cos²αcos²β-1/2(cos2αcos2β)
= sin²αsin²β+cos²αcos²β+2 sinαsinβcosαcosβ-2sinαsinβcosαcosβ-1/2(cos2αcos2β)
= (sinαsinβ+cosαcosβ)²+1/2sin2αsin2β-1/2(cos2αcos2β)
= cos²(α-β)+1/2(sin2αsin2β-cos2αcos2β)
= cos²(α-β)-1/2(cos2αcos2β-sin2αsin2β)
= cos²(α-β)-1/2cos(2α+2β)
=[1+ cos(2α-2β)]/2-1/2cos(2α+2β)
=1/2+1/2 cos(2α-2β)-1/2cos(2α+2β)
=1/2+1/2 [cos(2α-2β)-cos(2α+2β)]
=1/2+1/2 [cos(2α-2β)-cos(2α+2β)]
=1/2+1/2*(-2) sin[(2α-2β)+(2α+2β)]sin[(2α-2β)-(2α+2β)]
=1/2- sin(2α-2β+2α+2β)sin(2α-2β-2α-2β)
=1/2- sin(4α)sin(-4β)
=1/2+ sin4αsin4β

sin²αsin²β+cos²αcos²β-1/2(cos2αcos2β)=sin²αsin²β+cos²αcos²β-1/2(cos²α-sin²α)(cos²β-sin²β)=sin²αsin²β+cos²α...