[sin4α/(1+cosα)]×[cos2α/(1+cos2α)]×[cosα/(1+cosα)]×{cos(α/2)/[1+cos(α/2)]}=?应该是[sin4α/(1+cos4α)]×[cos2α/(1+cos2α)]×[cosα/(1+cosα)]×{cos(α/2)/[1+cos(α/2)]}=？打错了。对不起。

[sin4α/(1+cosα)]×[cos2α/(1+cos2α)]×[cosα/(1+cosα)]×{cos(α/2)/[1+cos(α/2)]}=?应该是[sin4α/(1+cos4α)]×[cos2α/(1+cos2α)]×[cosα/(1+cosα)]×{cos(α/2)/[1+cos(α/2)]}=？打错了。对不起。
[sin4α/(1+cosα)]×[cos2α/(1+cos2α)]×[cosα/(1+cosα)]×{cos(α/2)/[1+cos(α/2)]}=?
应该是[sin4α/(1+cos4α)]×[cos2α/(1+cos2α)]×[cosα/(1+cosα)]×{cos(α/2)/[1+cos(α/2)]}=？
打错了。对不起。

[sin4α/(1+cosα)]×[cos2α/(1+cos2α)]×[cosα/(1+cosα)]×{cos(α/2)/[1+cos(α/2)]}=?应该是[sin4α/(1+cos4α)]×[cos2α/(1+cos2α)]×[cosα/(1+cosα)]×{cos(α/2)/[1+cos(α/2)]}=？打错了。对不起。
[sin4α/(1+cos4α)][cos2α/(1+cos2α)]×[cosα/(1+cosα)]×{cos(α/2)/[1+cos(α/2)]}
=[2sin2αcos2α/2cos22α)× [cos2α/2cos2α)]×[cosα/2cos2（α/2)]×{cos(α/2)/[2cos2(α/4)]}
=【2sin2αcos2α×cos2α×cosα×cos(α/2)】/[2cos22α)×2cos2α×2cos2（α/2）×2cos2(α/4)】
=2sinαcosα/[8cosα×cos（α/2）×cos2(α/4)]
=2sin（α/2）/4cos2(α/4)=2sin(α/4)/[2cos(α/4)]=tan(α/4)
α≠4kπ+2π,2kπ+π,kπ+0.5π,0.5kπ+0.25π,k∈Z

sin4a=2sin2a*cos2a=4sina*cosa*cos2a=8sina/2*cosa/2*cosa*cos2a
1+cos4a=2cos^2(2a)
同理1+cos2a=2cos^2(a),1+cosa=2cos^2(a/2),
整理分子分母：分母为8cos^2(2a)cos^2(a)cos^2(a/2)(1+cos^2(a/2)
分子为8sina/2*cosa/2*cosa*cos2a*cos2a*cosa*cosa/2
约分整理得sina/2/(1+cosa/2)

[sin4α/(1+cos4α)]×[cos2α/(1+cos2α)]×[cosα/(1+cosα)]×{cos(α/2)/[1+cos(α/2)]}
=2sin2αcos2α/2(cos(2α))×[cos2α/(2(cosα)^2)]×[cosα/(2(cos(α/2))^2)]×{cos(α/2)/[2(cos(α/4))^2]}
=2sin2α/(8cosα×cos(α/2...

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[sin4α/(1+cos4α)]×[cos2α/(1+cos2α)]×[cosα/(1+cosα)]×{cos(α/2)/[1+cos(α/2)]}
=2sin2αcos2α/2(cos(2α))×[cos2α/(2(cosα)^2)]×[cosα/(2(cos(α/2))^2)]×{cos(α/2)/[2(cos(α/4))^2]}
=2sin2α/(8cosα×cos(α/2)×(cos(α/4))^2)
=sin(α/4)×cos(α/4)×cos(α/2)×cosα/(cosα×cos(α/2)×(cos(α/4))^2)
=tan(α/4)

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公式：cos2x=2(cosx)^2-1则，cos2x+1=2(cosx)^2 :sin2x=2sinxcosx
[sin4α/(1+cos4α)]×[cos2α/(1+cos2α)]×[cosα/(1+cosα)]×{cos(α/2)/[1+cos(α/2)]}
=[sin4α/2(cos2a)^2]×[cos2α/2(cosa)^2]×[cosα/2cosa/2)^2]×{cos(α/2)/2(cosa/4)^2}
=sin4a/[2cos2a×2cosa×2cos(a/2)×2(cosa/4)^2]
=tan(a/4)

答案如图：

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