[sin4α/(1+cosα)]×[cos2α/(1+cos2α)]×[cosα/(1+cosα)]×{cos(α/2)

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/10/06 02:34:43

[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

[sin4α/(1+cos4α)]×[cos2α/(1+cos2α)]×[cosα/(1+cosα)]=? 求证cos^8α-sin^8α 1/4sin2αsin4α=cos2α 化简:cos2α-cos2β/cosα-cosβ 化简(sinα-cosα)^2-1/-cos2α 已知sinα -2cosα=0,求(1)sinαcosα的值(2)sin4α+cos2α 化简:sin^2αsin^2β+cos^2αcos^2β-1/2cos2αcos2β 化解sin^2αsin^2β+cos^2αcos^2β-1/2cos2αcos2β 化简(sinα)^2*(sinβ)^2+(cosα)^2(cosβ)^2-1/2cos2αcos2β Sin²αSin²β+Cos²αCos²β-1/2Cos2αCos2β 化简：sin²αsin²β+cos²αcos²β-1/2cos2αcos2β sin²α*sin²β+cos²α*cos²β-1/2cos2αcos2β 化简：sin²αsin²β+cos²αcos²β—1/2cos2αcos2β