赞 zhanghhapyy 幼苗
共回答了20个问题采纳率：85% 举报
(1+(tan2α)^2)[cos(2α+π/3)+cos(2α-π/3)]
=[1+(2tanα/(1-tan^2α)^2][cos2αcos-sin2αsinπ/3+cos2αcos+sin2αsinπ/3]
=[(1+tan^2α)/(1-tan^2α)]^2(2cos2αcosπ/3)
=[1/(cos^2α-sin^2α)]^2(2cos2αcosπ/3)
=2cos2αcosπ/3/(cos2α)^2
=2cosπ/3/cos2α
=2cosπ/3/[2(cosα)^2-1]
=1/(-7/9)
=-9/7
不知对乎,供参考
也许我算错
=[1+(2tanα/(1-tan^2α)^2][cos2αcos-sin2αsinπ/3+cos2αcos+sin2αsinπ/3]
=[(1+tan^2α)/(1-tan^2α)]^2(2cos2αcosπ/3)
=[1/(cos^2α-sin^2α)]^2(2cos2αcosπ/3)
=2cos2αcosπ/3/(cos2α)^2
=2cosπ/3/cos2α
=2cosπ/3/[2(cosα)^2-1]
=1/(-7/9)
=-9/7
不知对乎,供参考
也许我算错
1年前 追问
3
2cosπ/3=2*1/2=1 (cosα)^2=1/3 [2(cosα)^2-1]=2*1/3-1=2/3-1=-1/3 2cosπ/3/[2(cosα)^2-1]=1/(-1/3)=-3 原来错了
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