问题 解答题
若α,β满足
cos2(α-β)-cos2(α+β)=
1
2
(1+cos2α)(1+cos2β)=
1
3
,求tanαtanβ的值.
答案

cos2(a-β)-cos2(a+β)

=

1+cos2(a-β)
2
-
1+cos2(a+β)
2

=

1
2
[cos(2a-2β)-cos(2a+2β)]

=sin2asin2β

=

1
2

又∵(1+cos2a)(1+cos2β)

=2cos2a2cos2β

=

1
3

sin2asin2β
2cos2a2sin2β

=

2sinacos2sinβcosβ
2cos2a2sin2β

=tanatanβ.

∴tanatanβ=

1
2
1
3
=
3
2

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