问题 解答题
已知
a
=(sinα,sinβ),
b
=(cos(α-β),-1)
c
=(cos(α+β),2)
α,β≠kπ+
π
2
(k∈Z)

(1)若
b
c
,求tanα•tanβ的值;
(2)求
a
2
+
b
c
的值.
答案

(1)∵

b
=(cos(α-β),-1),
c
=(cos(α+β),2),且
b
c

∴2cos(α-β)+cos(α+β)=0,即2(cosαcosβ+sinαsinβ)+cosαcosβ-sinαsinβ=0,

∴3cosαcosβ+sinαsinβ=0,又α,β≠kπ+

π
2
(k∈Z),

∴tanα•tanβ=-3;

(2)∵

a
=(sinα,sinβ),
b
=(cos(α-β),-1),
c
=(cos(α+β),2),

a
2+
b
c
=sin2α+sin2β+cos(α-β)cos(α+β)-2

=sin2α+sin2β+cos2αcos2β-sin2αsin2β-2

=sin2α+(1-sin2α)sin2β+cos2αcos2β-2

=sin2α+cos2αsin2β+cos2αcos2β-2

=sin2α+cos2α(sin2β+cos2β)-2

=sin2α+cos2α+2

=1-2

=-1.

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