问题
解答题
已知
(1)若
(2)求
|
答案
(1)∵
=(cos(α-β),-1),b
=(cos(α+β),2),且c
∥b
,c
∴2cos(α-β)+cos(α+β)=0,即2(cosαcosβ+sinαsinβ)+cosαcosβ-sinαsinβ=0,
∴3cosαcosβ+sinαsinβ=0,又α,β≠kπ+
(k∈Z),π 2
∴tanα•tanβ=-3;
(2)∵
=(sinα,sinβ),a
=(cos(α-β),-1),b
=(cos(α+β),2),c
∴
2+a
•b
=sin2α+sin2β+cos(α-β)cos(α+β)-2c
=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.