=dfrac (alpha +1)(2).

(B) alpha lt dfrac (1)(2). (C) alpha geqslant dfrac (1)(2), (D) alpha gt dfrac (

设alpha =(dfrac (1)(2),0,0,dfrac (1)(2)) ,=E-(alpha )^Talpha =E+2(alpha )^Talpha

例1(1)(2021·全国甲卷)若 alpha in (0,dfrac (pi )(2)), tan2α-|||-=dfrac (cos alpha )(2-s

[题目]若 sin alpha =dfrac (1)(3), 则 cos 2alpha = ()()-|||-

若 |X|lt x =alpha , 则x等于 () .-|||-(A) zg/2 (B) _(1)-dfrac (a)(2) (C) dfrac (21-a

设矩阵 =((alpha )_(1),(alpha )_(2),(alpha )_(3),(beta )_(1))，=((alpha )_(1),(alpha

dfrac (Fx)(sin alpha ) D. dfrac (Fx)(cos alpha )

【题目】-|||-设α1,a2,α3线性无关,证明: (alpha )_(1)+(alpha )_(2), (alpha )_(2)+(alpha )_(3),

向量方程((a)_(1)+alpha )-7((alpha )_(2)+alpha )+4(alpha )_(3)=0，其中((a)_(1)+alpha )-7

7.[福建2020适应性模拟]已知 tan alpha =2 ,则-|||-dfrac ({sin )^2alpha -(cos )^2alpha +2}(2{