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

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

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

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

4.已知 sin alpha =-dfrac (3)(5), 且α是第三象限的角,则 cos alpha = __ ,-|||-tan alpha = __ o

1.“sin^2 alpha +sin^2 beta =1”是“sin alpha +cos beta =0”的（）.A. 充分非必要条件B. 必要非充分条件C

6.(单选题)(三角函数基础118)已知 sin alpha -3cos alpha =0 且α是第三象限角,则 sin alpha -cos alpha 的

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

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

lim _(xarrow alpha )dfrac (sin x-sin alpha )(x-alpha )..

判断：若四阶行列式|A|=|(alpha )_(1),(alpha )_(2),(alpha )_(3),(alpha )_(4)|=2，则行列式|A|=|(a