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
7.[福建2020适应性模拟]已知 tan alpha =2 ,则-|||-dfrac ({sin )^2alpha -(cos )^2alpha +2}(2{
[题目]若 sin alpha =dfrac (1)(3), 则 cos 2alpha = ()()-|||-
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 )
已知sinα=-(3)/(5),且α是第三象限角,则cosα=( )A. -$\frac{2}{5}$B. -$\frac{4}{5}$C. $\frac{3}
lim _(xarrow alpha )dfrac (sin x-sin alpha )(x-alpha )..
求下列极限: lim _(xarrow alpha )dfrac (sin x-sin alpha )(x-alpha );求下列极限: ;
已知向量 a_1, a_2, a_3, a_4, a_5,且有 r(alpha _1, alpha _2, alpha _3, alpha _4)= 3,r(a
设向量组 alpha_1, alpha_2, alpha_3, alpha_4, alpha_5秩为 3,且满足 alpha_1 + alpha_3 - a