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 的值为() ()-|||-A dfrac (4sqrt {10)}(5)-|||-B dfrac (4sqrt {10)}(5)-|||-C dfrac (sqrt {10)}(5)-|||-D dfrac (sqrt {10)}(5)

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

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

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

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

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

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}

1.8 将下列各复数写成三角表示式:-|||-(1) -3+2i;-|||-(2) sin alpha +icos alpha ;-|||-(3) -sin d

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

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: ;