[题目]若 sin alpha =dfrac (1)(3), 则 cos 2alpha = ()()-|||-
lim _(xarrow alpha )dfrac (sin x-sin alpha )(x-alpha )..
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
求下列极限: lim _(xarrow alpha )dfrac (sin x-sin alpha )(x-alpha );求下列极限: ;
4.已知 sin alpha =-dfrac (3)(5), 且α是第三象限的角,则 cos alpha = __ ,-|||-tan alpha = __ o
(B) alpha lt dfrac (1)(2). (C) alpha geqslant dfrac (1)(2), (D) alpha gt dfrac (
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
=dfrac (alpha +1)(2).
设alpha =(dfrac (1)(2),0,0,dfrac (1)(2)) ,=E-(alpha )^Talpha =E+2(alpha )^Talpha