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

[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

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

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

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

11 判断设 -t(n) ,如果 (|X|geqslant k)=2alpha , 则 (Xlt k)=1-alpha .-|||-A.X-|||-B.

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

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

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

已知cosalpha=(1)/(2)，则cos(2pi-alpha)的值为A. $\frac{1}{2}$B. $-\frac{1}{2}$C. $\frac{