[题目]若 sin alpha =dfrac (1)(3), 则 cos 2alpha = ()()-|||-
已知cosalpha=(1)/(2),则cos(2pi-alpha)的值为A. $\frac{1}{2}$B. $-\frac{1}{2}$C. $\frac{
dfrac (Fx)(sin alpha ) D. dfrac (Fx)(cos alpha )
[案例分析题] 一低压照明系统,,问系统cosα为多大?若要将cosα提高到0.9,需要多大Qc,补偿前后系统无功功率如何变化?
判断题-|||-tan (2pi +alpha )=tan alpha .-|||-39.-|||-A 对-|||-B 错
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
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
16.设向量的方向余弦分别满足(1) cos alpha =0; (2) cos beta =1; (3) cos alpha =cos beta =0, 问这
4.已知 sin alpha =-dfrac (3)(5), 且α是第三象限的角,则 cos alpha = __ ,-|||-tan alpha = __ o