=-sqrt (2)(sin dfrac (pi )(6)+icos dfrac (pi )(6))的指数表达式为( )。A =-sqr
[题目]若 sin alpha =dfrac (1)(3), 则 cos 2alpha = ()()-|||-
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
1.9 利用复数的三角表示计算下列各式:-|||-(1) (1+i)(1-i) ;-|||-(2) (-2+3i)/(3+2i) ;-|||-(3) ((dfr
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
[例2](1)(2020·全国卷Ⅲ)已知 sin theta +sin (theta +dfrac (pi )(3))=1,-|||-则 sin (theta
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 )
lim _(xarrow alpha )dfrac (sin x-sin alpha )(x-alpha )..