5.若cosalpha=-(1)/(2)且alphain((pi)/(2),pi)，则alpha=(2)/(3)pi。

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

(int )_(-pi )^pi sqrt ({pi )^2-(x)^2}dx= )(int )_(-pi )^pi sqrt ({pi )^2-(x)^2}d

1357.已知 tan x=1, in (dfrac (pi )(2),dfrac (3pi )(2)), 则 =-|||-

3.要使函数φ(x )= ,dfrac {pi )(2)] (B)[π,2π] (C) [ 0,dfrac (pi )(2)] (D) [ dfrac

求函数 =x(y)^2+(z)^3-xyz 在点(1,1,2)处沿方向角为-|||-alpha =dfrac (pi )(3), beta =dfrac (pi

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

-3pi (R)^2-|||-C. -4pi (R)^2-|||-D. -5pi (R)^2

设 M=int_(-(pi)/(2))^(pi)/(2) (sin x)/(1+x^2) cos^4 x dx，N=int_(-(pi)/(2))^(pi)/(

函数 u=x^2+2y+3z^3 在点 (1,-1,1) 处沿方向角 alpha=(pi)/(4), beta=(pi)/(3), gamma=(pi)/(3)

设总体 sim pi (3) 1A2,···,A21 是来自sim pi (3) 1A2,···,A21的样本，则 sim pi (3) 1A2,···,A21