已知(sin )^2theta -(cos )^2theta =dfrac (2sqrt {2)}(3)，求(sin )^2theta -(cos )^2theta =dfrac (2sqrt {2)}(3).

6.求由曲线 rho =sqrt (2)sin theta 及 (rho )^2=cos 2theta 所围成图形的公共部分的面积.

( 10 ) (int )_(0)^dfrac (pi {4)}(tan )^2theta dtheta ;

设总体X的概率密度为(x;theta )= ) 2theta (x)^2theta -1, 0lt xlt 1 0, 其他, .．

若sin（α+β）+cos（α+β）=2sqrt(2)cos（α+(π)/(4)）sinβ，则（ ）A. tan（α-β）=1B. tan（α+β）=1C. t

求下列各曲线所围成图形的公共部分的面积:-|||-(1) rho =3cos theta 及 rho =1+cos theta ;-|||-(2) rho

[例2](1)(2020·全国卷Ⅲ)已知 sin theta +sin (theta +dfrac (pi )(3))=1,-|||-则 sin (theta

8.求下列各曲线所围成图形的公共部分的面积:-|||-(1) rho =3cos theta 及 rho =1+cos theta ;-|||-(2) rho

5.已知总体X的概率密度为-|||-f(x)= {e)^-dfrac ({x^2)(2theta )},xgt 0, 0,xleqslant 0, .-|

计算下列积分:-|||-(9) (int )_(-e)^-2dfrac (dx)(1+x);-|||-(10) (int )_(0)^dfrac (pi {4)

设总体X的概率密度为 (x;theta )= ^2), 0lt xlt 2theta 0, 是θ^2-|||-的无偏估计,则 = __