+sin dfrac (npi )(2)+... （ ）A 、是收敛的 ; B 、是收敛的，其和为sin dfrac (pi )(2)+sin dfrac (2pi )(2)+... +sin dfrac (npi )(2)+... ; C、 是发散的 ; D 、不一定发散

=-sqrt (2)(sin dfrac (pi )(6)+icos dfrac (pi )(6))的指数表达式为（ ）。A =-sqr

+sin dfrac (pi )({2)^n}+... ;

πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T

(sin x)=dfrac (1)({cos )^2x} in (0,dfrac (pi )(2)),则(sin x)=dfrac (1)({cos )^2x}

若数列_(n)=sin dfrac (n)(2)pi , 则数列_(n)=sin dfrac (n)(2)pi 发散正确错误若数列,则数列发散正确错误

(9) (int )_(-dfrac {pi )(2)}^dfrac (pi {2)}|sin x|arctan (e)^xdx

计算 lim _(xarrow dfrac {pi )(2)}dfrac (ln sin x)({(pi -2x))^2}

求lim _(xarrow dfrac {pi )(2)}dfrac (ln sin x)({(pi -2x))^2}求

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

( (int )_(dfrac {pi )(4)}^dfrac (pi {3)}dfrac (x)({sin )^2x}dx ;