1.一平面电磁波能表示成 _(x)=0, _(y)=2cos [ 2pi times (10)^14(dfrac (z)(c)-t)+dfrac (pi )(2)] _(2)=0, 则表明:-|||-①光波沿z轴正向传播。-|||-②原点的初位相是 π/2。-|||-③ 当 =2.5times (10)^-15s 时,原点处的电场矢量E沿y轴正向。-|||-④当 =2.5times (10)^-15s 时,原点处的磁感应强度矢量B沿x轴正向。

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

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

(B) dfrac (9)(2pi {c)_(0)(y)^2} - y↑-|||-(0,y)-|||-+q-|||-(C) dfrac (qa)(2pi {va

. _(Y)(y)=dfrac (1)(2sqrt {2pi )}(e)^-dfrac (y{2)} ,gt 0-|||-bigcirc ._(Y)(y)=d

5.证明 (z)=cos (z+dfrac (1)(z)) 用z的幂表出的洛朗展开式中的系数为-|||-_(n)=dfrac (1)(2pi )(int )_(

(B) dfrac (9)(2pi {c)_(0)(y)^2} - y 个-|||-P(0,y)-|||-+q-|||-(C) dfrac (qa)(2pi {

U=0-|||-(B) E=0 . =dfrac (q)(2pi {varepsilon )_(0)a}-|||-(C) =dfrac (q)(2pi {var

把复数=((cos dfrac {2pi )(9)+isin dfrac (2pi )(9))}^3转化为三角形式A =((cos dfrac {2pi )(9

一质点作简谐振动，其振动方程为=6.0times (10)^-2cos (dfrac (1)(3)pi t-dfrac (1)(4)pi )(SI)（1）当=6

(int )_(0)^dfrac (pi {4)}dfrac (x)(1+cos 2x)dx=( ) .(int )_(0)^dfrac (pi {4)}dfr