[单选题] sum _(n=0)^infty ((-1))^ndfrac (2n+3)((2n+1)!)=A.sin1+cos1B.2sin1+cos1C.2sin1+2cos1D.2sin1+3cos1

已知sin（α-β）=(1)/(3)，cosαsinβ=(1)/(6)，则cos（2α+2β）=（ ）A. $\frac{7}{9}$B. $\frac{1}{

(int )_(0)^12xcos ((x)^2+1)dx= ( )A.cos 2- cos 1B.cos 2+ cos 1C.sin 2- sin 1D.si

lim _(narrow infty )((2n+1))^2sin dfrac (1)(2{n)^2} 值是(··)。 ()A、3B、1C、2D、∞A、3B、1

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

证明：级数sum _(n=1)^infty (sin dfrac (1)({n)^2})收敛。证明：级数收敛。

+dfrac (sin n)({2)^n} ;-|||-(2) _(n)=1+dfrac (1)({2)^2}+dfrac (1)({3)^2}+... +df

计算：underset(lim)(n→∞)(1)/(n)[sin(π)/(n)+sin(2π)/(n)+…+sin(（n-1）π)/(n)]．计算：$\unde

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

数列极限lim _(narrow infty )(6(n)^2+2)sin dfrac (1)(3{n)^2+1}=( )A.B.C.2D.3数列极限

|sin 2|leqslant 1 D. (sin z)=cos z