已知 cos (dfrac (pi )(4)+x)=dfrac (3)(5), 求sin2x的值.

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

(int )_(-dfrac {pi )(2)}^dfrac (pi {2)}dfrac (|x|sin x)(1+{cos )^3x}dx=(int )_(-

设=(int )_(-dfrac {pi )(2)}^dfrac (pi {2)}dfrac (sin x)(1+{x)^2}(cos )^4xdx, =(in

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

2.求下列函数的极值:-|||-(6) (x)=sin x+cos x(-dfrac (pi )(2)leqslant xleqslant dfrac (pi

1.求下列极限:-|||-(1) lim _(xarrow 0)dfrac (sin 2x)(x);-|||-(3) lim _(xarrow dfrac {p

已知函数 (x)=dfrac (1)(2)sin (2x-dfrac (pi )(3)) x∈R,-|||-(1)求f(x)的最小正周期;-|||-(2)求f(

设(x)=sin x, 则(dfrac (pi )(2))= __-|||-_.-|||-A cos dfrac (pi )(2)-|||-B o-|||-C-

2.求下列函数在指定点的导数:-|||-(1)设 =f(x)=sin x-cos x, 求 (dfrac (pi )(4)), (dfrac (pi )(2))

求极限 lim _(xarrow 1)dfrac (ln [ cos (x-1)] )(1-sin dfrac {pi )(2)x}