化 tan dfrac (pi -C)(2)=-pi `D =dfrac (E+C)(2)=cotdfrac (4)(2)-|||-6三年 therefore dfrac (pi )(4)=dfrac (1)(3) _(0)=dfrac (1)(3) ()-|||-(A) dfrac (dfrac {sqrt {3)}(3)} 18 -dfrac (2sqrt {3)}(3)-|||-`c} dfrac (1)(3) (D -dfrac (1)(3)-|||-7.已 sin (y=-a)=dfrac (1)(3) .because 2+2+1dfrac (11pi )(2)=(a)_(1) 方于1-|||-(A) -dfrac (1)(2) IB dfrac (1)(2)-|||-(C) dfrac (sqrt {y)}(2) 10 -dfrac (sqrt {3)}(2)-|||-B.已 |tan (pi -alpha )=-2sin (dfrac (pi )(2)+alpha ) in acown n等于-|||-()-|||-(

dfrac (pi )(4) C. dfrac (pi )(2) D.π

(B) pi +dfrac (4)(3) (C) pi +2 . (D) pi +dfrac (8)(3)

dfrac (1)(2)pi .-|||-C.π-|||-D. dfrac (5)(4)pi .-|||-0

sqrt (2)pi -|||-C. dfrac (sqrt {2pi )}(4)-|||-D. dfrac (sqrt {2pi )}(8)

(int )_(0)^dfrac (pi {4)}(tan )^2xdx=1-dfrac (pi )(4) __-|||-__下列式子或叙述不正确的是A.B.设

3.要使函数φ(x )= ,dfrac {pi )(2)] (B)[π,2π] (C) [ 0,dfrac (pi )(2)] (D) [ dfrac

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

(B) dfrac (pi )(3) (C) π/6, (D) dfrac (pi )(2)A、AB、BC、CD、DA、AB、BC、CD、D

dfrac (q)(4pi varepsilon oR)-|||-C. dfrac (q)(4pi {varepsilon )_(0)(R)^2}-|||-D.

(B) dfrac (lambda )(2pi {varepsilon )_(0)a}. (C) dfrac (lambda )(4pi {varepsilon