(全国通用版)2018_2019高中数学第一章基本初等函数(Ⅱ)1.2任意角的三角函数1.2.4.2诱导公式2练习新人教B版必修4.doc

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1、1第 2 课时 诱导公式(2)课时过关能力提升1.若 sin( + )=, ,则 tan 等于( )(-2,0)A.- B.-32C.- D.-333解析: 由已知得 -sin = ,即 sin =-.又因为 ,(-2,0)所以 cos = ,于是 tan =- .32 33答案: D2.已知 |sin |= ,且 是第二象限的角,则 sin 等于 ( )(-2)A.- B. C.- D.223 223解析: 由已知得 sin = ,而 是第二象限的角,所以 sin = ,从而 cos =- =- ,13 1-2 223于是 sin =-sin =-cos = .(-2) (2-) 223答案

2、: D3.化简 tan(27- )tan(49- )tan(63+ )tan(139- )的结果为( )A.1 B.-1 C.2 D.-2解析: 原式 =tan(27- )tan(49- )tan90-(27- )tan90+(49- )=tan(27- )cot(27- )tan(49- )-cot(49- )=- 1.答案: B4.已知 sin 是方程 6x=1- 的根,则 的值等于( )(-5)(2-)(32+)(-)A. B. C.- D.520 1515 520 1802答案: A5.已知 cos 29=m,则 sin 241tan 151的值是( )A. B.1-2 1-2C. D

3、.-2-1 1-2解析: 由于 sin 241=sin(180+61)=-sin 61=-cos 29=-m,tan 151=tan(180-29)=-tan 29=- =- ,于是 sin 241tan 151=(-m) .2929 1-2 (- 1-2 )=1-2答案: B6.如果 cos = ,且 是第四象限的角,那么 cos = . (+2)解析: cos =-sin =- (- )= .(+2) 1-2 265答案:2657.sin 315-cos 135+2sin 570的值是 . 解析: sin 315-cos 135+2sin 570=-sin 45+cos 45+2sin 2

4、10=- +2sin(18022+22+30)=-1.答案: -18.若 f(x)=sin x,则 f(1)+f(3)+f(5)+f(2 015)+f(2 017)= . 6答案: 09.求证: .2(+)-11-22 =(9+)+1(+)-1证明 左边 =-2-12-2= ,-(+)2(+)(-)=+-右边 = ,+1-1=+- 左边 =右边, 原等式成立 .310.已知 f( )= .(-)(2-)(-+32)(-)(-)(1)化简 f( );(2)若 是第三象限的角,且 cos ,求 f( )的值;(-32)=15(3)若 =- ,求 f( )的值 .313解: (1)f( )= =-c

5、os .(-)(2) cos =-sin = , 是第三象限的角,(-32) 15 sin =- ,cos =- ,15 265f ( )= .265(3)- =-62 + ,313 53f ( )=f =-cos(-313) (-62+53)=-cos =-cos =- .53 3 1211.已知 sin(x+y)= 1,求证:tan(2 x+y)+tan y=0.证明 sin(x+y)=1,x+y= 2k + ,kZ .2x= 2k + -y,kZ,2 tan(2x+y)+tan y=tan +tan y2(2+2)-=tan(4k + -y)+tan y=tan( -y)+tan y=-tan y+tan y=0.

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