课时跟踪检测 (三十四) 同角三角函数的基本关系
课时跟踪检测 （三十四） 同角三角函数的基本关系
层级(一) “四基”落实练 1．sin α＝
5
5
，则sin 2α－cos 2α的值为( ) A ．－1
5
B ．－35
C.15
D ．35
解析：选B 因为sin α＝
55，所以cos 2α＝1－sin 2α＝45，则原式＝15－45＝－35
. 2．已知tan α＝－1
3，则sin α＋2cos α3cos α－sin α＝( )
A.7
8 B ．－35
C ．－23
D ．12
解析：选D sin α＋2cos α3cos α－sin α＝tan α＋23－tan α＝－13＋23＋13＝1
2.
3．已知α∈????π4，π2，sin 2α＝5
5，则tan 2α＝( ) A ．－2 B ．2 C.1
2
D ．－1
2
解析：选D ∵α∈????π4，π2，∴2α∈????π
2，π， 又sin 2α＝
5
5
， ∴cos 2α＝－
1－sin 22α＝－
1－15＝－255
.
∴tan 2α＝sin 2αcos 2α＝55－
25
5＝－1
2.
4．已知sin α－cos α＝－52，则tan α＋1tan α
的值为( ) A ．－4 B ．4 C ．－8
D ．8
解析：选C tan α＋1tan α＝sin αcos α＋cos αsin α＝1
sin αcos α.
∵sin αcos α＝1－(sin α－cos α)22＝－1
8，
∴tan α＋1
tan α
＝－8.
5．若α∈[0,2π)，且有1－cos 2 α＋ 1－sin 2 α＝sin α－cos α，则角α的取值范围为
( )
A.????0，π2 B ．????π
2，π C.????π
2，π D ．?
???π，3π2 解析：选B 因为
1－cos 2 α＋
1－sin 2 α＝sin α－cos α，
所以?????
sin α≥0，cos α≤0，
又α∈[0,2π)，所以α∈???
?π
2，π，故选B. 6．已知0＜α＜π2，tan α＝4
3，则sin α＋cos α＝________.
解析：∵0＜α＜π2，tan α＝4
3，
∴cos α＝
11＋tan 2α＝3
5
，sin α＝
1－cos 2α＝4
5
，
∴sin α＋cos α＝7
5.
答案：7
5
7．若tan α＋
1tan α＝3，则sin αcos α＝________，tan 2α＋1
tan 2α
＝________. 解析：∵tan α＋1tan α＝1
cos αsin α＝3，
∴sin αcos α＝1
3
.
∴tan 2α＋1
tan 2α＝????tan α＋1tan α2－2＝9－2＝7. 答案：1
3
7
8．已知α为第二象限角，则cos α1＋tan 2α＋sin α·
1＋
1
tan 2α
＝________.
解析：原式＝cos α
sin 2α＋cos 2α
cos 2α＋sin α
sin 2α＋cos 2αsin 2
α＝cos α·1
|cos α|
＋sin α·1|sin α|.因为α是第二象限角，所以sin α＞0，cos α＜0，所以cos α·1|cos α|＋sin α·1
|sin α|＝－1＋1＝0，即原式＝0.
答案：0
9．已知tan α＝2
3，求下列各式的值：
(1)cos α－sin αcos α＋sin α＋cos α＋sin αcos α－sin α； (2)1sin αcos α
； (3)sin 2α－2sin αcos α＋4cos 2α.
解：(1)原式＝1－tan α1＋tan α＋1＋tan α1－tan α＝1－231＋23＋1＋231－23＝26
5
.
(2)原式＝sin 2α＋cos 2αsin αcos α＝tan 2α＋1tan α＝13
6.
(3)原式＝sin 2α－2sin αcos α＋4cos 2α
sin 2α＋cos 2α
＝tan 2α－2tan α＋4tan 2α＋1
＝49－43＋449＋1＝2813.
10．化简下列各式： (1)sin α1＋sin α－sin α1－sin α； (2)????1sin α＋1tan α(1－cos α)．
解：(1)原式＝sin α(1－sin α)－sin α(1＋sin α)
(1＋sin α)(1－sin α)
＝－2sin 2α1－sin 2α
＝－2sin 2α
cos 2α＝－2tan 2
α.
(2)原式＝????1sin α＋cos αsin α(1－cos α) ＝1＋cos αsin α(1－cos α)＝sin 2α
sin α＝sin α.
层级(二) 素养提升练
1．已知sin θ＝m －3m ＋5，cos θ＝4－2m m ＋5????π
2<θ<π，则tan θ＝( ) A.4－2m
m －3 B ．±m －34－2m
C ．－
512
D ．－34或－512
解析：选C 由
sin 2θ＋cos 2θ＝1，有
? ????m －3m ＋52＋? ??
??
4－2m m ＋52＝1，化简得m 2－8m ＝0，解得m ＝0或m ＝8，由于θ在第二象限，所以sin θ>0，m ＝0舍去，故m ＝8，sin θ＝5
13，cos
θ＝－1213，得tan θ＝－512
.
2．已知P (－3，y )为角β的终边上的一点，且sin β＝1313，则2sin 2βsin 2β－cos 2β＝( )
A ．±1
2
B ．－211
C.36
D ．±2
解析：选B 因为r ＝
3＋y 2，故由正弦函数的定义可得y 3＋y 2
＝
1313，解得y ＝12
或y ＝－12(舍去)，所以tan β＝1
2
－3＝－36，所以2sin 2βsin 2β－cos 2β＝2tan 2βtan 2β－1＝2×???
?－
362??
??－362－1＝－2
11，
故选B.
3．化简：1－cos 4α－sin 4α
1－cos 6α－sin 6α＝________.
解析：原式＝(1－cos 4α)－sin 4α
(1－cos 6α)－sin 6α
＝
(1－cos 2α)(1＋cos 2α)－sin 4α
(1－cos 2α)(1＋cos 2α＋cos 4α)－sin 6α
＝
sin 2α(1＋cos 2α)－sin 4α
sin 2α(1＋cos 2α＋cos 4α)－sin 6α
＝1＋cos 2α－sin 2α
1＋cos 2
α＋cos 4
α－sin 4
α
＝2cos 2α
1＋cos 2α＋(cos 2α＋sin 2α)(cos 2α－sin 2α)
＝
2cos 2α
1＋cos 2α＋cos 2α－sin 2α
＝2cos 2α3cos 2α＝23. 答案：2
3
4．(1)已知sin α＋cos α＝2，求sin αcos α与sin 4α＋cos 4α的值； (2)已知sin α＋cos α＝1
3(0＜α＜π)，求sin α－cos α的值．
解：(1)∵sin α＋cos α＝2，
∴(sin α＋cos α)2＝1＋2sin αcos α＝2， 解得sin αcos α＝1
2
，
∴sin 4α＋cos 4α＝(sin 2α＋cos 2α)2－2sin 2αcos 2α ＝1－2×14＝1
2
.
(2)∵sin α＋cos α＝1
3(0＜α＜π)，
∴(sin α＋cos α)2＝1＋2sin αcos α＝1
9，
∴2sin αcos α＝－8
9
，
∴(sin α－cos α)2＝1－2sin αcos α＝1＋89＝17
9，
∵0＜α＜π，2sin αcos α＝－8
9，
∴sin α＞0，cos α＜0， ∴sin α－cos α＝
179＝173
. 5．已知关于x 的方程2x 2－(3＋1)x ＋m ＝0的两根分别是sin θ和cos θ，θ∈(0,2π)，
求：
(1)sin 2θsin θ－cos θ＋cos θ1－tan θ的值； (2)m 的值；
(3)方程的两根及此时θ的值． 解：(1)原式＝sin 2θsin θ－cos θ＋cos θ
1－
sin θcos θ
＝sin 2θsin θ－cos θ＋cos 2θcos θ－sin θ ＝sin 2θ－cos 2θsin θ－cos θ＝sin θ＋cos θ. 由条件知sin θ＋cos θ＝3＋1
2，
故sin 2θ
sin θ－cos θ＋cos θ
1－tan θ
＝3＋1
2.
(2)由已知，得sin θ＋cos θ＝3＋12，sin θcos θ＝m
2，
又1＋2sin θcos θ＝(sin θ＋cos θ)2，解得m ＝3
2
. (3)由???
??
sin θ＋cos θ＝3＋1
2，sin θcos θ＝3
4
，
得???
sin θ＝32，cos θ＝1
2
或???
sin θ＝1
2，
cos θ＝3
2
.
又θ∈(0,2π)，故θ＝π3或θ＝π
6
.
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课时跟踪检测（一） Welcome to the unit & Reading — Language points Ⅰ.单句语法填空 1．There are 30 people injured in the traffic accident but the injured were rescued at once.(injure) 2．She was confident of her becoming popular with the students. 3．I didn't want to add to my mother's worry, so I hurried home. 4．It is reported in the newspaper that talks between the two countries are making progress. 5．He appears careless, but in reality, he is a very careful person. Ⅱ.完成句子 1．如果你认为这样方便的话，我们把时间定在下午六点。 If you think that it_is_convenient_to_you,_we will fix the time at 6：00 pm. 2．我们都赞成他在会上提出的计划。 We all agreed to the plan he_had_put_forward at the meeting. 3．球队和球迷确信能够赢得即将到来的比赛。 The team and the fans are_confident_of_winning the coming match. 4．她吃了一惊，他居然还记得她的名字。 It_amazed_her that he still remembered her name. 5．需要采取措施使农田免遭洪水的侵袭。 Some measures are needed to_secure_the_farmland_against/from floods. 6．一看到这张聚会照片，她就陷入了对过去的美好回忆中。 No_sooner_had_she_seen the get-together photo than she got lost in the old sweet memories. 7．已经证明你的计划是实用的。 It_has_been_proved_that your plan is practical. Ⅲ.完形填空 Throughout history, many great leaders and explorers have tried hard to find objects or other ways to live forever. While it might sound __1__ to live forever, I believe that what makes life so __2__ is that it doesn't last forever. So, I don't __3__ that living forever is a good idea. For one thing, I think living forever would be very __4__. We always have to find ways to __5__ ourselves in a normal lifetime. __6__ you lived forever, you would still
人教版必修四 同角三角函数的基本关系教案
1.2.2同角三角函数的基本关系（3） 教学目的： 知识目标：根据三角函数关系式进行三角式的化简和证明； 能力目标：（1）了解已知一个三角函数关系式求三角函数（式）值的方法。 （2）灵活运用同角三角函数关系式的不同变形，提高三角恒等变形的能力； 德育目标：训练三角恒等变形的能力，进一步树立化归思想方法； 教学重点：同角三角函数的基本关系式 教学难点：如何运用公式对三角式进行化简和证明。 授课类型：新授课 教学模式：启发、诱导发现教学. 教 具：多媒体、实物投影仪 教学过程： 一、复习引入： 1．同角三角函数的基本关系式。 （1）倒数关系：sin csc 1αα?=，cos sec 1αα?=，tan cot 1αα?=． （2）商数关系： sin tan cos ααα=，cos cot sin ααα =． （3）平方关系：22sin cos 1αα+=，221tan sec αα+=，221cot csc αα+=． （练习）已知tan α43=，求cos α 2．tan αcos α= ，cot αsec α= ，（sec α+tan α）·（ ）=1 二、讲解新课： 例82tan α=-，试确定使等式成立的角α的集合。 =|1sin ||1sin |cos ||cos |αααα+-- =1sin 1sin |cos |ααα+-+=2sin |cos | αα． 2tan α-=-， ∴2sin |cos |αα2sin 0cos αα +=， 即得sin 0α=或|cos |cos 0αα=-≠． 所以，角α的集合为：{|k ααπ=或322,}22 k k k Z πππαπ+<<+∈． 例9．化简(1cot csc )(1tan sec )αααα-+-+． 解：原式=cos 1sin 1(1)(1)sin sin cos cos αααααα -+-+ 2sin cos 1cos sin 11(sin cos )sin cos sin cos αααααααααα-+-+--=?=?112sin cos 2sin cos αααα-+?==?． 说明：化简后的简单三角函数式应尽量满足以下几点： （1）所含三角函数的种类最少； （2）能求值（指准确值）尽量求值； （3）不含特殊角的三角函数值。 例10．求证： cos 1sin 1sin cos x x x x +=-． 证法一：由题义知cos 0x ≠，所以1sin 0,1sin 0x x +≠-≠．
课时跟踪检测(三十三) 数列求和
课时跟踪检测(三十三) 数 列 求 和 1．已知{a n }是首项为1的等比数列，S n 是{a n }的前n 项和，且9S 3＝S 6，则数列???? ?? 1a n 的 前5项和为( ) A.158或5 B.3116或5 C.3116 D.158 2．已知数列{a n }的前n 项和S n ＝an 2＋bn (a 、b ∈R )，且S 25＝100，则a 12＋a 14等于( ) A ．16 B ．8 C ．4 D ．不确定 3．数列112，314，518，7116，…，(2n －1)＋1 2n ，…的前n 项和S n 的值等于( ) A ．n 2＋1－1 2n B ．2n 2－n ＋1－1 2 n C ．n 2＋1－ 12 n －1 D ．n 2－n ＋1－1 2 n 4．(2019·“江南十校”联考)若数列{a n }为等比数列，且a 1＝1，q ＝2，则T n ＝1a 1a 2＋ 1 a 2a 3 ＋…＋ 1 a n a n ＋1 的结果可化为( ) A ．1－1 4n B ．1－1 2n C.2 3??? ?1－14n D.2 3? ???1－12n 5．(2019·珠海模拟)已知等差数列{a n }的前n 项和为S n ，a 5＝5，S 5＝15，则数列???? ? ?1a n a n ＋1的前100项和为( ) A.100101 B.99101 C.99100 D.101100 6．已知函数f (n )＝????? n 2(当n 为奇数时)， －n 2(当n 为偶数时)， 且a n ＝f (n )＋f (n ＋1)，则a 1＋a 2＋a 3＋…＋a 100 等于( ) A ．0 B ．100 C ．－100 D ．10 200 7．在等差数列{a n }中，S n 表示前n 项和，a 2＋a 8＝18－a 5，则S 9＝________. 8．对于数列{a n }，定义数列{a n ＋1－a n }为数列{a n }的“差数列”，若a 1＝2，{a n }的“差数列”的通项公式为2n ，则数列{a n }的前n 项和S n ＝________.
人教版高中英语必修一课时跟踪检测(一)
高中英语学习材料 madeofjingetieji 课时跟踪检测（一） Warming Up & Reading — Language Points Ⅰ.单句语法填空 1．It was in prison that he set down the story. 2．After going through the woods, you’ll find a hospital. 3．The house fell down before she was able to save her little daughter. 4．She joined an English club in order to improve her English. 5．The couple told us it was the fourth time that they had_visited (visit) the West Lake. 6．Before eating, please add some salt to the dish. 7．As far as I’m concerned (concern), it is not a good idea to go camping tonight. 8.While swimming (swim) in the river, we saw a fish jump out of the river. 9．It is no good staying up too late every day. Ⅱ.完成句子 1.He was_very_upset_about_the_fact (为此事很难过) that they couldn’t go to the beach. 2．The teacher made every effort to calm_his_student_down (使他的学生冷静下来) after the exam. 3．I got up early in_order/so_as_not_to_be_late_for_school (为了上学不迟到)． 4．The police went_through_our_luggage (检查了我们的行李) before we got on the plane. 5．A_series_of_wet_days (一连串的雨天) drove me crazy. 6．The time I spend on reading every day adds_up_to_two_hours (合计两个小时)．7．I won’t go there unless_(I_am)_invited (除非受到邀请)． 8．He treated me so badly that I couldn’t_stand_it_any_longer (不能再忍受了)． Ⅲ.阅读理解 A Dear Mr Black， I used to have a really good group of friends. Now they’re all getting into smoking and drinking. I want to find a new group of friends, but I’m shy. How can I know who are the types of people I should make friends with, and who will accept me? Yours， Mike
高中语文语文版必修四课时跟踪检测(十二)+师说+Word版含答案
课时跟踪检测（十二）师说 (时间：40分钟满分：60分) 一、文言基础专练(28分) 1．下列各句中加点的词语，解释不正确的一项是(3分)( ) A．吾从而师．之师：以……为师 B．吾未见其明．也明：明智 C．君子不齿．齿：并列 D．圣人无常．师常：经常 解析：选D D项，常：固定的。 2．下列加点词语的含义与现在的用法分析正确的一组是(3分)( ) ①古之学者 ．．受业解惑也③今之众人 ．．，其下圣人也亦远矣④小．．．必有师②师者，所以传道 学．而大遗，吾未见其明也⑤弟子不必 ．． ．．不如师，师不必贤于弟子⑥年十七，好古文A．全不相同 B．②③⑤和现在的用法相同 C．全都相同 D．①③⑥和现在的用法相同 解析：选A ①学者，古义：求学之人；今义：在学术上有一定成就的人。②传道，古义：传播道理，文中指传播儒家思想；今义：通常指传播宗教思想。③众人，古义：普通人；今义：大家。④小学，古义：在文中指对小的方面学习；今义：对儿童、少年实施初等教育的学校。⑤不必，古义：不一定；今义：用不着，不需要。⑥古文，古义：先秦两汉的文字；今义：相对于白话文的文言文。 3．从句式特征看，与“师者，所以传道受业解惑也”一句相同的一项是(3分)( ) A．道之所存，师之所存也B．句读之不知，惑之不解 C．不拘于时，学于余D．圣人无常师 解析：选A A项与例句同为判断句。B项是宾语前置；C项是被动句；D项是一般句式。 4．下列句子中，“师”字的用法不同于其他三项的一项是(3分)( ) A．于其身也，则耻师．焉 B．或师．焉，或不焉 C．爱其子，择师．而教之 D．师．道之不复，可知矣 解析：选C C项为名词，译为“老师”，其他三项为动词，意为“从师”。 5．下列加点词语含义相同的一组是(3分)( )
(精心整理)同角三角函数基本关系式练习题
任意角的三角函数 1.已知sin α=45 ，且α为第二象限角，那么tan α的值等于 （ ） (A)3 4 (B)43 - (C)4 3 (D)4 3- 2.若θ是第三象限角，且02 cos <θ，则2 θ是 （ ） A ．第一象限角 B ．第二象限角 C ．第三象限角 D ．第四象限 3.设是第二象限角,则sin cos αα ( ) (A) 1 (B)tan 2α (C) - tan 2α (D) 1- 4.若tan θ=3 1,π<θ<32 π,则sin θ·cos θ的值为 （ ） (A)±3 10 (B) 3 10 5 若α 是三角形的一个内角，且sin α+cos α=3 2 ,则三角形为 （ ） (A) 钝角三角形 (B)锐角三角形 (C)直角三角形 (D)等腰三角形 6.已知α的终边经过P （ππ6 5cos ,6 5sin ），则α可能是 （ ） A ．π6 5 B ． 6 π C ．3 π- D ．3 π 7．如果).cos(|cos |π+-=x x 则x 的取值范围是 （ ） A ．)(] 22 ,22 [Z k k k ∈++-ππππ B ．)() 22 3,22 (Z k k k ∈++ππππ C ．)(] 22 3,22 [Z k k k ∈++ππππ D ．)()2,2(Z k k k ∈++-ππππ 8.1tan sin )(++=x b x a x f ，满足.7)5(=f 则)5(-f 的值为 （ ） A ．5 B ．－5 C ．6 D ．－6 9. 扇形的周期是16，圆心角是2弧度，则扇形面积是______________
课时跟踪检测(四十四) 简单的三角恒等变换
课时跟踪检测（四十四） 简单的三角恒等变换 A 级——学考水平达标练 1．已知2sin α＝1＋cos α，则tan α 2＝( ) A ．1 2 B ．1 2或不存在 C ．2 D ．2或不存在 解析：选B 2sin α＝1＋cos α，即4sin α2cos α2＝2cos 2α2，当cos α2＝0时，tan α 2不存在， 当cos α2≠0时，tan α2＝1 2 . 2．若cos 2α＝－4 5，且α∈????π2，π，则sin α＝( ) A ．310 10 B ． 1010 C ．35 D ．－ 1010 解析：选A 因为α∈????π 2，π，所以sin α≥0，由半角公式可得sin α＝ 1－cos 2α 2 ＝310 10 . 3．设a ＝12cos 6°－3 2sin 6°，b ＝2sin 13°cos 13°，c ＝ 1－cos 50° 2 ，则有( ) A ．c ＜b ＜a B ．a ＜b ＜c C ．a ＜c ＜b D ．b ＜c ＜a 解析：选C 由已知可得a ＝sin 24°，b ＝sin 26°，c ＝sin 25°，所以a ＜c ＜b . 4．已知tan 2α＝－22，π4＜α＜π 2，则2cos 2α 2－sin α－1 2sin ????α＋π4＝( ) A ．－3＋2 2 B ．3－2 2 C ．－ 2 D ． 2 解析：选A 因为tan 2α＝－22，π4＜α＜π 2， 所以tan 2α＝2tan α 1－tan 2α ＝－22，解得tan α＝2，
所以2cos 2α 2 －sin α－12sin ????α＋π4＝cos α－sin αcos α＋sin α＝1－tan α1＋tan α＝1－2 1＋2＝－3＋2 2. 5．若sin θ＝35，5π 2＜θ＜3π，则tan θ2＋cos θ2＝( ) A ．3＋ 1010 B ．3－ 1010 C ．3＋310 10 D ．3－310 10 解析：选B 因为5π 2＜θ＜3π，所以cos θ＝－ 1－sin 2θ＝－45.因为5π4＜θ2＜3π 2，所以sin θ2 ＜0，cos θ2＜0，所以sin θ 2 ＝－ 1－cos θ2＝－310 10，cos θ2 ＝－ 1＋cos θ2＝－10 10 ，所以tan θ2＝sin θ 2cos θ2 ＝3.所以tan θ2＋cos θ2＝3－10 10. 6．若3sin x －3cos x ＝23sin(x ＋φ)，φ∈(－π，π)，则φ＝________. 解析：因为3sin x －3cos x ＝23 ??? ?32sin x －12cos x ＝23sin ??? ?x －π6， 又φ∈(－π，π)，所以φ＝－π 6. 答案：－π 6 7．若sin α1＋cos α＝1 2 ，则sin α＋cos α的值为________． 解析：∵sin α1＋cos α＝tan α2＝1 2，∴sin α＋cos α＝2tan α2 1＋tan 2α2＋1－tan 2α21＋tan 2 α2＝2×12＋1－ 1 41＋ 14＝75. 答案：7 5 8．已知等腰三角形的顶角的正弦值为5 13，则它的底角的余弦值为________． 解析：设等腰三角形的顶角为α，则底角为π－α2，由题意可知sin α＝5 13 ，所以cos α＝
高中语文(人教版)必修一同步练习：课时跟踪检测(八)(含解析)
课时跟踪检测（八） (时间：40分钟满分：50分) 一、基础巩固(15分) 1．下列词语中，加点字的注音全都正确的一组是(3分)() A．贫血．(xuè)血．淋淋(xiě) 叱．骂(chì) 拈．轻怕重(zhān) B．狗爪．(zhuǎ) 爪．牙(zhǎo) 解剖．(pāo) 舐．犊情深(shì) C．篱笆 ．．(lí ba) 拍摄．(niè) 罢黜．(chù) 罄．竹难书(qìng) D．镌．刻(juān) 譬．如(pì) 濒．危(bīn) 掷．地有声(zhì) 解析：选D A项，“拈”应读niān。B项，“剖”应读pōu。C项，“摄”应读shè。 2．下列词语中，没有错别字的一组是(3分)() A．租赁驾驭呻唤坛坛罐罐 B．熬煎寂寞歉意绿草如荫 C．羞耻责备款待坠落腐败 D．虫蛀内涵做揖逆来顺受 解析：选A B项，荫—茵。C项，坠—堕。D项，做—作。 3．依次填入下列各句横线处的词语，恰当的一项是(3分)() ①又有一次日本作家由起女士访问上海，来我家______，对日本产的包弟非常喜欢，她说她在东京家中也养了狗。 ②为了什么理由，好多平方英里的没有人迹的森林，遭人类______而为我所私有了吗？ ③如果雨下得太久，就会使地里的种子，低地的土豆烂掉，但它对高地草还是有好处的。______它对高地草很好，对我是______很幸运的了。 A．作客遗弃即使/也 B．作客丢弃既然/也 C．做客丢弃即使/也 D．做客遗弃既然/也 解析：选D“作客”，指寄居在别处。“做客”，访问别人，自己当客人。“遗弃”，抛弃，扔掉不要。“丢弃”，丢掉(原有的权利、主张、意见等)。“即使”，表示让步假设。“既然”，用于上句，指出已经成为现实的或已肯定的前提，下半句根据这个前提推出结论，常用“就”“也”“还”呼应。 4．下列各句中，没有语病、句意明确的一句是(3分)()
高中英语Unit2 课时跟踪检测四牛津译林版必修4
课时跟踪检测（四） Project Ⅰ.单句语法填空 1．Do you know how long it was before you realized that he was always helping me? 2．The school took the students' requirement into consideration that a party should_be_held (hold) to celebrate the victory. 3．Before you leave, make sure to keep the boat tied (tie) to the tree over there. 4．With the temperature falling so rapidly, we couldn't go on with the experiment. 5．Surprised (surprise) and happy, Tony stood up and accepted the prize. Ⅱ.完成句子 1．我们也建议人们多吃蔬菜和水果，减少动物脂肪的摄入，当然，也得控制体重。 We also advise people to eat a diet containing plenty of fruit and vegetables, to reduce their intake of animal fat, and of course, to_keep_their_weight_under_control. 2．趁他还没有退休，他正在尽力培养李明。 He is trying his best to train Li Ming before_he_retires. 3．妈妈让他弄掉鞋上的泥。 Mother told him to_remove_the_mud_from his shoes. 4．随着电脑越来越普及，手写信逐渐让位于电子邮件。 With computers becoming more and more popular nowadays, hand-written letters are gradually making_way_for e-mails. 5．你很难在这个城市找到一个满足你要求的旅馆。 It is hard for you to find a hotel in this city that can meet_your_expectations/requirements. 6．即使你申请加入他们的俱乐部，他们也不会同意的。 Even if you apply_to_join_their_club,_they will not approve it. Ⅲ.完形填空 It's commonly believed that school kids started taking summers off in the 19th century so they'd have time to work on the farm. Nice as that story is, it isn't __1__. Summer vacation has little to do with working in the __2__ and more to do with __3__，rich city kids playing hooky (逃学) and their parents. Before the Civil War, farm kids __4__ had summers off. They went to school during the hottest and __5__ months and stayed home during the spring and fall, when crops