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问题 解答题
已知tanα=2，试求值；
（1）
2sin(π-α)-cos(α+
π
2
)
cos(α-π)+sin(α-
π
2
)

（2）sin2α-sinα•cosα-2cos2α
答案

（1）

2sin(π-α)-cos(α+
π
2
)
cos(α-π)+sin(α-
π
2
)
=
2sinα+sinα
-cosα-cosα
=-
3
2
，tanα=-3

（2）sin2α-sinα•cosα-2cos2α=

sin 2α-sinα•cosα-2cos 2α 
sin2α+cos2α 
=
tan2α-tanα-2
tan2α+1
=0

听力题

单词拼写（共10小题；每小题   0.5分，满分5分）

小题1:.Many people hoped to attend the Guangzhou Asian Games Opening C________?

小题2:. It’s going to p          with rain. People in the street hurriedly run for shelters.

小题3:. The Englishmen are d          that the 2012 Olympic Games will be held in London.

小题4:. She is always o_____________, even when she has to face the failure in work and life.

小题5:. Science and technology are c_______________ improving, so we must work hard to keep pace with the development. Otherwise, we may fall behind.

小题6:. She doesn’t _____________(赞成)of my leaving school.

小题7:.I cannot _______________(集中) on my work when I’m hungry.

小题8:. Pictures of the suspect were _____________(公布)in all the daily papers.

小题9: My parents still __________________ (对待)me like a child.

小题10:. I have been reliably __________________ (通知) that the couple will marry next year.
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单项选择题

(31)到(35)题使用如下数据表。
“学生”表：学号C(8)，姓名C(8)，性别C(2)，系名(10)，出生日期D
“课程”表：课程编号C(4)，课程名称C(12)，开课系名C(10)
“成绩”表：学号C(8)，课程编号C(4)，成绩I

检索每门课程的总分，查询结果中包括课程名和总分，正确的语句是( )。

A．SELECT课程名称,COUNT(成绩)AS总分FROM课程,成绩； WHERE课程.课程编号=成绩.课程编号； ORDER BY成绩.课程编号

B．SELECT课程名称,COUNT(成绩)AS总分FROM课程,成绩； WHERE课程.课程编号=成绩.课程编号； GROUP BY成绩.课程编号

C．SELECT课程名称,SUM(成绩)AS总分FROM课程,成绩； WHERE课程.课程编号=成绩.课程编号； ORDERBY成绩.课程编号

D．SELECT课程名称,SUM(成绩)AS总分FROM课程,成绩； WHERE课程.课程编号=成绩.课程编号； GROUP BY成绩.课程编号
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