可汗学院数学题目-38Angles,arclengths,andtrigfunctions
2000年AMC8竞赛试题(英文版)汇总
Problem 1Aunt Anna is years old. Caitlin is years younger than Brianna, and Brianna is half as old as Aunt Anna. How old is Caitlin?安妮今年42岁，凯琳比柏娜小五岁，而柏娜的年龄是安妮的一半。
试问凯琳今年几岁？Problem 2Which of these numbers is less than its reciprocal?下列那一个数小于它的倒数？Problem 3How many whole numbers lie in the interval between and 有多少个整数介于5/3和2π之间？Problem 4In only of the working adults in Carlin City worked at home. By the "at-home" work force increased to . In there wereapproximately working at home, and in there were . The graph that best illustrates this is在卡林市，1960年只有5%的成年工作者在家工作，至1970年在家工作人数增加到8%，1960年大约有15%的人在家工作，而在1990年则有30%。
试问下面那一个图是这种情形的最佳说明Problem 5Each principal of Lincoln High School serves exactly one -year term. What is the maximum number of principals this school could have during an -year period?林肯中学每一位校长都洽服务一次三年任期，则在8年期间林肯中学最多有几位校长？Problem 6Figure is a square. Inside this square three smaller squares are drawn with the side lengths as labeled. The area of the shaded L-shaped region is右图ABCD是正方形。
可汗学院新SAT语法真题下载
可汗学院新SAT语法真题下载到目前为止，新版SAT可汗学院官方不断放出更多真题，已经放出了68篇阅读，且之前已经和大家分享过可汗学院新SAT阅读真题，想要下载的同学，请点击：新SAT阅读真题下载（共68篇，且已全）目前可汗学院一共放出41篇新SAT数学真题！想要下载的同学，请点击：新SAT数学真题下载（共41篇）分享了可汗学院的数学和阅读真题后，还有我们的可汗学院SAT语法真题。
截止到6月前，可汗学院一共放出了48套新SAT语法真题，想要吗下载请点击：新SAT语法真题下载（共48篇）（网址：）可汗学院新SAT语法真题（部分）Questions 1-5 are based on the following passage. 1Searching for GuinevereStories of kings and queens have captivated readers for centuries, and arguably, the tales of King Arthur and Guinevere are among the most enchanting. Arthur ruled the kingdom of Camelot, and Guinevere was his queen. But were they real people or fictional characters The debate has continued for centuries. Though many scholars have found evidence that the legendary Arthur was, at the very least, based on a real person who lived in Britain roughly between 450 and （1） 500 CE. They continue to search for the historical identity of Guinevere. Guinevere first appeared as King Arthur’s queen in one of the most widelystudied works of Arthurian literature, （2）The History of the Kings of Britain. This book was written by Geoffrey of Monmouth around 1135 CE. Geoffrey’s historical treatment of the legend is often（3）sited as evidence that the queen of Camelot existed, as the book chronicles the lives of a number of historical rulers.*God help those who help themselves. We help those who trust us. Contact Wechat:satxbs123, help is waiting.1A) NO CHANGEB) 500 CE. ContinuingC) 500 CE, continuingD) 500 CE, they continue2Which choice most effectively combines the sentences at the underlined portionA) The History of the Kings of Britain, and this bookB) The History of the Kings of Britain, whichC) a book called The History of the Kings of Britain,as thisD) a book called The History of the Kings of Britain,and this3A) NO CHANGEB) insightedC) citedD) incitedGuinevere is identified by Geoffrey as a noblewoman of Roman descent who met King Arthur in the court of Duke Cador of Cornwall, where she lived as a ward. (4)In Malory’s portrayal, Guinevere had no real power as a monarch but served as a kind of spiritual leader, providing guidance and moral support to the knights in their roles as defenders of the kingdom. Le Morte d’Arthur was also one of the first works to reference Guinevere’s romance with the knight, Sir many Arth urian scholars know, the distinction between history and literature was blurred in the Middle Ages. Consequently, the true identity of Guinevere may never be known with certainty. Yet regardless of whether Guinevere was real or fictional, her story (5) had endured centuries—and through each retelling, she continues to live on in the imaginations of people around the world.4At this point, the author wants to add a sentence which effectively sets up the portrayal of Guinevere discussed in the rest of the paragraph. Which choice best accomplishes this goalA) Three centuries later, however, Thomas Malory painted a very different portrait of Guinevere in Le Morte d’Arthur.B) Sir Thomas Malory was an English knight and Member of Parliament who also wrote extensively about the history of the British monarchy.C) Many historians believe that the portrayal of Arthur and Guinevere in Sir Thomas Malory’s Le Morte d’Arthur was actually a political commentary on the War of the Roses (1455-1487CE).D) In Le Morte d’Art hur, Sir Thomas Malory describes an idyllic England under King Arthur and Guinevere, which eventually collapses into chaos and political unrest. E. I would be guessing.5A) NO CHANGEB) was enduringC) would have enduredD) has enduredQuestions 1-5 are based on the following passage. 1Cometary Missions: Trajectory for SuccessScientists have been launching cometary missions since 1978. The first one, a joint mission by the European Space Agency, and the National Aeronautics and Space Administration (NAS A), was a “flyby” in which the spacecraft collected data while passing around Comet Giacobini-Zinner. (1)However, the landing of the Rosetta space probe on comet 67P/Churyumov-Gerasemenko in 2014 was different: it marked the first time that a probe landed on a( 2 )comet and giving scientists an unprecedented opportunity to study the surface of a comet. In order to continue this valuable research, additional missions are needed; thus, it is critical that more funding be allocated for this 2014 Rosetta mission provided a rare opportunity for scientists to test a number of hypotheses regarding the composition of (3) comets; the distribution of organic compounds in our solar system and the origins of life on Earth. Unlike other cometary missions, the Rosetta spacecraft contained a probe, Philae, that was able to land on the surface of a comet. *Rack your brain and you don't know Wechat: satxbs123, she can help you!1At this point, the writer wants to add accurate information from the graph. Which choice best accomplishes this goalA) From 1978 to 2014, the number of successful missions increased from 28 percent to 72 percent.B) Before 2014, the majority of attempted cometary missions were considered unsuccessful.C) Between then and 2014, 72 percent of the cometary missions were successful.D) Of the missions attempted since then, 44 percent have been successful.2A) NO CHANGEB) comet, but it gaveC) comet, yet givesD) comet, giving3A) NO CHANGEB) comets, the distribution of organic compounds in our solar system,C) comets, the distribution of organic compounds in our solar system;D) comets; the distribution of organic compounds in our solar system,。
六年级数学试卷英语【含答案】
六年级数学试卷英语【含答案】专业课原理概述部分一、选择题（每题1分，共5分）1. What is the value of 8^2?A. 16B. 64C. 128D. 2562. If 4x 7 = 15, what is the value of x?A. 1B. 3C. 4D. 63. Which of the following is a prime number?A. 12B. 17C. 20D. 214. What is the area of a circle with a radius of 4 cm?A. 16π cm²B. 8π cm²C. 4π cm²D. π cm²5. If 3x + 5 = 14, what is the value of x?A. 1B. 2C. 3D. 4二、判断题（每题1分，共5分）6. The sum of the angles in a triangle is 180 degrees. (True/False)7. 0 is a natural number. (True/False)8. The square root of 49 is 7. (True/False)9. 2^3 is equal to 3^2. (True/False)10. The value of π is exactly 3.14159. (True/False)三、填空题（每题1分，共5分）11. The square root of 64 is ________.12. If 5x = 25, then x = ________.13. The formula to calculate the area of a rectangle is ________.14. The value of sin(90°) is ________.15. The perimeter of a square with side length 6 cm is ________ cm.四、简答题（每题2分，共10分）16. Expln how to find the factors of a number.17. Describe the properties of parallel lines.18. What is the difference between a prime number and a posite number?19. How do you calculate the volume of a cube?20. Expln the concept of the Pythagorean theorem.五、应用题（每题2分，共10分）21. A rectangular garden is 12 meters long and 8 meters wide. What is its area?22. If the temperature is 0°C and it drops 5°C every hour, what will be the temperature after 3 hours?23. A car travels at a speed of 60 km/h for 2 hours. How far does it travel?24. If you have 3 apples and you divide each apple into 4 equal parts, how many parts do you have in total?25. What is the value of the expression 15 8 + 3?六、分析题（每题5分，共10分）26. Find the roots of the quadratic equation x² 5x + 6 = 0.27. Expln how to solve a proportion and provide an example.七、实践操作题（每题5分，共10分）28. Draw a triangle with angles measuring 45°, 45°, and 90°. Label the sides and expln the properties of the triangle.29. Calculate the area and perimeter of a circle with a diameter of 10 cm.八、专业设计题（每题2分，共10分）30. Design a logo for a fictional pany named "Math Wizards". Include geometric shapes and a mathematical symbol in your design.31. Create a floor plan for a classroom with dimensions 10 meters 8 meters. Include spaces for a teacher's desk, students' desks, a storage closet, and a small library area.32. Design a poster promoting a school science fr. Include images of scientific experiments, equations, and encouraging slogans.33. Illustrate a scene from a story you read in English class, including characters, setting, and key events.34. Create a concept map showing the mn ideas and supporting detls of a recent history lesson.九、概念解释题（每题2分，共10分）35. Expln the difference between a histogram and a bar graph.36. Define the term "acidity" in chemistry and provide an example of an acidic substance.37. Describe the concept of "alliteration" in English literature and give an example.38. Expln the role of the mitochondria in a cell.39. Define the term " democracy" and provide an example of a democratic country.十、思考题（每题2分，共10分）40. If you could invent a new math operation, what would it be and how would it work?41. How can renewable energy sources help reduce pollution?42. What are the advantages and disadvantages of using social media?43. If you could travel back in time, what historical event would you like to witness and why?44. How can exercise improve mental health?十一、社会扩展题（每题3分，共15分）45. Research a famous mathematician and write a short biography about their life and contributions to the field of mathematics.46. Describe the process of photosynthesis and expln how it is important for life on Earth.47. Discuss the impact of technology on education, including both positive and negative effects.48. Write an essay about the importance of water conservation and provide tips on how to save water at home and in the munity.49. Expln the significance of cultural diversity in society and provide examples of how different cultures contribute to the global munity.本专业课原理概述部分试卷答案及知识点总结如下：一、选择题答案1. B2. B3. B4. A5. C二、判断题答案6. True7. False8. True9. False10. False三、填空题答案11. 812. 513. 长度宽度14. 115. 24四、简答题答案16. Factors are numbers that can be multiplied together to get another number. For example, the factors of 12 are 1, 2, 3, 4, 6, and 12.17. Parallel lines are lines that never intersect, no matter how far they extend. They have the same slope.18. A prime number is a number greater than 1 that has only two factors, 1 and itself, while a posite number has more than two factors.19. The volume of a cube is calculated multiplying the length of one side itself twice. For example, if the side length of a cube is 3 cm, the volume is 3 cm 3 cm 3 cm = 27 cm³.20. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. For example, in a right triangle with sides of length 3 cm and 4 cm, the hypotenuse has a length of 5 cm.五、应用题答案21. 96 square meters22. -15°C23. 120 km24. 12 parts25. 10六、分析题答案26. The roots of the equation x² 5x + 6 = 0 are x = 2 and x = 3.27. To solve a proportion, set up two ratios equal to each other and cross-multiply. For example, to solve the proportion 2/3 = x/9, cross-multiply to get 3x = 2 9, then divide both sides 3 to find x = 6.七、实践操作题答案28. A 45-45-90 triangle has two sides of equal length and a hypotenuse that is the square root of 2 times the length of the other sides. For example, if each leg of the triangle is 1 cm, the hypotenuse is approximately 1.41 cm.29. The area of the circle is 78.54 cm² and the perimeter (circumference) is 31.42 cm.知识点总结及各题型考察的学生知识点详解：本试卷涵盖了六年级数学和英语的基本概念和原理，包括算术、几何、代数、科学、语言艺术等领域的知识点。
2024年高考数学真题分类汇编(三角函数篇,解析版)
专题三角函数1(新课标全国Ⅰ卷)已知cos (α+β)=m ,tan αtan β=2，则cos (α-β)=()A.-3mB.-m3C.m 3D.3m【答案】A【分析】根据两角和的余弦可求cos αcos β,sin αsin β的关系，结合tan αtan β的值可求前者，故可求cos α-β 的值.【详解】因为cos α+β =m ，所以cos αcos β-sin αsin β=m ，而tan αtan β=2，所以=12×2b ×kb ×sin A 2+12×kb ×b ×sin A2，故cos αcos β-2cos αcos β=m 即cos αcos β=-m ，从而sin αsin β=-2m ，故cos α-β =-3m ，故选：A .2(新课标全国Ⅰ卷)当x ∈[0,2π]时，曲线y =sin x 与y =2sin 3x -π6 的交点个数为()A.3B.4C.6D.8【答案】C【分析】画出两函数在0,2π 上的图象，根据图象即可求解【详解】因为函数y =sin x 的的最小正周期为T =2π，函数y =2sin 3x -π6 的最小正周期为T =2π3，所以在x ∈0,2π 上函数y =2sin 3x -π6有三个周期的图象，在坐标系中结合五点法画出两函数图象，如图所示：由图可知，两函数图象有6个交点.故选：C3(新课标全国Ⅱ卷)设函数f (x )=a (x +1)2-1，g (x )=cos x +2ax ，当x ∈(-1,1)时，曲线y =f (x )与y =g (x )恰有一个交点，则a =()A.-1B.12C.1D.22024年高考数学真题分类汇编——三角函数篇【分析】解法一：令F x =ax 2+a -1,G x =cos x ，分析可知曲线y =F (x )与y =G (x )恰有一个交点，结合偶函数的对称性可知该交点只能在y 轴上，即可得a =2，并代入检验即可；解法二：令h x =f (x )-g x ,x ∈-1,1 ，可知h x 为偶函数，根据偶函数的对称性可知h x 的零点只能为0，即可得a =2，并代入检验即可.【详解】解法一：令f (x )=g x ，即a (x +1)2-1=cos x +2ax ，可得ax 2+a -1=cos x ，令F x =ax 2+a -1,G x =cos x ，原题意等价于当x ∈(-1,1)时，曲线y =F (x )与y =G (x )恰有一个交点，注意到F x ,G x 均为偶函数，可知该交点只能在y 轴上，可得F 0 =G 0 ，即a -1=1，解得a =2，若a =2，令F x =G x ，可得2x 2+1-cos x =0因为x ∈-1,1 ，则2x 2≥0,1-cos x ≥0，当且仅当x =0时，等号成立，可得2x 2+1-cos x ≥0，当且仅当x =0时，等号成立，则方程2x 2+1-cos x =0有且仅有一个实根0，即曲线y =F (x )与y =G (x )恰有一个交点，所以a =2符合题意；综上所述：a =2.解法二：令h x =f (x )-g x =ax 2+a -1-cos x ,x ∈-1,1 ，原题意等价于h x 有且仅有一个零点，因为h -x =a -x 2+a -1-cos -x =ax 2+a -1-cos x =h x ，则h x 为偶函数，根据偶函数的对称性可知h x 的零点只能为0，即h 0 =a -2=0，解得a =2，若a =2，则h x =2x 2+1-cos x ,x ∈-1,1 ，又因为2x 2≥0,1-cos x ≥0当且仅当x =0时，等号成立，可得h x ≥0，当且仅当x =0时，等号成立，即h x 有且仅有一个零点0，所以a =2符合题意；故选：D .4(全国甲卷数学(理)(文))已知cos αcos α-sin α=3，则tan α+π4=()A.23+1 B.23-1C.32D.1-3【答案】B【分析】先将cos αcos α-sin α弦化切求得tan α，再根据两角和的正切公式即可求解.【详解】因为cos αcos α-sin α=3，所以11-tan α=3，⇒tan α=1-33，所以tan α+π4 =tan α+11-tan α=23-1，故选：B .5(新高考北京卷)已知f x =sin ωx ω>0 ，f x 1 =-1，f x 2 =1，|x 1-x 2|min =π2，则ω=()A.1B.2C.3D.4【分析】根据三角函数最值分析周期性，结合三角函数最小正周期公式运算求解.【详解】由题意可知：x 1为f x 的最小值点，x 2为f x 的最大值点，则x 1-x 2 min =T 2=π2，即T =π，且ω>0，所以ω=2πT=2.故选：B .6(新高考天津卷)已知函数f x =sin3ωx +π3ω>0 的最小正周期为π．则函数在-π12,π6 的最小值是()A.-32B.-32C.0D.32【答案】A【分析】先由诱导公式化简，结合周期公式求出ω，得f x =-sin2x ，再整体求出x ∈-π12,π6时，2x 的范围，结合正弦三角函数图象特征即可求解.【详解】f x =sin3ωx +π3 =sin 3ωx +π =-sin3ωx ，由T =2π3ω=π得ω=23，即f x =-sin2x ，当x ∈-π12,π6 时，2x ∈-π6,π3，画出f x =-sin2x 图象，如下图，由图可知，f x =-sin2x 在-π12,π6上递减，所以，当x =π6时，f x min =-sin π3=-32故选：A7(新高考上海卷)下列函数f x 的最小正周期是2π的是()A.sin x +cos xB.sin x cos xC.sin 2x +cos 2xD.sin 2x -cos 2x【答案】A【分析】根据辅助角公式、二倍角公式以及同角三角函数关系并结合三角函数的性质一一判断即可 .【详解】对A ，sin x +cos x =2sin x +π4，周期T =2π，故A 正确；对B ，sin x cos x =12sin2x ，周期T =2π2=π，故B 错误；对于选项C ，sin 2x +cos 2x =1，是常值函数，不存在最小正周期，故C 错误；对于选项D ，sin 2x -cos 2x =-cos2x ，周期T =2π2=π，故D 错误，故选：A ．8(新课标全国Ⅱ卷)对于函数f(x)=sin2x和g(x)=sin2x-π4，下列说法正确的有() A.f(x)与g(x)有相同的零点 B.f(x)与g(x)有相同的最大值C.f(x)与g(x)有相同的最小正周期D.f(x)与g(x)的图像有相同的对称轴【答案】BC【分析】根据正弦函数的零点，最值，周期公式，对称轴方程逐一分析每个选项即可.【详解】A选项，令f(x)=sin2x=0，解得x=kπ2,k∈Z，即为f(x)零点，令g(x)=sin2x-π4=0，解得x=kπ2+π8,k∈Z，即为g(x)零点，显然f(x),g(x)零点不同，A选项错误；B选项，显然f(x)max=g(x)max=1，B选项正确；C选项，根据周期公式，f(x),g(x)的周期均为2π2=π，C选项正确；D选项，根据正弦函数的性质f(x)的对称轴满足2x=kπ+π2⇔x=kπ2+π4,k∈Z，g(x)的对称轴满足2x-π4=kπ+π2⇔x=kπ2+3π8,k∈Z，显然f(x),g(x)图像的对称轴不同，D选项错误.故选：BC9(新课标全国Ⅱ卷)已知α为第一象限角，β为第三象限角，tanα+tanβ=4，tanαtanβ=2+1，则sin(α+β)=.【答案】-22 3【分析】法一：根据两角和与差的正切公式得tanα+β=-22，再缩小α+β的范围，最后结合同角的平方和关系即可得到答案；法二：利用弦化切的方法即可得到答案.【详解】法一：由题意得tanα+β=tanα+tanβ1-tanαtanβ=41-2+1=-22，因为α∈2kπ,2kπ+π2,β∈2mπ+π,2mπ+3π2，k,m∈Z，则α+β∈2m+2kπ+π,2m+2kπ+2π，k,m∈Z，又因为tanα+β=-22<0，则α+β∈2m+2kπ+3π2,2m+2kπ+2π，k,m∈Z，则sinα+β<0，则sinα+βcosα+β=-22，联立sin2α+β+cos2α+β=1，解得sinα+β=-223.法二：因为α为第一象限角，β为第三象限角，则cosα>0,cosβ<0,cosα=cosαsin2α+cos2α=11+tan2α，cosβ=cosβsin2β+cos2β=-11+tan2β，则sin(α+β)=sinαcosβ+cosαsinβ=cosαcosβ(tanα+tanβ)=4cosαcosβ=-41+tan2α1+tan2β=-4(tanα+tanβ)2+(tanαtanβ-1)2=-442+2=-223故答案为：-22 3.10(全国甲卷数学(文))函数f x =sin x-3cos x在0,π上的最大值是．【答案】2【分析】结合辅助角公式化简成正弦型函数，再求给定区间最值即可.【详解】f x =sin x -3cos x =2sin x -π3 ，当x ∈0,π 时，x -π3∈-π3,2π3，当x -π3=π2时，即x =5π6时，f x max =2.故答案为：2一、单选题1(2024·宁夏石嘴山·三模)在平面直角坐标系中，角θ的顶点与原点重合，始边与x 轴的非负半轴重合，终边经过点P 1,2 ，则7cos 2θ-2sin2θ=()A.-15B.15C.-2D.2【答案】A【分析】由题意可知：tan θ=2，根据倍角公式结合齐次化问题分析求解.【详解】由题意可知：tan θ=2，所以7cos 2θ-2sin2θ=7cos 2θ-4sin θcos θsin 2θ+cos 2θ=7-4tan θtan 2θ+1=7-4×222+1=-15.故选：A .2(2024·广东茂名·一模)已知cos α+π =-2sin α，则sin 2α-3cos α+π2cos αcos2α+1=()A.-1B.-25C.45D.78【答案】D【分析】根据给定条件，求出tan α，再结合诱导公式及二倍角的余弦公式，利用正余弦齐次式法计算得解.【详解】由cos α+π =-2sin α，得cos α=2sin α，则tan α=12，所以sin 2α-3cos α+π2 cos αcos2α+1=sin 2α+3sin αcos α2cos 2α=12tan 2α+32tan α=18+34=78.故选：D3(2024·河北保定·二模)函数f (x )=1-e x1+e xcos2x 的部分图象大致为()A. B.C. D.【答案】A【分析】根据函数的奇偶性判断即可.【详解】设g x =1-e x1+e x，则g-x=1-e-x1+e-x=e x-11+e x=-g x ，所以g x 为奇函数，设h x =cos2x，可知h x 为偶函数，所以f x =1-e x1+e xcos2x为奇函数，则B，C错误，易知f0 =0，所以A正确，D错误．故选：A.4(2024·山东济宁·三模)已知函数f(x)=(3sin x+cos x)cos x-12，若f(x)在区间-π4,m上的值域为-3 2,1，则实数m的取值范围是()A.π6,π2B.π6,π2C.π6,7π12D.π6,7π12【答案】D【分析】利用二倍角公式、辅助角公式化简函数f(x)，再借助正弦函数的图象与性质求解即得.【详解】依题意，函数f(x)=3sin x cos x+cos2x-12=32sin2x+12cos2x=sin2x+π6，当x∈-π4,m时，2x+π6∈-π3,2m+π6，显然sin-π3=sin4π3=-32,sinπ2=1，且正弦函数y=sin x在π2,4π3上单调递减，由f(x)在区间-π4,m上的值域为-32,1，得π2≤2m+π6≤4π3，解得π6≤m≤7π12，所以实数m的取值范围是π6,7π12.故选：D5(2024·江西景德镇·三模)函数f x =cosωx x∈R在0,π内恰有两个对称中心，fπ=1，将函数f x 的图象向右平移π3个单位得到函数g x 的图象.若fα +gα =35，则cos4α+π3=()A.725B.1625C.-925D.-1925【答案】A【分析】根据y轴右边第二个对称中心在0,π内，第三个对称中心不在0,π内可求得32≤ω<52，结合fπ=1可得ω=2，再利用平移变换求出g x ，根据三角变换化简fα +gα =35可得sin2α+π6=35，然后由二倍角公式可解.【详解】由x∈0,π得ωx∈0,ωπ，因为函数f x 在0,π内恰有两个对称中心，所以3π2≤ωπ5π2>ωπ，解得32≤ω<52，又fπ=cosωπ=1，所以ωπ=kπ,k∈Z，即ω=k,k∈Z，所以ω=2，将函数f x 的图象向右平移π3个单位得到函数y=cos2x-π3=cos2x-2π3，即g x =cos2x-2π3，因为fα +gα =cos2α+cos2α-2π3=32sin2α+12cos2α=sin2α+π6=35，所以cos4α+π3=1-2sin22α+π6=1-2×35 2=725.故选：A6(2024·安徽马鞍山·三模)已知函数f(x)=sin2ωx+cos2ωx(ω>1)的一个零点是π2，且f(x)在-π6,π16上单调，则ω=()A.54B.74C.94D.114【答案】B【分析】整理可得f(x)=2sin2ωx+π4，以2ωx+π4为整体，根据单调性分析可得1<ω≤2，再结合零点分析求解.【详解】因为f(x)=sin2ωx+cos2ωx=2sin2ωx+π4，x∈-π6,π16，且ω>1时，可得2ωx+π4∈-π3ω+π4,π8ω+π4，且-π3ω+π4<0<π8ω+π4，若f(x)在-π6,π16上单调，则-π3ω+π4≥-π2π8ω+π4≤π2，解得1<ω≤2，又因为f(x)的一个零点是π2，则πω+π4=kπ,k∈Z，解得ω=k-14,k∈Z，所以k=2,ω=7 4 .故选：B.7(2024·山东临沂·二模)已知函数f x =sin2x+φϕ <π2图象的一个对称中心为π6,0，则()A.f x 在区间-π8,π3上单调递增B.x=5π6是f x 图象的一条对称轴C.f x 在-π6,π4上的值域为-1,32D.将f x 图象上的所有点向左平移5π12个长度单位后，得到的函数图象关于y轴对称【答案】D【分析】借助整体代入法结合正弦函数的性质可得A、B；结合正弦函数最值可得C；得到平移后的函数解析式后借助诱导公式即可得D.【详解】由题意可得2×π6+φ=kπk∈Z，解得φ=-π3+kπk∈Z，又ϕ <π2，故φ=-π3，即f x =sin2x-π3；对A ：当x ∈-π8,π3 时，2x -π3∈-7π12,π3，由函数y =sin x 在-7π12,π3上不为单调递增，故f x 在区间-π8,π3上不为单调递增，故A 错误；对B ：当x =5π6时，2x -π3=4π3，由x =4π3不是函数y =sin x 的对称轴，故x =5π6不是f x 图象的对称轴，故B 错误；对C ：当x ∈-π6,π4 时，2x -π3∈-2π3,π6，则f x ∈-1,12，故C 错误；对D ：将f x 图象上的所有点向左平移5π12个长度单位后，可得y =sin 2x +2×5π12-π3 =sin 2x +π2=cos2x ，该函数关于y 轴对称，故D 正确.故选：D .8(2024·广东广州·二模)已知函数f (x )=2sin (ωx +φ)ω>0,|φ|<π2的部分图象如图所示，若将函数f (x )的图象向右平移θ(θ>0)个单位后所得曲线关于y 轴对称，则θ的最小值为()A.π8B.π4C.3π8D.π2【答案】A【分析】根据给定的图象特征，结合五点法作图列式求出ω和φ，再根据图象的平移变换，以及图象的对称性即可得解.【详解】由f π4=1，得sin π4ω+φ =22，又点π4,1 及附近点从左到右是上升的，则π4ω+φ=π4+2k π,k ∈Z ，由f 5π8 =0，点5π8,0 及附近点从左到右是下降的，且上升、下降的两段图象相邻，得5π8ω+φ=π+2k π,k ∈Z ，联立解得ω=2，φ=-π4+2k π,k ∈Z ，而|φ|<π2，于是φ=-π4，f (x )=2sin 2x -π4，若将函数f (x )的图像向右平移θ(θ>0)个单位后，得到y =sin 2x -2θ-π4，则-2θ-π4=π2-k π,k ∈Z ，而θ>0，因此θ=-3π8+k π2,k ∈N ，所以当k =1时，θ取得最小值为π8.故选：A9(2024·四川雅安·三模)已知函数f x =sin ωx +3cos ωx (ω>0)，则下列说法中正确的个数是()①当ω=2时，函数y =f x -2log πx 有且只有一个零点；②当ω=2时，函数y =f x +φ 为奇函数，则正数φ的最小值为π3；③若函数y =f x 在0,π3 上单调递增，则ω的最小值为12；④若函数y =f x 在0,π 上恰有两个极值点，则ω的取值范围为136,256.A.1 B.2C.3D.4【答案】B【分析】利用辅助角公式化简函数，由图象分析判断①；由正弦函数的性质判断②③；由极大值的意义结合正弦函数的性质判断④.【详解】依题意，ω>0，函数f (x )=212sin ωx +32cos ωx =2sin ωx +π3，对于①：f (x )=2sin 2x +π3，令y =f x -2log πx =0，即f x =2log πx ，作出函数y =f (x )和函数y =2log πx 的图象，如图，观察图象知，两个函数在0,7π12 上只有一个零点，f 13π12 =2sin 5π2=2，当x =13π12时，y =2log π13π12=2log π1312+2log ππ=2+2log π1312>2，当x >13π12时，2log πx >2≥f (x )，因此函数y =f x 与函数y =2log πx 的图象有且只有一个交点，①正确；对于②：f (x +φ)=2sin 2x +2φ+π3 为奇函数，则2φ+π3=k π,k ∈Z ，φ=-π6+k π2,k ∈Z ，即正数φ的最小值为π3，②正确；对于③：当x ∈0,π3 时，ωx +π3∈π3,π(ω+1)3，由y =f x 在0,π3 上单调递增，得π(ω+1)3≤π2ω>0，解得0<ω≤12，正数ω有最大值12，③错误；对于④：当x ∈(0,π)时，ωx +π3∈π3,ωπ+π3，而y =f x 在(0,π)上恰有两个极值点，由正弦函数的性质得3π2<ωπ+π3≤5π2，解得76<ω≤136，因此ω的取值范围是76,136，④错误.综上，共2个正确，故选：B .10(2024·河北保定·二模)已知tan α=3cos αsin α+11，则cos2α=()A.-78B.78C.79D.-79【答案】B【分析】利用切化弦和同角三角函数的关系，解出sin α，再结合二倍角公式即可求解.【详解】因为sin αcos α=3cos αsin α+11，所以4sin 2α+11sin α-3=0，解得sin α=14或sin α=-3(舍去)，所以cos2α=1-2sin 2α=78．故选：B .11(2024·河北衡水·三模)已知sin (3α-β)=m sin (α-β)，tan (2α-β)=n tan α，则m ，n 的关系为()A.m =2nB.n =m +1mC.n =m m -1D.n =m +1m -1【答案】D【分析】利用和差角的正弦公式化简，结合已知列出方程即可求解.【详解】依题意，sin (3α-β)=sin [(2α-β)+α]=sin (2α-β)cos α+cos (2α-β)sin α，sin (α-β)=sin [(2α-β)-α]=sin (2α-β)cos α-cos (2α-β)sin α，则sin (2α-β)cos α+cos (2α-β)sin α=m sin (2α-β)cos α-m cos (2α-β)sin α，即sin (2α-β)cos αcos (2α-β)sin α=m +1m -1，即tan (2α-β)tan α=m +1m -1=n .故选：D12(2024·辽宁沈阳·三模)已知tan α2=2，则sin 2α2+sin α的值是()A.25B.45C.65D.85【答案】D【分析】利用二倍角公式和同角之间的转化，进行求解判断选项【详解】当tan α2=2，则sin 2α2+sin α=sin 2α2+2sin α2cos α2sin 2α2+cos 2α2=tan 2α2+2tan α2tan 2α2+1=22+2×222+1=85故选：D13(2024·贵州黔东南·二模)已知0<α<β<π，且sin α+β =2cos α+β ，sin αsin β-3cos αcos β=0，则tan α-β =()A.-1 B.-32C.-12D.12【答案】C【分析】找出tan α和tan β的关系，求出tan α和tan β即可求解.【详解】∵sin αsin β-3cos αcos β=0，∴sin αsin β=3cos αcos β，∴tan αtan β=3①，∵sin α+β =2cos α+β ，∴tan α+β =2⇒tan α+tan β1-tan αtan β=2⇒tan α+tan β1-3=2，∴tan α+tan β=-4②，由①②解得tan α=-1tan β=-3或tan α=-3tan β=-1 ，∵0<α<β<π，∴tan α<tan β，∴tan α=-3tan β=-1 ，∴tan α-β =tan α-tan β1+tan αtan β=-12.故选：C .二、多选题14(2024·河北张家口·三模)已知函数f (x )=23cos 2x +2sin x cos x ，则下列说法正确的是()A.函数f (x )的一个周期为2πB.函数f (x )的图象关于点π3,0 对称C.将函数f (x )的图象向右平移φ(φ>0)个单位长度，得到函数g (x )的图象，若函数g (x )为偶函数，则φ的最小值为5π12D.若f 12α-5π24 -3=12，其中α为锐角，则sin α-cos α的值为6-308【答案】ACD【分析】利用三角恒等变换公式化简，由周期公式可判断A ；代入验证可判断B ；根据平移变化求g (x )，由奇偶性可求出φ，可判断C ；根据已知化简可得sin α-π12 =14，将目标式化为2sin α-π12 -π6 ，由和差角公式求解可判断D .【详解】对于A ，因为f (x )=31+cos2x +sin2x =2sin 2x +π3+3，所以f (x )的最小值周期T =2π2=π，所以2π是函数f (x )的一个周期，A 正确；对于B ，因为f π3 =2sin 2×π3+π3 +3=3，所以，点π3,0 不是函数f (x )的对称中心，B 错误；对于C ，由题知，g x =f (x -φ)=2sin 2(x -φ)+π3 +3=2sin 2x +π3-2φ +3，若函数g (x )为偶函数，则π3-2φ=π2+k π,k ∈Z ，得φ=-π12-k π2,k ∈Z ，因为φ>0，所以φ的最小值为5π12，C 正确；对于D ，若f 12α-5π24-3=2sin 212α-5π24 +π3 =2sin α-π12 =12，则sin α-π12 =14，因为α为锐角，-π12<α-π12<5π12，所以cos α-π12 =154，所以sin α-cos α=2sin α-π4 =2sin α-π12 -π6=232sin α-π12 -12cos α-π12=232×14-12×154=6-308，D 正确.故选：ACD 15(2024·辽宁鞍山·模拟预测)已知函数f x =sin x ⋅cos x ，则()A.f x 是奇函数B.f x 的最小正周期为2πC.f x 的最小值为-12D.f x 在0,π2上单调递增【答案】AC【分析】首先化简函数f x =12sin2x ，再根据函数的性质判断各选项.【详解】f x =sin x ⋅cos x =12sin2x ，函数的定义域为R ，对A ，f -x =-12sin2x =-f x ，所以函数f x 是奇函数，故A 正确；对B ，函数f x 的最小正周期为2π2=π，故B 错误；对C ，函数f x 的最小值为-12，故C 正确；对D ，x ∈0,π2 ，2x ∈0,π ，函数f x 不单调，f x 在0,π4 上单调递增，在π4,π2上单调递减，故D 错误.故选：AC16(2024·安徽·三模)已知函数f x =sin x -3cos x ，则()A.f x 是偶函数B.f x 的最小正周期是πC.f x 的值域为-3,2D.f x 在-π,-π2上单调递增【答案】AC【分析】对于A ，直接用偶函数的定义即可验证；对于B ，直接说明f 0 ≠f π 即可否定；对于C ，先证明-3≤f x ≤2，再说明对-3≤u ≤2总有f x =u 有解即可验证；对于D ，直接说明f -5π6>f -2π3 即可否定.【详解】对于A ，由于f x 的定义域为R ，且f -x =sin -x -3cos -x =-sin x -3cos x =sin x -3cos x =f x ，故f x 是偶函数，A 正确；对于B ，由于f 0 =sin0 -3cos0=-3，f π =sinπ -3cosπ=3，故f 0 ≠f π ，这说明π不是f x 的周期，B 错误；对于C ，由于f x =sin x -3cos x ≤sin x +3cos x =sin x +3cos x 2≤sin x +3cos x 2+3sin x -cos x 2=sin 2x +3cos 2x +23sin x cos x +3sin 2x +cos 2x -23sin x cos x =4sin 2x +4cos 2x =4=2，且f x =sin x -3cos x ≥-3cos x ≥-3，故-3≤f x ≤2.而对-3≤u ≤2，有f 0 =-3≤u ，f 5π6 =2≥u ，故由零点存在定理知一定存在x ∈0,5π6使得f x =u .所以f x 的值域为-3,2 ，C 正确；对于D ，由于-π<-5π6<-2π3<-π2，f -5π6 =2>3=f -2π3 ，故f x 在-π,-π2上并不是单调递增的，D 错误.故选：AC .17(2024·山西太原·模拟预测)已知函数f x =sin 2x +φ 0<φ<π2 的图象关于直线x =π12对称，且h x =sin2x -f x ，则()A.φ=π12B.h x 的图象关于点π6,0中心对称C.f x 与h x 的图象关于直线x =π4对称 D.h x 在区间π6,5π12内单调递增【答案】BCD【分析】根据正弦函数的对称性求解φ判断A ，先求出h x =sin 2x -π3，然后利用正弦函数的对称性求解判断B ，根据对称函数的性质判断C ，结合正弦函数的单调性代入验证判断D .【详解】由题意得2×π12+φ=π2+k π，k ∈Z ，解得φ=π3+k π，k ∈Z ，又因为0<φ<π2，所以φ=π3，A 错误；由φ=π3可知f x =sin 2x +π3，则h x =sin2x -sin 2x +π3 =12sin2x -32cos2x =sin 2x -π3，令2x -π3=k π，k ∈Z ，解得x =π6+k π2，k ∈Z ，令k =0，得x =π6，所以点π6,0 是曲线y =h x 的对称中心，B 正确；因为f π2-x =sin 2π2-x +π3 =sin 4π3-2x =sin 2x -π3=h x ，所以f x 与h x 的图象关于直线x =π4对称，C 正确；当x ∈π6,5π12 时，2x -π3∈0,π2 ，故h x 在区间π6,5π12内单调递增，D 正确.故选：BCD 18(2024·浙江金华·三模)已知函数f x =sin2ωx cos φ+cos2ωx sin φω>0,0<φ<π2的部分图象如图所示，则()A.φ=π6B.ω=2C.f x +π6为偶函数 D.f x 在区间0,π2的最小值为-12【答案】ACD【分析】先由正弦展开式，五点法结合图象求出f x =sin 2x +π6，可得A 正确，B 错误；由诱导公式可得C 正确；整体代入由正弦函数的值域可得D 正确.【详解】由题意得f x =sin 2ω+φ ，由图象可得f 0 =12⇒sin φ=12，又0<φ<π2，所以φ=π6，由五点法可得ω×4π3+π6=3π2⇒ω=1，所以f x =sin 2x +π6 .A ：由以上解析可得φ=π6，故A 正确；B ：由以上解析可得ω=1，故B 错误；C ：f x +π6 =sin 2x +π6 +π6=cos2x ，故C 正确；D ：当x ∈0,π2 ⇒2x +π6∈π6,7π6 时，sin 2x +π6 ∈-12,1，所以最小值为-12，故D 正确；故选：ACD .19(2024·浙江温州·二模)已知角α的顶点为坐标原点，始边与x 轴的非负半轴重合，P -3,4 为其终边上一点，若角β的终边与角2α的终边关于直线y =-x 对称，则()A.cos π+α =35B.β=2k π+π2+2αk ∈Z C.tan β=724D.角β的终边在第一象限【答案】ACD【分析】根据三角函数的定义，可求角α的三角函数，结合诱导公式判断A 的真假；利用二倍角公式，求出2α的三角函数值，结合三角函数的概念指出角2α的终边与单位圆的交点，由对称性确定角β终边与单位圆交点，从而判断BCD 的真假.【详解】因为角α的顶点为坐标原点，始边与x 轴的非负半轴重合，终边经过点P -3,4 ，所以：OP =5，所以sin α=45，cos α=-35，所以cos π+α =-cos α=35，故A 对；又sin2α=2sin α⋅cos α=2×45×-35 =-2425，cos2α=cos 2α-sin 2α=-35 2-45 2=-725，所以2α的终边与单位圆的交点坐标为：-725,-2425 ，因为角β的终边与角2α的终边关于直线y =-x 对称，所以角β的终边与单位圆的交点为2425,725，所以tan β=724，且β的终边在第一象限，故CD 正确；又因为终边在直线y =-x 的角为：k π-π4,k ∈Z ，角2α的终边与角β的终边关于y =-x 对称，所以2α+β2=k π-π4⇒β=2k π-π2-2αk ∈Z ，故B 错误.故选：ACD20(2024·广东佛山·二模)已知函数f x =sin x +cos2x 与g x =sin2x +cos x ，记h x =λf x +μg x ，其中λ，μ∈R 且λ2+μ2≠0．下列说法正确的是()A.h x 一定为周期函数B.若λ⋅μ>0，则h x 在0,π2上总有零点C.h x 可能为偶函数 D.h x 在区间0,2π 上的图象过3个定点【答案】ABD【分析】对于A ：计算h x +2π ，化简即可；对于B ：求出h x ，然后计算h 0 h π2的正负即可；对于C ：计算h x ,h -x 是否恒相等即可；对于D ：令f x =0g x =0，求解x 即可.【详解】对于A ，∀x ∈R ，h x +2π =λf x +2π +μg x +2π =λf x +μg x =h x ，A 正确；对于B ，h x =λcos x -2sin2x +μ2cos2x -sin x ，则h 0 =λ+2μ，h π2=-3μ，因为λμ>0，即λ，μ同号，所以h 0 h π2<0，由零点存在定理知h x 在0,π2上总有零点，故B 正确；对于C ，h x =λsin x +λcos2x +μsin2x +μcos x ，h -x =-λsin x +λcos2x -μsin2x +μcos x ，由h x =h -x 得2λsin x +2μsin2x =2λsin x +2μ⋅2sin x cos x =2sin x λ+2μcos x =0对x ∈R 恒成立，则λ=μ=0与题意不符，故C 错误；对于D ，令f x =0g x =0 ，则sin x +cos2x =1-2sin 2x +sin x =-sin x -1 2sin x +1 =0sin2x +cos x =cos x 2sin x +1 =0 ⇒sin x =1或sin x =-12cos x =0或sin x =-12，即x ∈-π6+2k π,π2+2k π,7π6+2k π ，k ∈Z ，故所有定点坐标为-π6+2k π,0 ，π2+2k π,0 ，7π6+2k π,0 ，k ∈Z ，又因为x ∈0,2π ，所以函数h x 的图象过定点π2,0 ，7π6,0 ，11π6,0 ，故D 正确；故选：ABD ．21(2024·湖南·二模)已知函数f x =12cos 2x -π3 ，把y =f x 的图象向右平移π3个单位长度，得到函数y =g x 的图象，以下说法正确的是()A.x =π6是y =f x 图象的一条对称轴B.f x 的单调递减区间为k π+π6,k π+2π3k ∈Z C.y =g x 的图象关于原点对称D.f x +g x 的最大值为12【答案】ABD【分析】根据题意，求得g x =-12cos2x 的图象，结合三角函数的图象与性质，以及两角差的正弦公式，逐项判定，即可求解.【详解】将函数f x =12cos 2x -π3 的图象向右平移π3个单位长度，得到函数y =g x =12cos 2x -π =-12cos2x 的图象，对于A 中，令x =π6，求得f x =12，即为函数y =f x 最大值，所以直线x =π6是函数f x 图象的一条对称轴，所以A 正确；对于B 中，令2k π≤2x -π3≤2k π+π,k ∈Z ，解得k π+π6≤x ≤k π+2π3,k ∈Z ，可得f x 的单调减区间为k π+π6,k π+2π3,k ∈Z ，所以B 正确.对于C 中，由于g x =-12cos2x 是偶函数，可得函数g x 的图象关于y 轴对称，所以C 错误.对于D 中，由f x +g x =12cos 2x -π3 +-12cos2x =1212cos2x +32sin2x -12cos2x =34sin2x -14cos2x =12sin 2x -π6 ≤12，即f x +g x 的最大值为12，所以D 正确.故选：ABD .22(2024·广东江门·一模)已知函数f (x )=sin 2ωx +π3 +sin 2ωx -π3+23cos 2ωx -3(ω>0)，则下列结论正确的是()A.若f x 相邻两条对称轴的距离为π2，则ω=2B.当ω=1，x ∈0,π2时，f x 的值域为-3,2 C.当ω=1时，f x 的图象向左平移π6个单位长度得到函数解析式为y =2cos 2x +π6D.若f x 在区间0,π6上有且仅有两个零点，则5≤ω<8【答案】BCD【分析】根据三角恒等变换化简f x =2sin 2ωx +π3，进而根据周期可判断A ，根据整体法求解函数的值域判断B ，根据函数图象的平移可判断C ，根据零点个数确定不等式满足的条件可判断D .【详解】f (x )=sin 2ωx +π3 +sin 2ωx -π3+23cos 2ωx -3=sin2ωx cos π3+cos2ωx sin π3+sin2ωx cos π3-cos2ωx sin π3+3cos2ωx=sin2ωx +3cos2ωx =2sin 2ωx +π3，对于A ，若f x 相邻两条对称轴的距离为π2，则T =2×π2=π=2π2ω,故ω=1，A 错误，对于B ，当ω=1，f x =2sin 2x +π3 ，当x ∈0,π2 时，2x +π3∈π3,4π3，则f x 的值域为-3,2 ，B 正确，对于C ，当ω=1，f x =2sin 2x +π3，f x 的图象向左平移π6个单位长度得到函数解析式为f x +π6 =2sin 2x +π6 +π3 =2sin 2x +2π3 =2cos 2x +π6，C 正确，对于D ，当x ∈0,π6 时，2ωx +π3∈π3,2ωπ6+π3，若f x 在区间0,π6 上有且仅有两个零点，则2π≤2ωπ6+π3<3π，解得5≤ω<8，故D 正确，故选：BCD 三、填空题23(2024·北京·三模)已知函数f (x )=sin x cos ωx ,x ∈R ．①若ω=1，则f (x )的最小正周期是；，②若ω=2，则f (x )的值域是．【答案】π[-1,1]【分析】把ω=1代入，t 明智二倍角的正弦，结合正弦函数的周期求出f (x )的最小正周期；把ω=2代入，利用二倍角的余弦公式，借助换元法，利用导数求出f (x )的值域.【详解】当ω=1时，f (x )=sin x cos x =12sin2x ，函数f (x )的最小正周期为2π2=π；当ω=2时，f (x )=sin x cos2x =sin x (1-2sin 2x )，令sin x =t ∈[-1,1]，g (t )=t (1-2t 2)=-2t 3+t ，求导得g (t )=-6t 2+1，当-1≤t <-66或66<t ≤1时，g (t )<0，当-66<t <66时，g (t )>0，函数g (t )在-1,-66 ，66,1 上单调递减，在-66,66上单调递增，g (-1)=1,g 66 =69，g (1)=-1,g -66 =-69，所以g (t )min =-1,g (t )max =1，f (x )的值域是[-1,1].故答案为：π；[-1,1]24(2024·北京·模拟预测)已知函数f (x )=sin ωx -2cos ωx (ω>0)，且f α+x =f α-x .若两个不等的实数x 1,x 2满足f x 1 f x 2 =5且x 1-x 2 min =π，则sin4α=.【答案】-45/-0.8【分析】利用辅助角公式化简f (x )的解析式，再由题意可得函数关于x =α对称，且最小正周期T =π，即可求出ω的值，从而得到2α=φ+π2+k π,k ∈Z ，再由二倍角公式及同角三角函数的基本关系计算可得．【详解】因为f (x )=sin ωx -2cos ωx =5sin ωx -φ ，其中tan φ=2，由f α+x =f α-x ，可得f x 关于x =α对称，又两个不等的实数x 1,x 2满足f x 1 f x 2 =5且x 1-x 2 min =π，所以f x 的最小正周期T =π，又ω>0，所以2πω=π，解得ω=2，所以f x =5sin 2x -φ ，所以2α-φ=π2+k π,k ∈Z ，则2α=φ+π2+k π,k ∈Z ，所以sin4α=sin2φ+π2+k π =sin 2φ+π+2k π =-sin2φ=-2sin φcos φsin 2φ+cos 2φ=-2tan φtan 2φ+1=-2×222+1=-45.故答案为：-4525(2024·湖北荆州·三模)设0<α<β<π2，tan α=m tan β，cos α-β =35，若满足条件的α与β存在且唯一，则m =，tan αtan β=.【答案】191【分析】由tan α=m tan β得到sin αcos β=m cos αsin β，再结合cos α-β =35，利用sin α-β =-45，得到cos αsin β=-45m -1 ，sin αcos β=-4m5m -1 ，从而sin α+β =-4m +1 5m -1，再由满足条件的α与β存在且唯一，得到α+β唯一，从而sin α+β =-4m +15m -1=1，求得m 即可.【详解】解：由tan α=m tan β，得sin αcos α=m sin βcos β，即sin αcos β=m cos αsin β，因为0<α<β<π2，tan α=m tan β，所以-π2<α-β<0，0<m <1，又cos α-β =35，所以sin α-β <0，从而sin α-β =sin αcos β-cos αsin β=m -1 cos αsin β=-45，所以cos αsin β=-45m -1，所以sin αcos β=m cos αsin β=-4m5m -1，所以sin α+β =sin αcos β+cos αsin β=-4m +15m -1，因为α,β∈0,π2，所以α+β∈0,π ，因为满足条件的α与β存在且唯一，所以α+β唯一，所以sin α+β =-4m +1 5m -1=1，所以m =19，经检验符合题意，所以tan α=19tan β，则tan α-β =-43=tan α-tan β1+tan αtan β=tan α-9tan α1+9tan 2α，解得tan α=13，所以tan αtan β=9tan 2α=1.故答案为：19，1【点睛】关键点点睛：关键是结合已知得出sin α+β =-4m +15m -1 =1，求出m ，由此即可顺利得解.。
可汗学院新SAT数学真题下载
可汗学院新SAT数学真题下载到目前为止，新版SAT可汗学院官方不断放出更多真题，已经放出了68篇阅读，且之前已经和大家分享过可汗学院新SAT阅读真题，想要下载的同学，请点击：新SAT阅读真题下载（共68篇，且已全）今天主要和大家分享可汗学院新SAT数学真题！目前可汗学院一共放出41篇新SAT数学真题！想要下载的同学，请点击：新SAT数学真题下载（共41篇）另外，想要下载OG的同学，请点击：新SAT官方指南下载以下是部分可汗学院新SAT数学真题。
Solving linear equations1. Solve for k:k + 22 = 29k =Correct answer: 7 Difficulty level: 12. Solve for n:-8 + n = 23n =Correct answer: 31 Difficulty level: 13. Solve for x:x – 9 = 1x =Correct answer: 10 Difficulty level: 14. Solve for n:18 = n - 18n =Correct answer: 36 Difficulty level: 15. Solve for k:k + 10 = 27k =Correct answer: 17 Difficulty level: 16. Solve for t:t + 25 = 26t =7. Solve for k:-30 + k = 22k =Correct answer: 52 Difficulty level: 18.Solve for k:19 = 14 + kk =Correct answer: 5 Difficulty level: 19. Solve for q:3 = -8 + qq =Correct answer: 11 Difficulty level: 110. Solve for p:10 = -19 + pp =Correct answer: 29 Difficulty level: 111. Solve for r:-12 + r = 7r =Correct answer: 19 Difficulty level: 112. Solve for p:p + 12 = 30p =Correct answer: 18 Difficulty level: 113. Solve for p:p – 18 = 3p =Correct answer: 21 Difficulty level: 114. Solve for k:30 = k + 23k =15. Solve for y:25 = -14 + yy =Correct answer: 39 Difficulty level: 116. Solve for r:20 = r + 11r =Correct answer: 9 Difficulty level: 117. Solve for z:-26 + z = 15z =Correct answer: 41 Difficulty level: 118. Solve for p:29 = p + 20p =Correct answer: 9 Difficulty level: 119. Solve for y:24 = y + 19y =Correct answer: 5 Difficulty level: 120. Solve for n:27 = 15 + nn =Correct answer: 12 Difficulty level: 121. Solve for k:2 = -30 + kk =Correct answer: 32 Difficulty level: 122. Solve for n:19 + n = 30n =。
可汗学院数学题目-18Polynomialfactorsandgraphs
Polynomial factors and graphs1.P(x) =2x3−18xGiven the polynomial function P defined above, what are its zeros?A. {−9,−6,2,3}B. {−9,0,2}C. {−3,3}D. {−3,0,3}Correct Answer: D Difficult Level: 22. Which of the following functions could represent the graph BELOW in the xy-plane, where y=P(x)?A. P(x) =x2+6x+8B. P(x) =x3+6x2+8xC. P(x) =x2−6x+8D. P(x) =x3−6x2+8xCorrect Answer: B Difficult Level: 23. A polynomial has zeros at−9,2, and0. Which of the following could be the polynomial?A. x2−7x−18B.x3+7x2−18xC.x3+7x2−18xD.x3+6x2−25x+18Correct Answer: B Difficult Level: 24. (x−7) (x+5) (2x−3)Given the polynomial above, what are its zeros?A.｛-7, 5, -3｝B. {7, - 5, 3}C. {-7, 5, -3}2D. {7, -5, 3}2Correct Answer: D Difficult Level: 25. 2(x+55) (x−17)Given the polynomial above, what are its zeros?A. x=−55and x=17B. x=−55, x=−2, and x=17C. x=−17and x=55D. x=−17,x=2, and x=55Correct Answer: A Difficult Level: 26. G (z) = (z−1)5−(z−1)4 The polynomial function G is defined above. What is the product of the zeros of G? Fill in the blank:Correct Answer: 2 Difficult Level: 27. The polynomial function P has zeros at3and6. Which of the following could be the definition of P?A. P(x)=x2+9x+18B. P(x)=x2−9x+18C. P(x)=x2+3x+6D. P(x)=x2+6x+3Correct Answer: B Difficult Level: 28. Which of the following equations could represent the graph below in the xy-plane?I. y=(x−3)(x+3)II. y=(x−4)2III. y=(x+2)(x+7)A. ⅠonlyB. ⅡonlyC. I and ⅢonlyD. I,Ⅱand ⅢCorrect Answer: A Difficult Level: 29. h (t)=(t−8)1(t−4)2(t−2)3(t−1)4The polynomial function h is defined above. How many distinct zeros does h have? Correct Answer: 4 Difficult Level: 210.x3 +25x2+50x−1000The polynomial above has(x−5)and (x+10)as factors. What is the remaining factor?A. (x+2)B.(x-2)C. (x+2)D. (x-20)Correct Answer: C Difficult Level: 211.x3+7x2−36The polynomial above has zeros at −6 and 2. If the remaining zero is z, then what is the value of −z?Correct Answer: 3 Difficult Level: 312. The function p is a polynomial of t such that (t−10), (22−t),(t+10), and (20+t)are all factors of p(t). Which of the following could be the graph of y=p(t)in the ty-plane?A. B.C. D.Correct Answer: A Difficult Level: 313. Which of the following graphs appears to represent a polynomial function with a double zero?A. B.C. D.Correct Answer: D Difficult Level: 314.g(x)=x4−4x3+6x2−4x+1The function g is defined above. Given that all zeros of g are integers between−1 and 1inclusive, how many distinct zeros does g have?Correct Answer: 1 Difficult Level: 315.x2−ax+24If one of the zeros of the polynomial above is 8, what is the other zero?Correct Answer: 3 Difficult Level: 316. For a function g, the graph of y=g(x)y=g(x)is shown BELOW. When g(x)is divided by(x+10), the remainder is−20. Which of the following is closest to the remainder when g(x)is divided by(x−10)?A. -28B. -2C. 2D. 28Correct Answer: C Difficult Level: 317. The polynomial function f is defined as f(c)=(c−k)(c2−4c+4) where k is a constant. The value 2 is a zero of f. What is the remainder of f(c) when divided by (c−2)?Correct Answer: 0 Difficult Level: 318. Which of the following graphs in the xy-plane have−3and5as all of their distinct zeros for −6≤x≤6?I. II.III.A. I onlyB. I and II onlyC. II and III onlyD. I. II and III.Correct Answer: B Difficult Level: 319. Given some rational constant a, which polynomial equation could represent the graph BELOW in the xy-plane?A.y=a(x+1)2(x−5)B.y=a(x+1) (x−5)C.y=a(x+1) (2x+1)(x−5)D.y=a(x−1) (x+5)2Correct Answer: B Difficult Level: 320. The equation s=(t+3)2(t+2)(t+1)(t)(t−1) is graphed on the st-plane. What is the product of the t-intercepts of the graph?Correct Answer: 0 Difficult Level: 321. q(v)=(v−8)(v−5)(v−4)(v+5)(v+10)The function q is defined above. If the sum of the zeros of q is s, what is the value of s?Correct Answer: 2 Difficult Level: 322. (x−√3)2(x−√7)Given the polynomial above, what are its zeros?A.x=−√3and x=−√7B. x=√3and x=√7C.x=3and x=√7D.x=−3and x=−√7Correct Answer: B Difficult Level: 323. The graph of the polynomial equation y=α(t) is shown BELOW. Which of the following must be true?** leading coefficient is positive.** sum of the distinct t intercepts is negative.** constant coefficient is positive.** product of the distinct t intercepts is negative.Correct Answer: D Difficult Level: 324. The graph shown at left could represent which of the following equations?A. h=−(b−10)(b−20)(b+20)B. h=(b−10)(b−20)(b+20)C. h=−(b+10)(b−20)(b+20)D. h=(b+10)(b−20)(b+20)Correct Answer: A Difficult Level: 325.x3+7x2−36The polynomial above has zeros at −6 and 2. If the remaining zero is z, then what is the value of−z?Correct Answer: 3 Difficult Level: 326. The polynomial function f is defined as f(m)=(m3−m2−17m−15)(m+1). Whenf(m) is divided by (m+1), what is the remainder?Correct Answer: 0 Difficult Level: 427.p(n)=(n3−12n2+47n−60)(n−4)q(n)=(n+13)(n−4)The functions p and q are defined above. One of the functions has a zero at n=5. What is (p+q)(5)?Correct Answer: 18 Difficult Level: 428. Which of the following could be the equation corresponding to the graph BELOW?A.s=a⋅a⋅a⋅aB.s=(a−1)(a−1)C.s=a⋅a⋅a⋅a⋅aD.s=(a−1)(a−1)(a−1)Correct Answer: C Difficult Level: 429.p=(w−30)(w2+178w+7921)Given that −89 is a double zero of the polynomial equation above, which of the following could be the graph of the equation in the pw-plane?A. B.C. D.Correct Answer: D Difficult Level: 430.ℓ(x)=x4+36x2−10,000The polynomial function ℓ is defined above. What is the remainder of ℓ(x) when divided by (x+10)?Correct Answer: 3600 Difficult Level: 431.g (w)=(w+13)3(w+19)2The polynomial function g is defined above. When g (w) is divided by (w+16), the remainder is r. What is the value of ∣r∣?Correct Answer: 243 Difficult Level: 432. A function p p is defined as p(x) = (x−a)(x−15)(x−20)+15 where a is a constant. Given that p (7)=15, what is the value of a?Correct Answer: 7 Difficult Level: 433. A function w is a defined as w(x)=ax2+bx+c where a, b, and c are constants.If a=3 and w(3)=w(15)=0, then what is the absolute value of b?Correct Answer: 54 Difficult Level: 434. A function s is defined as s(x)=(x−4)(x−5)2. A function h is defined ash(x)=(x−a)s(x). For some constant a, (x−a)3 is a factor of h. What is s(a)?Correct Answer: 0 Difficult Level: 435.12x2+ax+2, what is the other zero?If one of the zeros of the above polynomial is 23Correct Answer: 1Difficult Level: 44** any polynomial function h, the polynomial function g is defined to be g(w)=(w−25 )(w−1)h(w). If h has zeros at 1 and 3 only, what is the sum of the distinct zeros of g?Correct Answer: 36 Difficult Level: 437. The polynomial function f f is defined as f(x)=(x−c1)(x−c2)(x−c3)⋯(x−cn)for some positive integer n. Each of the values c1,c2,c3,,,cn is a real number. The graph of y=f(x)is shown BELOW. Which of the following could be the value of n?A. 0B. 1C. 2D. 3Correct Answer: D Difficult Level: 438. The polynomial function f is defined as f(m)=(m3−m2−17m−15)(m+1). When f(m) is divided by (m+1), what is the remainder?Correct Answer: 0 Difficult Level: 4。
2024年高考数学三角函数空间向量历年真题详解
2024年高考数学三角函数空间向量历年真题详解为了帮助广大高中生备战2024年高考数学考试，本文将详细解析数学科目中的三角函数和空间向量部分的历年真题。
通过对每道题目的解析和详细讲解，希望能够帮助大家更好地掌握这一部分知识点，并提供备考的指导。
1. 试题一第一道题目涉及三角函数的求解，题目如下：已知三角函数$\sin\theta = \frac{1}{2}$，求$\cos\theta$的值。
解析：根据已知信息，我们可以利用三角函数的定义来求解。
由于$\sin\theta = \frac{1}{2}$，我们可以根据单位圆上的特点得出$\sin$对应的坐标为$\left(\frac{1}{2}, \frac{\pi}{6}\right)$。
根据单位圆上的关系，$\cos\theta$的值为$\frac{\sqrt{3}}{2}$。
因此，$\cos\theta$的值为$\frac{\sqrt{3}}{2}$。
2. 试题二第二道题目涉及空间向量的投影问题，题目如下：已知向量$\overrightarrow{AB} = 3\overrightarrow{i} +4\overrightarrow{j} + 5\overrightarrow{k}$，向量$\overrightarrow{AC} = \overrightarrow{i} + 2\overrightarrow{j} + \overrightarrow{k}$，求向量$\overrightarrow{AB}$在向量$\overrightarrow{AC}$上的投影。
解析：首先，我们需要计算向量$\overrightarrow{AC}$的模长。
根据向量的定义，$\overrightarrow{AC}$的模长为$\sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}$。
接下来，我们需要计算向量$\overrightarrow{AB}$在向量$\overrightarrow{AC}$上的投影。
国际高中数学测试卷(4)英文卷
Unit Two Test Choose:(20’)1.What would be the next two number in the following set?{5, 7, 9, 11}( ).A)10, 12 B)13, 14 C)12, 13 D)13, 152.What shape and how many of this shape will be next in the pattern below?( ).A)1 Circle B)3 Circles C)1 Triangle D)3 Triangles3.Identify the if-then form of the given statement."A regular polygon has all its sides equal."( )A)If a polygon is regular, then it has all its sides equal. B)A polygon is regular, if it has all its sides equal.C)A regular polygon has all its sides equal. D)If a polygon has all its sides equal, then it is regular.4.Which of the following connectors is the logical disjunction?( ).A)OR B)AND C)BUT D)Both 5.Which of the following is a counterexample of the conjecture below?Conjecture: The product of two positive numbers is always greater than either number( ).A)2,2 B)21,2 C)3,10 D)2,-1 6.If ∠CFE is an obtuse angle,what is true.( ).A)∠CFE is greater than 90 degrees and less than 180 degrees. B)∠CFE is supplementary.C)∠CFE is less than 90 degrees. D)∠CFE can not be 110 degrees.7.What s a conjecture?( ).A) a guess B) unproven statement based on observationsC) switching the hypothesis and conclusion D)a false statement8.A specific case of when a conjecture is false is called a(n)?( )A)Counterexample B)Inverse C)Wrong answer D)Negation9.write the converse of the statement in symbolic form.( )A ) p → qB ) t → ~ wC ) ~ m → pD ) q → p 10.write the contrapositive of the statement in symbolic form.( )A ) p → qB ) t → ~ wC ) ~ m → pD ) ~ q → ~ pFor the following problems, Find the hypothesis and conclusion in each conditional statement.(8’)1. If today is Monday, then tomorrow is Tuesday.Hypothesis:_________________________; Conclusion:_________________________.2.If a truck weighs 2 tons, then it weighs 4000 pounds.Hypothesis:_________________________; Conclusion:_________________________.3.All chimpanzees love bananas.Hypothesis:_________________________; Conclusion:_________________________.4.Collinear points lie on the same line.Hypothesis:_________________________; Conclusion:_________________________.Determine if each statement is true or false. If false, provide a counterexample. (4’)1、If an animal can swim, then it is a fish.___________________________________________________________________________.2、If two angles are congruent, then they are right angles.___________________________________________________________________________.Write the contrapositive of the statement in words. (6’)1)If you care enough to send the best, then you send Trademark cards.___________________________________________________________________________.2)If you use Trickle deodorant, then you won’t have body odor.___________________________________________________________________________.Supply the word,phrase or symbol that can be placed in each blank to make the resulting statement true.(6’)(1)When p is true and q is false ,then p^~q is _____________.(2)When p ˅~q is false ，then p is ___________ and q is ______.(3)If the conclusion q is true ,then p→q must be ______.Use the following to answer questions 1-5:(10’)p: I love Geometry q:Mrs. Davis is a great teacherr: Mr. Tucker likes Algebra s:Algebra stinksWrite the symbolic representation.1、If I love geometry, then algebra stinks2、If Mrs. Davis is a great teacher or Mr. Tucker does not like Algebra, then I love Geometry.3、Algebra stinks if I love Geometry and Mrs. Davis is not a great teacher if Mr. Tucker likes Algebra.Write the words for each.4、r p →~5、q r s →∧)(Write the inverse of the statement in words.(6’) 1) If you buy Goal toothpaste, then your children will brush longer.___________________________________________________________________________2) When you serve imported sparkling water, it shows that you have good taste.___________________________________________________________________________ 3) The man who wears Curtis clothes is well dressed.___________________________________________________________________________Inductive or Deductive Reasoning(6’)1、Marcus learns in Social Studies that a presidential election happens every four years. He knows that the last presidential election was in 2004, so he concludes that the next presidential election will be in 2008.2、The United States Census Bureau collects data on the earnings of American citizens. Using data for the three years from 2001 to 2003, the bureau concluded that the national average median income for a four-person family was $43,527.3、Every cross country practice begins with a warm-up jog and stretching. Practice is about to start and the team assumes they will jog 2 warmup laps and stretch before their workout starts.Extra Questions(15’)Point O is the center of the regular octagon ABCDEFGH , and X is the midpoint of the side AB What fraction of the area of the octagon is shaded?One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?。
小外考试真题,2024年数学,英文
小外考试真题,2024年数学,英文2024 Math Test Sample QuestionsPart 1: Multiple Choice1. What is the value of x when 2x + 5 = 15?A) 5B) 7C) 8D) 102. What is the square root of 144?A) 10B) 12C) 14D) 163. If a triangle has angles measuring 30 degrees, 60 degrees, and 90 degrees, what type of triangle is it?A) EquilateralB) IsoscelesC) ScaleneD) Right4. What is the value of 5^3?A) 5B) 25C) 125D) 6255. If a circle has a radius of 6 cm, what is its circumference?A) 12π cmB) 18π cmC) 24π cmD) 36π cmPart 2: Short Answer1. Solve the equation 2x - 3 = 9 for x.2. Find the missing side length of a right triangle with legs measuring 5 and 12.3. What is the volume of a cube with a side length of 4?4. Calculate the area of a rectangle with dimensions 8 x 12.5. If a car travels at a speed of 60 miles per hour, how far will it travel in 3 hours?Part 3: Word Problems1. A recipe calls for 3 cups of flour to make 24 cookies. How much flour is needed to make 48 cookies?2. If a company sells a product for $20 and makes a profit of 25%, how much did it cost to produce the product?3. A garden is in the shape of a rectangle with dimensions 10 feet by 20 feet. If each square foot of the garden requires 1 gallon of water, how many gallons of water are needed to water the entire garden?4. Michelle has 3 times as many marbles as Sarah. If Michelle has 24 marbles, how many marbles does Sarah have?5. A train travels at a speed of 50 miles per hour. If it takes 3 hours to reach its destination, how far did the train travel?Part 4: Extended Response1. Solve the following system of equations:2x + y = 103x - 2y = 42. If a basketball team scores an average of 80 points per game over 10 games, how many total points did they score?3. A square and a triangle have the same perimeter. The square has a side length of 5 and the triangle has a base of 10 and a height of4. Which shape has a greater area?4. Evaluate the expression:(5 + 3) x (4 - 2)5. A company sells a product for $50 and has a profit margin of 20%. How much does it cost to produce the product?Good luck on your math test!。
初二数学几何原理英语阅读理解20题
初二数学几何原理英语阅读理解20题1.Geometry is the study of shapes and their properties. Triangles have three sides. Which of the following is NOT a property of triangles?A.They have three angles.B.They can be equilateral.C.They always have a right angle.D.They can be isosceles.答案：C。
解析：三角形有三个角，选项 A 正确；三角形可以是等边三角形，选项B 正确；三角形不一定总有直角，所以选项C 错误；三角形可以是等腰三角形，选项D 正确。
2.In geometry, a quadrilateral is a polygon with four sides. Which of the following is an example of a quadrilateral?A.TriangleB.CircleC.SquareD.Cube答案：C。
解析：三角形有三条边，不是四边形，选项A 错误；圆不是多边形，不是四边形，选项B 错误；正方形有四条边，是四边形，选项C 正确；立方体是立体图形，不是四边形，选项D 错误。
3.Triangles can be classified by their sides and angles. A triangle with all sides equal is called _____.A.right triangleB.isosceles triangleC.equilateral triangleD.scalene triangle答案：C。
解析：有一个角是直角的三角形是直角三角形，选项A 错误；有两条边相等的三角形是等腰三角形，选项B 错误；三条边都相等的三角形是等边三角形，选项C 正确；三条边都不相等的三角形是不等边三角形，选项D 错误。
数学英语练习题2
Math Problems in English (2)Directions: Each of the questions has four answer choices. For each of these questions, select the best of the answer choices given.41. A necklace is made by stringing N individual beads together in the repeating pattern red bead, greenbead, white bead, blue bead, and yellow bead. If the necklace design begins with a red bead and ends with a white bead, then N could equalA. 54B. 68C. 76D. 8242. John was assigned some math exercises for homework. He answered half of them in study hall. Afterschool he answered 7 more exercises. If he still has 11 exercises to do, how many exercises were assigned?A. 36B. 24C. 12D. 843. The average of 3 different positive integers is 100 and the largest of these integers is 120, what is theleast possible value of the smallest of these integers?A. 1B. 10C. 61D. 7144. If when a certain integer x is divided by 5 the remainder is 2, then each of the following could also be aninteger EXCEPTA. x/17B. x/11C. x/10D. x/645. Over the last three years a scientist had an average yearly income of $45,000. The scientist earned 1.5times as much the second year as the first year and 2.5 times as much the third year as the first year.What was the scientist’s income the second year?A. $9,000B. $13,500C. $27,000D. $40,50046. Salesperson A’s compensation for any week is $360 plus 6% of the proportion of A’s total sales above$1,000 for that week. Salesperson B’s compensation for any week is 8% of B’s total sales for that week.For what amount of total weekly sales would both salespeople earn the same compensation?A. $21,000B. $18,000C. $15,000D. $4,50047. An instructor scored a student’s test of 50 questions by subtracting 2 times the number of incorrectanswers from the number of correct answers. If the student answered all the questions and received a sore of 38, how many questions did that student answer correctly?A. 46B. 47C. 48D. 4948. When you add 5 to a certain number, then subtract -10, multiply by 4, and divide by 6, you get 12. What isthe number?A. -33B. 1/3C. 3D. 13 2/349. If k is an integer greater than 44 and less than 51, then which of the following could be the product of 11and k?A. 572B. 550C. 484D. 44050. If x and y are prime integers, which of the following CANNOT be the sum of x and y?A. 23B. 16C. 13D. 951. To fill a number of vacancies, an employer must hire 3 programmers from among 6 applicants, and 2managers from among 4 applicants. What is the total number of ways in which she can make her decision?A. 1,940B. 132C. 120D. 6052. A $500 investment and a $1,500 investment have a combined yearly return of 8.5% of the total of the twoinvestments. If the $500 investment has a yearly return of 7%, what percent yearly return does the $1,500 investment have?A. 9%B. 10%C. 10.5%D. 11%53. All of the following have the same number of distinct prime factors EXCEPTA. 20B. 21C. 24D. 3054. A haberdasher sells neckties for $7 each and shirts for $12 each. If he sells $95 worth of ties and shirts,what is the least amount of ties he could have sold?A. 3B. 4C. 5D. 655. A machine costs x dollars per day to maintain and y cents for each unit it produces. If the machine isoperated 7 days a week and produces n units in a week, which of the following is the total cost, in dollars, of operating the machine for a week?A. 7x+100ynB. 7x+ynC. (700x+yn)/100D. (7x+100yn)/10056. Coins are to be put into 7 pockets so that each pocket contains at least one coin. At most 3 of the pocketsare to contain the same number of coins, and no two of the remaining pockets are to contain an equal number of coins. What is the least possible number of coins needed for the pocket?A. 7B. 13C. 17D. 2257. Jack is standing 30 yards due north of point P. Sue is standing 72yards due east of point P. What is the shortest distance betweenJack and Sue?A. 60 yardsB. 78 yardsC. 90 yardsD. 100yards58. If n is a prime number greater than 3, what is the remainder when n2 is divided by 12?A. 0B. 1C. 2D. 359. In each production lot for a certain toy, 25 % of the toys are red and 75% of the toys are blue. Half the toysare size A and half are size B. If 10 out of a lot of 100 toys are red and size A, how many of the toys are blue and size B?A. 15B. 25C. 30D. 3560. A certain copy machine produces 13 copies every 10 seconds. If the machine operates withoutinterruption, how many copies would it produce in an hour?A. 4,680B. 4,690C. 4,710D. 4,73261. What is the twenty-first term of the sequence given by x n = 4n - 3 ?A. 6B. 72C. 81D. 8762. In a marketing survey for products A, B, and C, 1000 people were askedwhich of the products, if any, they use. The three circular regions in thediagram represent the numbers of people who use products A, B, and C,according to the survey results. Of the people surveyed, a total of 400 use A, a total of 400 use B, and a total of 450 use C. How many of the people surveyed use exactly one of the products?A. 325B. 250C. 150D. 10075100125125B AC63. The contents of a certain box consists of 14 apples and 23 oranges. How many oranges must be removedfrom the box so that 70 percent of the pieces of fruit in the box will be apples?A. 3B. 6C. 14D. 1764. A rectangle with dimensions 24 inches by 42 inches is to be divided into squares of equal size. Which ofthe following could be a length of the squares?A. 4 inchesB. 6 inchesC. 7 inchesD. 8 inchesaverage of the student’s two lowest scores ?A. 129-2aB. 258-2aC. 129-aD. 258+2a67. The lengths of the three sides of a right triangle are given by three consecutive even integers. Find thelengths of the three sides.A. 4, 6, 8B. 6, 8, 10C. 8, 10, 12D. 10, 12, 1468. A prize of $240 is divided between two persons. If one person receives $180, then what is the differencebetween the amounts received by the persons ?A. $30B. $60C. $120D.$21069. A salesman makes a profit of 25% on all sales. How many fax machines will he sell for $375 each to makea total commission of at least $500 ?A. 4B. 5C. 6D. 1570. Joe works two part-time jobs. One week Joe worked 8 hours at one job, earning $150, and 4.5 hours atother job, earning $90. What were his average hourly earnings for the week ?A. $8.00B. $9.60C. $16.00D. $19.20A. 10B. 13C. 16D. 1873. If a three-digit integer is selected at random from the integers 100 through 199, inclusive, what is theprobability that the first digit and the last digit of the integer are each equal to one more than the middle digit ?A. 2/225B. 1/111C. 1/110D. 1/10074. The width of a rectangle is 6 cm less than the length. If the perimeter of the rectangle is 48 cm, what is thelength of the rectangle in centimeters ?A. 15B. 12C. 9D. 675. In a certain company, the ratio of the number of women employees to the number of men employees is 3to 2. If the total number of employees is 240, then how many of the employees are men ?A. 40B. 48C. 96D. 14476. If x, y, and z are positive integers and 3x=4y=7z, then the least possible value of x+y+z isA. 33B. 40C. 49D. 6177. A number of bricks were purchased to build a fireplace, at a cost of 40 cents each, but only 3/4 of themwere needed. If the unused 190 bricks were returned and their cost refunded, what was the cost of the bricks used to make the fireplace ?A. $228B. $304C. $414D. $57078. If the width of a rectangle is increased by 25% while the length remains constant, the resulting area iswhat percent of the original area ?A. 25%B. 75%C. 125%D. 225%79. A four-character password consists of one letter of the alphabet and three different digits between 0 and 9,inclusive. The letter must appear as the second or third character of the password. How many different passwords are possible ?A. 5,040B. 18,720C. 26, 000D. 37,44080. In the figure, PQRS is a square and each of the four circles has aradius of r. What fractional part of the area of the square isunshaded ?A. (π-4)/2B. (4-π)/4C. π/4D. 4/π数学术语angle 角area 面积average 平均数arithmetic mean 算术平均数base 底面circle 圆形circumference 圆周长cubic foot 立方英尺consecutive even integers连续偶整数consecutive positive integers连续正整数cylinder 圆柱体denominator 分母difference 差digit 数字even integer 偶整数fraction 分数height 高度hexagon 六边形integer 整数length 长度median 中数multiple 倍数numerator 分子odd number 奇数parallelogram 平行四边形perimeter 周长point 点positive integer 正整数prime factor 质因子prime number 质数probability 概率product 乘积quadrilateral 四边形radius 半径reciprocal 倒数rectangle 长方形rectangular solid 长方体remainder 余数right triangle 直角三角形sequence 数列side 边square 正方形square inch 平方英寸standard deviation 标准差sum 和term 项triangle 三角形vertex 顶点volume 体积width 宽度其他生词和词组due north 正北duo 二重唱necktie 领带string 使连成一串tie 平局trio 三重唱。
新SAT语法试卷解析可汗学院最新试题详解
新SAT语法试卷解析可汗学院最新试题详解可汗学院新放出来的几套题难度比较偏低，适合基础阶段练习使用。
大家千万记得，切忌考前用难度系数偏低的题当成模考题，以免影响整体做题的速度。
The Two Faces of Tlatilco这篇考古学文章在讲关于Tlatilco culture中一些发现。
重点研究的是下面这个不知道是什么的大概N多个眼睛鼻子嘴的面具。
然而这些都不重要，毕竟是语法题。
我们只需要知道本篇文章argue的是关于这个东东的象征意义就可以了。
Q1：标点符号题，难度系数：低。
考查主从句之间的连接。
比较容易错的点是正确选项中将关系代词当成插入语使用，容易引起学生的误解。
Q2：句子结构题，难度系数：低。
正确选项是唯一句子结构完整的选项。
Q3：句间逻辑关系，难度系数：低。
上一句讲述关于mythologies的两面性，下一句解释这个名词本身就是有两个意思。
Q4 :过渡，难度系数：低。
用于衔接上下文。
Q5：标点符号，难度系数：低。
本题考查标点符号中unnecessary punctuation，主语和动词之间不加标点符号。
Q6 : 过渡句，难度系数：中。
本题考查上下文之间的transition，但是需要阅读的信息偏多，并且需要做到承上启下，容易有遗漏信息或者是句子的中心偏移的情况存在，容易影响做题速度。
Q7：effective language use，难度系数：低。
选择不包含重复信息的简洁有效选项即可。
Q8：时态，难度系数：中。
不少同学在这道题中都容易选出B的错误选项。
本题中出现了it’s possible that..明确表推测，用may have been.Q9：主谓一致，难度系数：低。
判断准确的主语，depictions，答案选择动词复数形式。
Q10：人称&用语正式性，难度系数：中排除错误的人称，根据正式性进行判断。
Q11：conclusion总结性信息概括结果，选择添加。
Comedy in Contrast：The Style of Flannery O’Connor这篇文章是关于Flannery O’Connor的小说风格。
“可汗学院”逻辑问题带答案
“可汗学院”（Khan Academy）提出的考验智商的谜题。
你可能没听说过“可汗学院”，但“可汗学院”的谜题被苹果采用一定是有其道理的。
可汗学院由孟加拉裔美国人萨尔曼•可汗（Salman Kahan）创立，是一家由谷歌和比尔&梅琳达•盖茨基金会背后支持的教育性非营利组织，主旨在于利用网络影片进行免费授课，目前已经有关于数学、历史、金融、物理、化学、生物、天文学等科目的内容。
幸运的是，这些问题虽然刁钻，但却都有唯一的答案，所以你只要有备而来，还是可以应对自如的，下面是8个苹果面试过程中求职者可能遇到的问题，以及已经被各路聪明的求职者破解的答案。
问题一：“你面前有两扇门，其中一扇门内藏着宝藏，但如果你不小心闯入另一扇门，只能痛苦地慢慢死掉……”！这一听就是那种经典的最令人头痛的一类问题，但其实与其他问题相比，这只是个热身。
在这两扇门后面，有两个人，这两个人都知道哪扇门后有宝藏，哪扇门擅闯者死，而这两个人呢，一个人只说真话，一个人只说假话。
谁说真话谁说假话那就要看你有没有智慧自己找出来了，游戏规则是，你只能问这两个人每人一个问题。
那么，你问什么问题问哪个人根据他们的回答，你又该怎么做求职者的最佳答案：1、随便问其中一个人：“如果我问另一个人，他会跟我说哪扇门后是宝藏如果你问的恰好是讲真话的那个人，那他指给你的答案就是那扇通向死亡的门，因为他会诚实地告诉你那个说谎的人会怎么说。
&如果你问的是那个只说谎话的，你得到的也是错误的答案，因为另一个人是讲真话的，说谎话的人会告诉你与讲真话的人相反的答案。
所以你只要随便问一个人上述问题，然后选择与他们说的相反的门就行了。
问题二：“你前面站了5个人，他们中间只有一个人讲真话……”这个问题比上个问题难就难在，你只知道他们五个中有一个只讲真话，但其余四个，他们有时候讲真话，有时候讲假话，只有一点可以确定，这四个人将真话和假话有个规律：如果这次讲了真话，下次就会讲假话，如果这次讲假话，下次就讲真话。
2024年高考数学三角函数几何证明历年真题讲解
2024年高考数学三角函数几何证明历年真题讲解随着时间的推移，高考成为每年千千万万考生所追逐的目标。
在2024年高考数学考试中，三角函数和几何证明一直是考生们最头疼的科目之一。
为了帮助考生更好地复习和应对高考数学考试，本文将针对2024年高考数学试卷中的几个与三角函数和几何证明相关的历年真题进行讲解。
1. 高考真题一：（题目具体内容省略）解析：该题目涉及到三角恒等式的证明。
首先，我们可以利用基本的正弦函数和余弦函数的定义，将左边的式子化简为sinx+cosx 的形式。
接着，我们可以利用三角函数的和差化积公式对右边进行展开。
通过比较左右两边的式子，我们可以得出结论，从而完成证明。
2. 高考真题二：（题目具体内容省略）解析：该题目涉及到平面几何中的相似三角形证明。
首先，我们可以利用已知条件，得到两个三角形的对应边长比相等。
接着，通过边长比的性质，我们可以得到两个角的比值相等。
再利用三角函数的定义，我们可以得到两个角的正弦值、余弦值和正切值相等。
最后，我们根据三角恒等式，将两个角的正弦值和余弦值表示为同一函数，从而完成证明。
3. 高考真题三：（题目具体内容省略）解析：该题目涉及到向量的证明。
首先，我们可以设定一个与向量有关的等式，然后根据向量运算的性质，对式子进行分解和重组。
接着利用向量的模和夹角的定义，我们可以得到等式两边向量模的关系以及夹角的关系。
最后，通过数学推理和恒等式，我们可以证明该等式成立。
通过以上三个例子的讲解，我们可以看出，数学中的三角函数和几何证明是高考数学中的重点内容之一。
为了在2024年的高考中取得好成绩，考生们需要掌握三角函数的定义、性质和常用的恒等式，同时需要熟练掌握几何证明的基本方法和技巧。
最后，提醒考生，在备考过程中要注重练习，多做历年真题和模拟试题，加深对数学知识和解题方法的理解。
同时，要注重思维的拓展和应用能力的培养，灵活运用所学数学知识解决实际问题。
相信通过不懈的努力和正确的备考方法，考生们一定能在2024年高考数学考试中取得优异的成绩！。
2024年高考数学一模《三角函数》题型及答案
三角函数题型01任意角的三角函数题型02两角和与差的三角函数题型03三角函数的图象与性质题型04解三角形题型01任意角的三角函数1(2024·辽宁沈阳·统考一模)sin x =1的一个充分不必要条件是.2(2024·重庆·统考一模)英国著名数学家布鲁克·泰勒(Taylor Brook )以微积分学中将函数展开成无穷级数的定理著称于世泰勒提出了适用于所有函数的泰勒级数，泰勒级数用无限连加式来表示一个函数，如：sin x =x -x 33!+x 55!-x 77!+⋯，其中n !=1×2×3×⋯×n ．根据该展开式可知，与2-233!+255!-277!+⋯的值最接近的是()A.sin2°B.sin24.6°C.cos24.6°D.cos65.4°3(2024·福建厦门·统考一模)若sin α+π4 =-35，则cos α-π4 =．4(2024·山东济南·山东省实验中学校考一模)下列说法正确的是()A.cos2sin3<0B.若圆心角为π3的扇形的弧长为π，则扇形的面积为3π2C.终边落在直线y =x 上的角的集合是α α=π4+2k π,k ∈Z D.函数y =tan 2x -π6 的定义域为x x ≠π3+k π2,k ∈Z ，π为该函数的一个周期5(2024·山东济南·山东省实验中学校考一模)已知函数f (x )=cos xx，若A ，B 是锐角△ABC 的两个内角，则下列结论一定正确的是()A.f (sin A )>f (sin B )B.f (cos A )>f (cos B )C.f (sin A )>f (cos B )D.f (cos A )>f (sin B )6(2024·河北·校联考一模)在△ABC 中，若A =nB n ∈N * ，则()A.对任意的n ≥2，都有sin A <n sin BB.对任意的n ≥2，都有tan A <n tan BC.存在n ，使sin A >n sin B 成立D.存在n ，使tan A >n tan B 成立题型02两角和与差的三角函数7(2024·广西南宁·南宁三中校联考一模)若cos α+π4 =35，则sin2α=()A.725B.-725C.925D.-9258(2024·黑龙江齐齐哈尔·统考一模)已知cos α+π6 =14，则sin 2α-π6 =()A.78B.-78C.38D.-389(2024·辽宁沈阳·统考一模)已知sin π2-θ +cos π3-θ =1，则cos 2θ-π3=()A.13B.-13C.33D.-3310(2024·浙江·校联考一模)已知α是第二象限角，β∈0,π2 ,tan α+π4 =-14，现将角α的终边逆时针旋转β后得到角γ，若tan γ=17，则tan β=．11(2024·安徽合肥·合肥一六八中学校考一模)已知tan α-11+tan α=2，则sin 2α+π6的值为()A.-4+3310B.-4-3310C.4+3310D.4-331012(2024·江西吉安·吉安一中校考一模)已知α∈0,π ，且3tan α=10cos2α，则cos α可能为()A.-1010B.-55C.1010D.5513(2024·吉林延边·统考一模)已知函数f x =12-sin 2ωx +32sin2ωx ,ω>0 的最小正周期为4π．(1)求ω的值，并写出f x 的对称轴方程；(2)在△ABC 中角A ,B ,C 的对边分别是a ,b ,c 满足2a -c cos B =b ⋅cos C ，求函数f A 的取值范围．题型03三角函数的图象与性质14(2024·福建厦门·统考一模)已知函数f (x )=2sin 2x -π3，则()A.f (x )的最小正周期为π2B.f (x )的图象关于点2π3,0 成中心对称C.f (x )在区间0,π3上单调递增D.若f (x )的图象关于直线x =x 0对称，则sin2x 0=1215(2024·吉林延边·统考一模)将函数f x =sin ωx +π6 (ω>0)的图象向左平移π2个单位长度后得到曲线C ，若C 关于y 轴对称，则ω的最小值是()A.13B.23C.43D.5316(2024·黑龙江齐齐哈尔·统考一模)已知函数f x =cos2x +a cos x +2，则下列说法正确的有()A.当a =0时，f x 的最小正周期为πB.当a =1时，f x 的最小值为78C.当a =3时，f x 在区间0,2π 上有4个零点D.若f x 在0,π3上单调递减，则a ≥217(2024·湖南长沙·雅礼中学校考一模)已知函数f (x )=sin ωx +3cos ωx (ω>0)满足:f π6=2,f 2π3=0,则()A.曲线y =f (x )关于直线x =7π6对称 B.函数y =f x -π3是奇函数C.函数y =f (x )在π6,7π6单调递减 D.函数y =f (x )的值域为[-2,2]18(2024·辽宁沈阳·统考一模)如图，点A ,B ,C 是函数f x =sin ωx +φ (ω>0)的图象与直线y =32相邻的三个交点，且BC -AB =π3,f -π12=0，则()A.ω=4B.f 9π8 =12C.函数f x 在π3,π2上单调递减D.若将函数f x 的图象沿x 轴平移θ个单位，得到一个偶函数的图像，则θ 的最小值为π2419(2024·重庆·统考一模)已知f x =2a sin ωx ⋅cos ωx +b cos2ωx ω>0,a >0,b >0 的部分图象如图所示，当x ∈0,3π4时，f x 的最大值为．20(2024·云南曲靖·统考一模)函数f x =A sin ωx +φ (其中A >0，ω>0，φ ≤π2)的部分图象如图所示，则()A.f 0=-1B.函数f x 的最小正周期是2πC.函数f x 的图象关于直线x=π3对称D.将函数f x 的图象向左平移π6个单位长度以后，所得的函数图象关于原点对称21(2024·浙江·校联考一模)已知函数y=2sinωx+φ，该图象上最高点与最低点的最近距离为5，且点1,0是函数的一个对称点，则ω和φ的值可能是()A.ω=-π3,φ=-π3B.ω=-π3,φ=2π3C.ω=π3,φ=π3D.ω=π3,φ=2π322(2024·广东深圳·校考一模)已知函数f x =cosωx+π3+1(ω>0)的最小正周期为π，则f x 在区间0,π2上的最大值为()A.12B.1 C.32D.223(2024·山西晋城·统考一模)若函数f(x)=cosωx(0<ω<100)在π,5π2上至少有两个极大值点和两个零点，则ω的取值范围为．24(2024·广西南宁·南宁三中校联考一模)在物理学中，把物体受到的力(总是指向平衡位置)正比于它离开平衡位置的距离的运动称为“简谐运动”．在适当的直角坐标系下，某个简谐运动可以用函数f x =A sinωx+φA>0,ω>0,φ <π的部分图象如图所示，则下列结论正确的是()A.ω=2，频率为1π，初相为π6B.函数f x 的图象关于直线x=-π6对称C.函数f x 在π12,13π24上的值域为0,2D.若把f x 图像上所有点的横坐标缩短为原来的23倍，纵坐标不变，再向左平移π12个单位，则所得函数是y=2sin3x+π12题型04解三角形25(2024·河南郑州·郑州市宇华实验学校校考一模)如图，为了测量某湿地A，B两点间的距离，观察者找到在同一直线上的三点C，D，E．从D点测得∠ADC=67.5°，从C点测得∠ACD=45°，∠BCE=75°，从E点测得∠BEC=60°．若测得DC=23，CE=2(单位：百米)，则A，B两点的距离为()A.6B.22C.3D.2326(2024·广东深圳·校考一模)在△ABC 中，角A ，B ，C 的对边分别为a ，b ，c ，若a =3，b =5，c =2a cos A ，则cos A =() A.13B.24C.33D.6327(2024·河南郑州·郑州市宇华实验学校校考一模)在锐角△ABC 中，角A ,B ,C 所对的边分别为a ,b ,c ，且c -b =2b cos A ，则下列结论正确的有()A.A =2BB.B 的取值范围为0,π4C.ab的取值范围为2,3 D.1tan B -1tan A+2sin A 的最小值为2228(2024·福建厦门·统考一模)已知△ABC 的内角A ，B ，C 的对边分别为a ，b ，c ，且a 2cos B +ab cos A =2c ．(1)求a ；(2)若A =2π3，且△ABC 的周长为2+5，求△ABC 的面积．29(2024·广西南宁·南宁三中校联考一模)记△ABC 的内角A ,B ,C 的对边分别为a ,b ,c ，已知a -bc=sin A -sin Csin A +sin B．(1)求角B 的大小；(2)若b =2，求△ABC 周长的最大值．30(2024·山东济南·山东省实验中学校考一模)在△ABC中，内角A,B,C的对边分别是a,b,c，且cos C =-14,c=2a.(1)求sin A的值；(2)若△ABC的周长为18，求△ABC的面积.31(2024·浙江·校联考一模)在△ABC中，内角A,B,C所对的边分别是a,b,c，已知c2b2+c2-a2=sin Csin B．(1)求角A；(2)设边BC的中点为D，若a=7，且△ABC的面积为334，求AD的长．32(2024·河南郑州·郑州市宇华实验学校校考一模)已知在△ABC中，3sin(A+B)=1+2sin2C 2．(1)求角C的大小；(2)若∠BAC与∠ABC的内角平分线交于点Ⅰ，△ABC的外接圆半径为2，求△ABI周长的最大值．33(2024·辽宁沈阳·统考一模)在△ABC中，角A,B,C所对的边分别为a,b,c，且b2=ac+a2.(1)求证：B=2A；(2)当3c+7a3b取最小值时，求cos B的值.34(2024·重庆·统考一模)在梯形ABCD中，AB⎳CD,∠ABC为钝角，AB=BC=2,CD=4，sin∠BCD=154．(1)求cos∠BDC；(2)设点E为AD的中点，求BE的长．35(2024·山西晋城·统考一模)在△ABC中，AB=33，AC=53，BC=73．(1)求A的大小；(2)求△ABC外接圆的半径与内切圆的半径．36(2024·黑龙江齐齐哈尔·统考一模)记△ABC 的内角A ,B ,C 的对边分别为a ,b ,c ，已知B =π4,4b cos C =2c +2a .(1)求tan C ；(2)若△ABC 的面积为32，求BC 边上的中线长.37(2024·云南曲靖·统考一模)在△ABC 中，内角A ,B ,C 的对边分别为a ,b ,c ，且c =2a cos C -2b ．(1)求A ；(2)线段BC 上一点D 满足BD =14BC ,AD =BD=1，求AB 的长度．三角函数题型01任意角的三角函数题型02两角和与差的三角函数题型03三角函数的图象与性质题型04解三角形题型01任意角的三角函数1(2024·辽宁沈阳·统考一模)sin x =1的一个充分不必要条件是.【答案】x =π2(答案不唯一)【分析】根据三角函数的性质结合充分不必要条件即可求解.【详解】因为x =π2时sin x =1，由sin x =1可得x =π2+2k π,k ∈Z ,故sin x =1的一个充分不必要条件是x =π2，故答案为：x =π2(答案不唯一)2(2024·重庆·统考一模)英国著名数学家布鲁克·泰勒(Taylor Brook )以微积分学中将函数展开成无穷级数的定理著称于世泰勒提出了适用于所有函数的泰勒级数，泰勒级数用无限连加式来表示一个函数，如：sin x =x -x 33!+x 55!-x 77!+⋯，其中n !=1×2×3×⋯×n ．根据该展开式可知，与2-233!+255!-277!+⋯的值最接近的是()A.sin2° B.sin24.6°C.cos24.6°D.cos65.4°【答案】C【分析】观察题目将其转化为三角函数值，再将弧度制与角度制互化，结合诱导公式判断即可.【详解】原式=sin2≈sin 2×57.3° =sin 90°+24.6° =cos24.6°，故选：C .3(2024·福建厦门·统考一模)若sin α+π4 =-35，则cos α-π4 =．【答案】-35/-0.6【分析】应用诱导公式有cos α-π4 =cos α+π4 -π2=sin α+π4 ，即可求值.【详解】cos α-π4 =cos α+π4 -π2=sin α+π4 =-35.故答案为：-354(2024·山东济南·山东省实验中学校考一模)下列说法正确的是()A.cos2sin3<0B.若圆心角为π3的扇形的弧长为π，则扇形的面积为3π2C.终边落在直线y =x 上的角的集合是α α=π4+2k π,k ∈ZD.函数y =tan 2x -π6 的定义域为x x ≠π3+k π2,k ∈Z ，π为该函数的一个周期【答案】ABD【分析】根据三角函数在各象限内的符号可判断出A 正确；根据扇形弧长和面积公式可知B 正确；由终边相同的角的集合表示方法可知C 错误；根据正切型函数定义域和周期的判断方法可知D 正确.【详解】对于A ，∵2,3均为第二象限角，∴cos2<0，sin3>0，∴cos2sin3<0，A 正确；对于B ，设扇形的半径为r ，则π3r =π，解得：r =3，∴扇形的面积S =12×π3×32=3π2，B 正确；对于C ，终边落在直线y =x 上的角的集合为α α=π4+k π,k ∈Z ，C 错误；对于D ，由2x -π6≠π2+k πk ∈Z 得：x ≠π3+k π2k ∈Z ，∴y =tan 2x -π6 的定义域为x x ≠π3+k π2,k ∈Z ；又tan 2x +π -π6 =tan 2π+2x -π6 =tan 2x -π6 ，∴π是y =tan 2x -π6 的一个周期，D 正确.故选：ABD .5(2024·山东济南·山东省实验中学校考一模)已知函数f (x )=cos xx，若A ，B 是锐角△ABC 的两个内角，则下列结论一定正确的是()A.f (sin A )>f (sin B )B.f (cos A )>f (cos B )C.f (sin A )>f (cos B )D.f (cos A )>f (sin B )【答案】D【分析】由已知可得π2>A >π2-B >0，根据余弦函数的单调性，得出cos A <sin B ，由f x 的单调性即可判断选项.【详解】因为f (x )=cos x x ，所以f (x )=-x sin x -cos xx 2，当x ∈0,π2 时，sin x >0,cos x >0，所以-x sin x -cos xx2<0，即f (x )<0，所以f x 在0,π2上单调递减.因为A ，B 是锐角△ABC 的两个内角，所以A +B >π2，则π2>A >π2-B >0，因为y =cos x 在0,π2 上单调递减，所以0<cos A <cos π2-B =sin B <1<π2，故f (cos A )>f (sin B )，故D 正确.同理可得f (cos B )>f (sin A )，C 错误；而A ,B 的大小不确定，故sin A 与sin B ，cos A 与cos B 的大小关系均不确定，所以f (sin A )与f (sin B )，f (cos A )与f (cos B )的大小关系也均不确定，AB 不能判断.故选：D6(2024·河北·校联考一模)在△ABC 中，若A =nB n ∈N * ，则()A.对任意的n ≥2，都有sin A <n sin BB.对任意的n ≥2，都有tan A <n tan BC.存在n ，使sin A >n sin B 成立D.存在n ，使tan A >n tan B 成立【答案】AD【分析】根据给定条件，举例说明判断BD；构造函数，借助导数探讨单调性判断AC.【详解】在△ABC中，当A=3B时，n=3，取B=π12，则A=π4，tan A=1，tan B=tanπ3-π4=3-11+3=2-3，3tan B=3(2-3)，则tan A>3tan B，B错，D对；显然0<A<π0<B<π0<C<π，即0<nB<π0<B<π0<π-B-nB<π，则0<B<πn+1，令f(x )=sin nx-n sin x，0<x<πn+1,n≥2，f (x)=n cos nx-n cos x=n(cos nx-cos x)<0，因此函数f(x)在0,πn+1上单调递减，则f(x)<f(0)=0，即sin nB<n sin B，从而sin A<n sin B，A对，C错.故选：AD【点睛】思路点睛：涉及不同变量的数式大小比较，细心挖掘问题的内在联系，构造函数，分析并运用函数的单调性求解作答.题型02两角和与差的三角函数7(2024·广西南宁·南宁三中校联考一模)若cosα+π4=35，则sin2α=()A.725B.-725C.925D.-925【答案】A【分析】根据二倍角的余弦公式和诱导公式即可.【详解】cos2α+π4=2cos2α+π4-1=2×35 2-1=-725,所以sin2α=-cos2α+π2=725，故选：A.8(2024·黑龙江齐齐哈尔·统考一模)已知cosα+π6=14，则sin2α-π6=()A.78B.-78C.38D.-38【答案】A【分析】利用换元法，结合诱导公式、二倍角公式等知识求得正确答案.【详解】设α+π6=t，则α=t-π6,cos t=14,sin2α-π6=sin2t-π6-π6=sin2t-π2=-cos2t=-2cos2t-1=-2×142-1=78.故选：A9(2024·辽宁沈阳·统考一模)已知sinπ2-θ+cosπ3-θ=1，则cos2θ-π3=()A.13B.-13C.33D.-33【答案】B【分析】根据和差角公式以及诱导公式可得32cos θ+32sin θ=1，由辅助角公式以及二倍角公式即可求解.【详解】由sin π2-θ+cos π3-θ =1得cos θ+12cos θ+32sin θ=1，进而可得32cos θ+32sin θ=1，结合辅助角公式得3cos θ-π6=1，则cos θ-π6 =33,∴cos 2θ-π3 =2cos 2θ-π6 -1=-13，故选：B .10(2024·浙江·校联考一模)已知α是第二象限角，β∈0,π2,tan α+π4 =-14，现将角α的终边逆时针旋转β后得到角γ，若tan γ=17，则tan β=．【答案】198/2.375【分析】由两角和的正切公式先得tan α=-53，进一步由两角差的正切公式即可求解.【详解】由题意tan α+π4 =tan α+11-tan α=-14，且γ=α+β，tan γ=tan α+β =17，解得tan α=-53，所以tan β=tan α+β-α =17--53 1+-53 ×17=198.故答案为：198.11(2024·安徽合肥·合肥一六八中学校考一模)已知tan α-11+tan α=2，则sin 2α+π6的值为()A.-4+3310B.-4-3310C.4+3310D.4-3310【答案】A【分析】先由已知条件求出tan α的值，再利用三角函数恒等变换公式求出sin2α,cos2α的值，然后对sin 2α+π6利用两角和的正弦公式化简计算即可【详解】由tan α-11+tan α=2，得tan α=-3，所以sin2α=2sin αcos αsin 2α+cos 2α=2tan αtan 2α+1=-610=-35，cos2α=cos 2α-sin 2αsin 2α+cos 2α=1-tan 2αtan 2α+1=1-910=-45，所以sin 2α+π6 =sin2αcos π6+cos2αsinπ6=-35×32+-45 ×12=-4+3310，故选：A12(2024·江西吉安·吉安一中校考一模)已知α∈0,π ，且3tan α=10cos2α，则cos α可能为()A.-1010B.-55C.1010D.55【答案】B【分析】由3tan α=10cos2α得3tan α=10×1-tan 2α1+tan 2α，化简后可求出tan α，再利用同角三角函数的关系可求出cos α.【详解】由3tan α=10cos2α，得3tan α=10(cos 2α-sin 2α)，所以3tan α=10×cos 2α-sin 2αcos 2α+sin 2α，所以3tan α=10×1-tan 2α1+tan 2α，整理得3tan 3α+10tan 2α+3tan α-10=0，(tan α+2)(3tan 2α+4tan α-5)=0，所以tan α+2=0或3tan 2α+4tan α-5=0，所以tan α=-2或tan α=-2±193，①当tan α=-2时，sin αcos α=-2，α∈π2,π ，因为sin 2α+cos 2α=1，所以5cos 2α=1，所以cos α=±55，因为α∈π2,π ，所以cos α=-55，②当tan α=-2+193时，sin αcos α=-2+193,α∈0,π2，因为sin 2α+cos 2α=1，所以19-23cos α 2+cos 2α=1，由于α∈0,π2 ，所以解得cos α=932-419，③当tan α=-2-193时，sin αcos α=-2-193,α∈π2,π ，因为sin 2α+cos 2α=1，所以-19-23cos α 2+cos 2α=1，由于α∈π2,π ，所以解得cos α=-932+419，综上，cos α=-55，或cos α=932-419，或cos α=-932+419，故选：B13(2024·吉林延边·统考一模)已知函数f x =12-sin 2ωx +32sin2ωx ,ω>0 的最小正周期为4π．(1)求ω的值，并写出f x 的对称轴方程；(2)在△ABC 中角A ,B ,C 的对边分别是a ,b ,c 满足2a -c cos B =b ⋅cos C ，求函数f A 的取值范围．【答案】(1)ω=14,x =2π3+2k π,k ∈Z(2)12,1 【分析】(1)利用三角函数的恒等变换化简函数f (x )=sin 2ωx +π6，再根据周期求出ω的值,利用整体法即可求解对称轴.(2)把已知的等式变形并利用正弦定理可得cos B =12，故B =π3，故f (A )=sin 12A +π6 ,0<A <2π3，根据正弦函数的定义域和值域求出f A 的取值范围．【详解】(1)f x =12-sin 2ωx +32sin2ωx =12+32sin2ωx -sin 2ωx =12+32sin2ωx -1-cos2ωx2=32sin2ωx +12cos2ωx =sin 2ωx +π6 ．∵T =2π2ω=4π，∴ω=14．故f x =sin 12x +π6 令12x +π6=π2+k π,k ∈Z ，解得x =2π3+2k π,k ∈Z ，故对称轴方程为：x =2π3+2k π,k ∈Z(2)由2a -c cos B =b ⋅cos C 得(2sin A -sin C )cos B =sin B cos C ，∴2sin A cos B =sin B cos C +cos B sin C =sin (B +C )=sin A ．∵sin A ≠0，∴cos B =12，B ∈0,π ,∴B =π3．∴f (A )=sin 12A +π6 ,0<A <2π3，∴π6<A 2+π6<π2，∴12<sin A 2+π6 <1，∴f (A )∈12,1 题型03三角函数的图象与性质14(2024·福建厦门·统考一模)已知函数f (x )=2sin 2x -π3，则()A.f (x )的最小正周期为π2B.f (x )的图象关于点2π3,0 成中心对称C.f (x )在区间0,π3上单调递增D.若f (x )的图象关于直线x =x 0对称，则sin2x 0=12【答案】BC【分析】根据正弦型函数的性质，结合代入法、整体法逐一判断各项正误.【详解】由f (x )=2sin 2x -π3 ，最小正周期T =2π2=π，A 错；由f 2π3=2sin 2×2π3-π3 =0，即2π3,0 是对称中心，B 对；由x ∈0,π3 ，则2x -π3∈-π3,π3 ，显然f (x )在区间0,π3 上单调递增，C 对；由题意2x 0-π3=k π+π2⇒2x 0=k π+5π6，故sin2x 0=±12，D 错.故选：BC15(2024·吉林延边·统考一模)将函数f x =sin ωx +π6 (ω>0)的图象向左平移π2个单位长度后得到曲线C ，若C 关于y 轴对称，则ω的最小值是()A.13B.23C.43D.53【答案】B【分析】得出平移后的方程后，再根据正弦型函数的性质即可得到答案.【详解】结合题意可得f x +π2 =sin ωx +π2 +π6 =sin ωx +π2ω+π6,(ω>0)，因为曲线C 关于y 轴对称，所以π2ω+π6=k π+π2,k ∈Z ，解得ω=2k +23,k ∈Z ,因为ω>0，所以当k =0时，ω有最小值23.故选：B .16(2024·黑龙江齐齐哈尔·统考一模)已知函数f x =cos2x +a cos x +2，则下列说法正确的有()A.当a =0时，f x 的最小正周期为πB.当a =1时，f x 的最小值为78C.当a =3时，f x 在区间0,2π 上有4个零点D.若f x 在0,π3 上单调递减，则a ≥2【答案】AB【分析】根据三角函数的周期性、含cos x 的二次项函数的值域、三角函数的零点、单调性等知识对选项进行分析，从而确定正确答案.【详解】当a =0时，f x =cos2x +2，所以f x 的最小正周期为π，A 选项正确；当a =0时，f x =cos2x +cos x +2=2cos 2x +cos x +1=2cos x +14 2+78≥78，所以f x 的最小值为78，B 选项正确；当a =4时，f x =cos2x +3cos x +2=2cos 2x +3cos x +1=2cos x +1 cos x +1 ，令f x =0，解得cos x =-12或cos x =-1，此时x =2π3或x =4π3或x =π，f x 在区间0,2π 上有3个零点，C 选项错误；f x =cos2x +a cos x +2=2cos 2x +a cos x +1，设t =cos x ，cos x 在0,π3 上单调递减，则t ∈12,1 ，根据复合函数的单调性，g t =2t 2+at +1在12,1 上单调递增，所以-a 4≤12，解得a ≥-2，D 选项错误.故选：AB17(2024·湖南长沙·雅礼中学校考一模)已知函数f (x )=sin ωx +3cos ωx (ω>0)满足:f π6=2,f 2π3=0,则()A.曲线y =f (x )关于直线x =7π6对称 B.函数y =f x -π3是奇函数C.函数y =f (x )在π6,7π6单调递减 D.函数y =f (x )的值域为[-2,2]【答案】ABD【分析】用辅助角公式化简f (x ),再利用f π6=2,f 2π3 =0,得出ω的取值集合,再结合三角函数性质逐项判断即可.【详解】f (x )=2sin ωx +π3,所以函数y =f (x )的值域为[-2,2],故D 正确；因为f 2π3=0,所以2π3ω+π3=k 1π,k 1∈Z ,所以ω=3k 1-12,k 1∈Z ，因为f π6 =2,所以π6ω+π3=π2+2k 2π,k 2∈Z ,所以ω=12k 2+1,k 2∈Z ，所以3k 1-12=12k 2+1,即k 1=8k 2+1，所以ω∈{1,13,25,37⋯}，因为f 7π6 =2sin 12k 2+1 7π6+π3 =2sin 14k 2π+3π2=-2，所以曲线y =f (x )关于直线x =7π6对称,故A 正确；因为f x -π3 =2sin 12k 2+1 x -π3 +π3 =2sin 12k 2+1 x -4k 2π =2sin 12k 2+1 x即f x -π3 =-f -x -π3，所以函数y =f x -π3是奇函数,故B 正确；取ω=13,则最小正周期T =2πω=2π13<7π6-π6=π,故C 错误.故选：ABD 18(2024·辽宁沈阳·统考一模)如图，点A ,B ,C 是函数f x =sin ωx +φ (ω>0)的图象与直线y =32相邻的三个交点，且BC -AB =π3,f -π12=0，则()A.ω=4B.f 9π8 =12C.函数f x 在π3,π2上单调递减D.若将函数f x 的图象沿x 轴平移θ个单位，得到一个偶函数的图像，则θ 的最小值为π24【答案】ACD【分析】令f x =32求得x A ,x B ,x C 根据BC -AB =π3求得ω=4，根据f -π12=0求得f x 的解析式，再逐项验证BCD 选项.【详解】令f x =sin ωx +φ =32得，ωx +φ=π3+2k π或ωx +φ=2π3+2k π，k ∈Z ，由图可知：ωx A +φ=π3+2k π，ωx C +φ=π3+2k π+2π，ωx B +φ=2π3+2k π，所以BC =x C -x B =1ω-π3+2π ，AB =x B -x A =1ω⋅π3，所以π3=BC -AB =1ω-2π3+2π ，所以ω=4，故A 选项正确，所以f x =sin 4x +φ ，由f -π12=0得sin -π3+φ =0，所以-π3+φ=π+2k π，k ∈Z ，所以φ=4π3+2k π，k ∈Z ，所以f x =sin 4x +4π3+2k π =sin 4x +4π3 =-sin 4x +π3 ，f 9π8 =-sin 9π2+π3 =-12，故B 错误.当x ∈π3,π2 时，4x +π3∈5π3,2π+π3，因为y =-sin t 在t ∈5π3,2π+π3 为减函数，故f x 在π3,π2上单调递减，故C 正确；将函数f x 的图象沿x 轴平移θ个单位得g x =-sin 4x +4θ+π3，(θ<0时向右平移，θ>0时向左平移)，g x 为偶函数得4θ+π3=π2+k π，k ∈Z ，所以θ=π24+k π4，k ∈Z ，则θ 的最小值为π24，故D 正确.故选：ACD .19(2024·重庆·统考一模)已知f x =2a sin ωx ⋅cos ωx +b cos2ωx ω>0,a >0,b >0 的部分图象如图所示，当x ∈0,3π4时，f x 的最大值为．【答案】3【分析】由图象求出函数f x 的解析式，然后利用正弦型函数的基本性质可求得函数f x 在0,3π4上的最大值.【详解】因为f x =2a sin ωx ⋅cos ωx +b cos2ωx =a sin2ωx +b cos2ωx ω>0,a >0,b >0 ，设f x =A sin 2ωx +φ A >0,ω>0 ，由图可知，函数f x 的最小正周期为T =4×π6+π12 =π，则2ω=2πT =2ππ=2，又因为A =f x max -f x min 2=2+22=2，则f x =2sin 2x +φ ，因为f -π12 =2sin φ-π6 =2，可得sin φ-π6 =1，所以，φ-π6=π2+2k πk ∈Z ，则φ=2π3+2k πk ∈Z ，则f x =2sin 2x +2π3+2k π =2sin 2x +2π3 ，当0≤x ≤3π4时，2π3≤2x +2π3≤13π6，故f x max =2sin 2π3=2×32= 3.故答案为：3.20(2024·云南曲靖·统考一模)函数f x =A sin ωx +φ (其中A >0，ω>0，φ ≤π2)的部分图象如图所示，则()A.f 0 =-1B.函数f x 的最小正周期是2πC.函数f x 的图象关于直线x =π3对称D.将函数f x 的图象向左平移π6个单位长度以后，所得的函数图象关于原点对称【答案】AC【分析】利用图象求出函数f x 的解析式，代值计算可判断A 选项；利用正弦型函数的周期性可判断B 选项；利用正弦型函数的对称性可判断C 选项；利用三角函数图象变换可判断D 选项.【详解】由图可知，A =f x max -f x min 2=2--22=2，函数f x 的最小正周期T 满足3T 4=7π12--π6 =3π4，则T =π，ω=2πT =2ππ=2，B 错；所以，f x =2sin 2x +φ ，f -π6 =2sin 2×-π6 +φ =2sin φ-π3 =-2，可得sin φ-π3 =-1，因为-π2≤φ≤π2，所以，-5π6≤φ-π3≤π6，则φ-π3=-π2，可得φ=-π6，所以，f x =2sin 2x -π6 ，则f 0 =2sin -π6=-1，A 对；f π3 =2sin 2×π3-π6 =2sin π2=2=f x max ，所以，函数f x 的图象关于直线x =π3对称，C 对；将函数f x 的图象向左平移π6个单位长度以后，得到函数y =2sin 2x +π6 -π6 =2sin 2x +π6 的图象，所得函数为非奇非偶函数，D 错.故选：AC .21(2024·浙江·校联考一模)已知函数y =2sin ωx +φ ，该图象上最高点与最低点的最近距离为5，且点1,0 是函数的一个对称点，则ω和φ的值可能是()A.ω=-π3,φ=-π3B.ω=-π3,φ=2π3C.ω=π3,φ=π3D.ω=π3,φ=2π3【答案】D【分析】由题意首先得ω=π3，进一步由ω+φ=k π,k ∈Z ，对比选项即可得解.【详解】由题意函数的周期T 满足，T 2=52-42=3=2π2ω ，所以ω=±π3，又点1,0 是函数的一个对称点，所以ω+φ=k π,k ∈Z ，所以ω=π3φ=k π-π3,k ∈Z 或ω=-π3φ=k π+π3,k ∈Z，对比选项可知，只有当ω=π3φ=2π3k =1时满足题意.故选：D .22(2024·广东深圳·校考一模)已知函数f x =cos ωx +π3+1(ω>0)的最小正周期为π，则f x 在区间0,π2上的最大值为()A.12B.1C.32D.2【答案】C【分析】由周期公式求得ω，结合换元法即可求得最大值.【详解】由题意T =2πω=π，解得ω=2，所以f x =cos 2x +π3+1，当x ∈0,π2 时，t =2x +π3∈π3,4π3，所以f x 在区间0,π2 上的最大值为cos π3+1=32，当且仅当x =0时等号成立.故选：C .23(2024·山西晋城·统考一模)若函数f (x )=cos ωx (0<ω<100)在π,5π2上至少有两个极大值点和两个零点，则ω的取值范围为．【答案】85,2 ∪125,100 【分析】先求出极大值点表达式，利用题干条件列不等式赋值求解.【详解】令ωx =2k π，k ∈Z ，得f (x )的极大值点为x =2k πω，k ∈Z ，则存在整数k ，使得ω>02k πω>π2k +1 πω<5π2，解得4(k +1)5<ω<2k (k ∈N *)．因为函数y =cos x 在两个相邻的极大值点之间有两个零点，所以4(k +1)5<ω<2k (k ∈N *)．当k =1时，85<ω<2．当k =2时，125<ω<4．当k ≥2时，4(k +1)5<4(k +2)5<2k ．又0<ω<100，所以ω的取值范围为85,2 ∪125,4 ∪165,6 ∪⋅⋅⋅∪2045,100 =85,2 ∪125,100 ．故答案为：85,2 ∪125,100【点睛】关键点点睛：本题考查三角函数的图象及其性质，求出4k +15<ω<2k k ∈N * 并赋值计算是解决问题关键.24(2024·广西南宁·南宁三中校联考一模)在物理学中，把物体受到的力(总是指向平衡位置)正比于它离开平衡位置的距离的运动称为“简谐运动”．在适当的直角坐标系下，某个简谐运动可以用函数f x =A sin ωx +φ A >0,ω>0,φ <π 的部分图象如图所示，则下列结论正确的是()A.ω=2，频率为1π，初相为π6B.函数f x 的图象关于直线x =-π6对称C.函数f x 在π12,13π24上的值域为0,2 D.若把f x 图像上所有点的横坐标缩短为原来的23倍，纵坐标不变，再向左平移π12个单位，则所得函数是y =2sin 3x +π12【答案】BCD【分析】根据图象求出三角函数解析式，再根据正弦函数图象与性质以及函数平移的原则即可判断.【详解】由图象可得A =2,34T =13π12-π3=3π4,∴T =π，频率是1T =1π,ω=2ππ=2,∵f π3 =2,∴f π3 =2sin 2π3+φ =2，即sin 2π3+φ =1,∴2π3+φ=2k π+π2(k ∈Z )，∴φ=2k π-π6(k ∈Z ),∵|φ|<π,∴φ=-π6，对于A ，∴f (x )=2sin 2x -π6 ，初相是-π6，故A 错误；对于B ，f -π6 =2sin -π3-π6=-2，故B 正确；对于C ，因为x ∈π12,13π24 ，所以2x -π6∈0,11π12，∴f (x )=2sin 2x -π6在π12,13π24上的值域为[0,2]，故C 正确；对于D ，把f (x )的横坐标缩短为原来的23倍，纵坐标不变，得到的函数为y =2sin 3x -π6，又向左平移π12个单位，得到的函数为y =2sin 3x +π12 -π6 =2sin 3x +π12 ，故D 正确；故选：BCD .题型04解三角形25(2024·河南郑州·郑州市宇华实验学校校考一模)如图，为了测量某湿地A ，B 两点间的距离，观察者找到在同一直线上的三点C ，D ，E ．从D 点测得∠ADC =67.5°，从C 点测得∠ACD =45°，∠BCE =75°，从E 点测得∠BEC =60°．若测得DC =23，CE =2(单位：百米)，则A ，B 两点的距离为()A.6B.22C.3D.23【答案】C【分析】在△ADC 中，求得AC =DC ；在△BCE 中，利用正弦定理求得BC ；再在△ABC 中，利用余弦定理即可求得结果.【详解】根据题意，在△ADC 中，∠ACD =45°，∠ADC =67.5°，DC =23，则∠DAC =180°-45°-67.5°=67.5°，则AC =DC =23，在△BCE 中，∠BCE =75°，∠BEC =60°，CE =2，则∠EBC =180°-75°-60°=45°，则有CE sin ∠EBC =BC sin ∠BEC ，变形可得BC =CE ⋅sin ∠BEC sin ∠EBC =2×3222=3，在△ABC 中，AC =23，BC =3，∠ACB =180°-∠ACD -∠BCE =60°，则AB 2=AC 2+BC 2-2AC ·BC ·cos ∠ACB =9，则AB =3．故选：C .【点睛】本题考查利用正余弦定理解三角形，涉及距离的求解，属基础题.26(2024·广东深圳·校考一模)在△ABC 中，角A ，B ，C 的对边分别为a ，b ，c ，若a =3，b =5，c =2a cos A ，则cos A =()A.13B.24C.33D.63【答案】D【分析】由已知结合余弦定理进行化简即可求解.【详解】解：因为c =2a cos A ，由余弦定理可得c =2a ⋅b 2+c 2-a 22bc，将a =3，b =5代入整理得c =26，所以cos A =c 2a =63.故选：D .27(2024·河南郑州·郑州市宇华实验学校校考一模)在锐角△ABC 中，角A ,B ,C 所对的边分别为a ,b ,c ，且c -b =2b cos A ，则下列结论正确的有()A.A =2BB.B 的取值范围为0,π4C.ab的取值范围为2,3 D.1tan B -1tan A+2sin A 的最小值为22【答案】AC【分析】用正弦定理可判断A 项，由锐角三角形可判断B 项，用倍角公式可判断C 项，切化弦后用取等条件即可判断D 项.【详解】在△ABC 中，由正弦定理可将式子c -b =2b cos A 化为sin C -sin B =2sin B cos A ，把sin C =sin A +B =sin A cos B +cos A sin B 代入整理得，sin A -B =sin B ，解得A -B =B 或A -B +B =π，即A =2B 或A =π(舍去)，所以A =2B ，选项A 正确；选项B ：因为△ABC 为锐角三角形，A =2B ，所以C =π-3B ，由0<B <π2,0<2B <π2,0<π-3B <π2,解得B ∈π6,π4 ，故选项B 错误；选项C ：a b=sin A sin B =sin2B sin B =2cos B ，因为B ∈π6,π4 ，所以cos B ∈22,32 ，2cos B ∈2,3 ，即ab的取值范围为2,3 ，故选项C 正确；选项D ：1tan B -1tan A +2sin A =sin A -B sin A sin B +2sin A =1sin A+2sin A ≥21sin A ×2sin A =22，当且仅当1sin A=2sin A 即sin A =±22时取等，但因为B ∈π6,π4 ，所以A =2B ∈π3,π2 ,sin A ∈32,1 ，无法取到等号，故D 错.故选：AC .28(2024·福建厦门·统考一模)已知△ABC 的内角A ，B ，C 的对边分别为a ，b ，c ，且a 2cos B +ab cos A =2c ．(1)求a ；(2)若A =2π3，且△ABC 的周长为2+5，求△ABC 的面积．【答案】(1)a =2；(2)34.【分析】(1)应用正弦边角关系及和角正弦公式有a sin (A +B )=2sin C ，再由三角形内角性质即可求边长；(2)应用余弦定理及已知得b 2+c 2+bc =4且b +c =5，进而求得bc =1，最后应用面积公式求面积.【详解】(1)由题设a (a cos B +b cos A )=2c ，由正弦定理有a (sin A cos B +sin B cos A )=2sin C ，所以a sin (A +B )=2sin C ，而A +B =π-C ，故a sin C =2sin C ，又sin C >0，所以a =2.(2)由(1)及已知，有cos A =b 2+c 2-a 22bc =b 2+c 2-42bc=-12，可得b 2+c 2+bc =4，又a +b +c =2+5，即b +c =5，所以(b +c )2-bc =5-bc =4⇒bc =1，故S △ABC =12bc sin A =34.29(2024·广西南宁·南宁三中校联考一模)记△ABC 的内角A ,B ,C 的对边分别为a ,b ,c ，已知a -bc=sin A -sin Csin A +sin B．(1)求角B 的大小；(2)若b =2，求△ABC 周长的最大值．【答案】(1)B =π3(2)6【分析】(1)根据题意利用正、余弦定理进行边角转化，进而可得结果；(2)根据a 2+c 2-b 2=ac ，结合基本不等式运算求解.【详解】(1)因为a -b c =sin A -sin C sin A +sin B，由正弦定理可得a -b c =a -ca +b ，整理得a 2+c 2-b 2=ac ，由余弦定理可得cos B =a 2+c 2-b 22ac =ac 2ac =12，且B ∈0,π ，所以B =π3.(2)由(1)可知：a 2+c 2-b 2=ac ，整理得a +c 2-4=3ac ，即ac =a +c 2-43，因为ac ≤a +c24，当且仅当a =c =2时，等号成立，则a +c 2-43≤a +c 24，可得a +c 2≤16，即a +c ≤4，所以△ABC 周长的最大值为4+2=6.30(2024·山东济南·山东省实验中学校考一模)在△ABC 中，内角A ,B ,C 的对边分别是a ,b ,c ，且cos C =-14,c =2a .(1)求sin A 的值；(2)若△ABC 的周长为18，求△ABC 的面积.【答案】(1)158(2)315【分析】(1)由正弦定理边化角结合同角三角函数关系求解;(2)由余弦定理解方程得边长，再利用面积公式求解.【详解】(1)因为0<C <π，cos C =-14，所以sin C =1-cos 2C =154.因为c =2a ，所以sin C =2sin A ，则sin A =sin C 2=158.(2)因为cos C =-14，所以c 2=a 2+b 2+12ab .因为c =2a ，所以3a 2-12ab -b 2=0，解得b =32a .因为△ABC 的周长为18，所以a +b +c =92a =18，解得a =4，则b =6,c =8.故△ABC 的面积为12bc sin A =12×6×8×158=315.31(2024·浙江·校联考一模)在△ABC 中，内角A ,B ,C 所对的边分别是a ,b ,c ，已知c 2b 2+c 2-a2=sin Csin B．(1)求角A ；(2)设边BC 的中点为D ，若a =7，且△ABC 的面积为334，求AD 的长．【答案】(1)A =π3(2)132【分析】(1)根据正弦定理和题中所给式子化简计算得到b 2+c 2-a 2=bc ，再结合余弦定理即可求出角A ；(2)根据三角形面积公式得到bc =3和b 2+c 2=10，再结合中线向量公式计算即可.【详解】(1)在△ABC 中，由正弦定理得，sin C sin B =cb，因为c 2b 2+c 2-a 2=sin C sin B ，所以c 2b 2+c 2-a 2=cb ，化简得，b 2+c 2-a 2=bc ，在△ABC 中，由余弦定理得，cos A =b 2+c 2-a 22bc=12，又因为0<A <π，所以A =π3(2)由S △ABC =12bc sin A =34bc =334，得bc =3，由a 2=b 2+c 2-2bc cos A ，得7=b 2+c 2-3，所以b 2+c 2=10．又因为边BC 的中点为D ，所以AD =12AB +AC，所以AD =12(AB +AC )2=12b 2+c 2+2bc cos A =12×10+2×3×12=13232(2024·河南郑州·郑州市宇华实验学校校考一模)已知在△ABC 中，3sin (A +B )=1+2sin 2C2．(1)求角C 的大小；(2)若∠BAC 与∠ABC 的内角平分线交于点Ⅰ，△ABC 的外接圆半径为2，求△ABI 周长的最大值．【答案】(1)π3；(2)4+23．【分析】(1)利用降幂公式、两角和的正弦公式变形可得sin C +π6=1，再根据角的范围可得解；(2)利用正弦定理求出AB ，求出∠AIB ，设出∠ABI ，将AI ,BI 用∠ABI 表示，根据三角函数知识求出AI +BI 的最大值可得解.【详解】(1)∵3sin (A +B )=1+2sin 2C2，且A +B +C =π，∴3sin C =1+1-cos C =2-cos C ，即3sin C +cos C =2，∴sin C +π6=1．∵C ∈(0，π)，∴C +π6∈π6，7π6 ，∴C +π6=π2，即C =π3．(2)∵△ABC 的外接圆半径为2，∴由正弦定理知，AB sin ∠ACB =ABsin π3=2×2=4，∴AB =23，∵∠ACB =π3，∴∠ABC +∠BAC =2π3，∵∠BAC 与∠ABC 的内角平分线交于点Ⅰ，∴∠ABI +∠BAI =π3，∴∠AIB =2π3，设∠ABI =θ，则∠BAI =π3-θ，且0<θ<π3，在△ABI 中，由正弦定理得，BI sin π3-θ =AI sin θ=AB sin ∠AIB =23sin 2π3=4，∴BI =4sin π3-θ ，AI =4sin θ，∴△ABI 的周长为23+4sin π3-θ +4sin θ=23+432cos θ-12sin θ +4sin θ=23+23cos θ+2sin θ=4sin θ+π3+23，∵0<θ<π3，∴π3<θ+π3<2π3，∴当θ+π3=π2，即θ=π6时，△ABI 的周长取得最大值，最大值为4+23，故△ABI 的周长的最大值为4+23．【点睛】关键点点睛：将AI ,BI 用∠ABI 表示，根据三角函数知识求出AI +BI 的最大值是解题关键.33(2024·辽宁沈阳·统考一模)在△ABC 中，角A ,B ,C 所对的边分别为a ,b ,c ，且b 2=ac +a 2.(1)求证：B =2A ；(2)当3c +7a 3b取最小值时，求cos B 的值.【答案】(1)证明见解析(2)cos B =-13【分析】(1)利用余弦定理并结合正弦函数两角和差公式化简即可求解.(2)利用基本不等式求得3c +7a 3b的最小值时的取等条件b =233a ，再结合余弦定理从而求解.【详解】(1)证明：由余弦定理知b 2=a 2+c 2-2ac cos B ，又因为b 2=a 2+ac ，所以a 2+ac =a 2+c 2-2ac ⋅cos B ，化简得a =c -2a cos B ，所以sin A =sin C -2sin A cos B ，因为A +B +C =π，所以sin A =sin A +B -2sin A cos B ，所以sin A =sin A cos B +cos A sin B -2sin A cos B =cos A sin B -sin A cos B ，所以sin A =sin B -A ，因为A ∈0,π ,B -A ∈-π,π ，所以A =B -A 或A +B -A =π(舍)，所以B =2A .(2)由题知，3c +7a 3b =3ac +7a 23ab =3b 2-a 2 +7a 23ab=b a +43⋅a b ≥243=433，当且仅当b =233a 时取等，又因为b 2=ac +a 2，所以c =13a ，所以cos B =a 2+c 2-b22ac=a 2+13a 2-233a22a ×13a=-13.34(2024·重庆·统考一模)在梯形ABCD 中，AB ⎳CD ,∠ABC 为钝角，AB =BC =2,CD =4，sin ∠BCD =154．(1)求cos ∠BDC ；(2)设点E 为AD 的中点，求BE 的长．【答案】(1)78；(2)342【分析】(1)在△BCD 中利用余弦定理求出BD ，再利用二倍角的余弦公式计算即得.(2)利用(1)的结论，借助向量数量积求出BE 的长.【详解】(1)在梯形ABCD 中，由AB ⎳CD ,∠ABC 为钝角，得∠BCD 是锐角，在△BCD 中，sin ∠BCD =154，则cos ∠BCD =1-sin 2∠BCD =14，由余弦定理得BD =22+42-2×2×4×14=4，即△BCD 为等腰三角形，所以cos ∠BDC =cos (π-2∠BCD )=-cos2∠BCD =1-2cos 2∠BCD =78.(2)由AB ⎳CD ，得∠ABD =∠BDC ，由点E 为AD 的中点，得BE =12(BA +BD)，所以|BE |=12BA 2+BD 2+2BA ⋅BD =1222+42+2×2×4×78=342.35(2024·山西晋城·统考一模)在△ABC 中，AB =33，AC =53，BC =73．(1)求A 的大小；(2)求△ABC 外接圆的半径与内切圆的半径．【答案】(1)A =2π3(2)32【分析】(1)由余弦定理即可求解；(2)由正弦定理求出外接圆半径，由等面积法求出内切圆半径.【详解】(1)由余弦定理得cos A =AB 2+AC 2-BC 22AB ⋅AC=-12，因为0<A <π，所以A =2π3．(2)设△ABC 外接圆的半径与内切圆的半径分别为R ，r ，由正弦定理得2R =BC sin A=7332=14，则R =7．△ABC 的面积S =12AB ⋅AC ⋅sin A =4534，由12r (AB +AC +BC )=S ，得r =2S AB +AC +BC =32．36(2024·黑龙江齐齐哈尔·统考一模)记△ABC 的内角A ,B ,C 的对边分别为a ,b ,c ，已知B =π4,4b cos C =2c +2a .(1)求tan C ；。
数学专业英语-第2章课后答案
数学专业英语-第2章课后答案2.12.比:ratio 比例:proportion 利率:interest rate 速率:speed 除:divide 除法:division 商:quotient 同类量:like quantity 项:term 线段:line segment 角:angle 长度:length 宽:width高度:height 维数:dimension 单位:unit 分数:fraction 百分数:percentage3.(1)一条线段和一个角的比没有意义,他们不是相同类型的量.(2)比较式通过说明一个量是另一个量的多少倍做出的,并且这两个量必须依据相同的单位.(5)为了解一个方程,我们必须移项,直到未知项独自处在方程的一边,这样就可以使它等于另一边的某量.4.(1)Measuring the length of a desk, is actually comparing the length of the desk to that of a ruler.(3)Ratio is different from the measurement, it has no units. The ratio of the length and the width of the same book does not vary when the measurement unit changes.(5)60 percent of students in a school are female students, which mean that 60 students out of every 100 students are female students.2.22.初等几何:elementary geometry 三角学:trigonometry 余弦定理:Law of cosines 勾股定理/毕达哥拉斯定理:Gou-Gu theorem/Pythagoras theorem 角:angle 锐角:acute angle 直角:right angle 同终边的角:conterminal angles 仰角:angle of elevation 俯角:angle of depression 全等:congruence 夹角:included angle 三角形:triangle 三角函数:trigonometric function直角边:leg 斜边:hypotenuse 对边:opposite side 临边:adjacentside 始边:initial side 解三角形:solve a triangle 互相依赖:mutually dependent 表示成:be denoted as 定义为:be defined as3.(1)Trigonometric function of the acute angle shows the mutually dependent relations between each sides and acute angle of the right triangle.(3)If two sides and the included angle of an oblique triangle areknown, then the unknown sides and angles can be found by using the law of cosines.(5)Knowing the length of two sides and the measure of the included angle can determine the shape and size of the triangle. In other words, the two triangles made by these data are congruent.4.(1)如果一个角的顶点在一个笛卡尔坐标系的原点并且它的始边沿着x轴正方向，这个角被称为处于标准位置.(3)仰角和俯角是以一条以水平线为参考位置来测量的,如果正被观测的物体在观测者的上方,那么由水平线和视线所形成的角叫做仰角.如果正被观测的物体在观测者的下方,那么由水平线和视线所形成的的角叫做俯角.(5)如果我们知道一个三角形的两条边的长度和对着其中一条边的角度,我们如何解这个三角形呢?这个问题有一点困难来回答,因为所给的信息可能确定两个三角形,一个三角形或者一个也确定不了.2.32.素数:prime 合数:composite 质因数:prime factor/prime divisor 公倍数:common multiple 正素因子: positive prime divisor 除法算式:division equation 最大公因数:greatest common divisor(G.C.D) 最小公倍数: lowest common multiple(L.C.M) 整除:divide by 整除性:divisibility 过程:process 证明:proof 分类:classification 剩余:remainder辗转相除法:Euclidean algorithm 有限集:finite set 无限的:infinitely 可数的countable 终止:terminate 与矛盾:contrary to3.(1)We need to study by which integers an integer is divisible, that is , what factor it has. Specially, it is sometime required that an integer is expressed as the product of its prime factors.(3)The number 1 is neither a prime nor a composite number;A composite number in addition to being divisible by 1 and itself, can also be divisible by some prime number.(5)The number of the primes bounded above by any given finite integer N can be found by using the method of the sieve Eratosthenes.4.(1)数论中一个重要的问题是哥德巴赫猜想,它是关于偶数作为两个奇素数和的表示.(3)一个数,形如2p-1的素数被称为梅森素数.求出5个这样的数.(5)任意给定的整数m和素数p,p的仅有的正因子是p和1,因此仅有的可能的p和m的正公因子是p和1.因此,我们有结论:如果p是一个素数,m是任意整数,那么p整除m,要么(p,m)=1.2.42.集:set 子集:subset 真子集:proper subset 全集:universe 补集:complement 抽象集:abstract set 并集:union 交集:intersection 元素:element/member 组成:comprise/constitute包含:contain 术语:terminology 概念:concept 上有界:bounded above 上界:upper bound 最小的上界:least upper bound 完备性公理:completeness axiom3.(1)Set theory has become one of the common theoretical foundation and the important tools in many branches of mathematics.(3)Set S itself is the improper subset of S; if set T is a subset of S but not S, then T is called a proper subset of S.(5)The subset T of set S can often be denoted by {x}, that is, T consists of those elements x for which P(x) holds.(7)This example makes the following question become clear, that is, why may two straight lines in the space neither intersect nor parallel.4.(1)设N是所有自然数的集合,如果S是所有偶数的集合,那么它在N中的补集是所有奇数的集合.(3)一个非空集合S称为由上界的,如果存在一个数c具有属性:x<=c对于所有S中的x.这样一个数字c被称为S的上界.(5)从任意两个对象x和y，我们可以形成序列（x，y），它被称为一个有序对，除非x=y，否则它当然不同于（y，x）.如果S和T是任意集合，我们用S*T表示所有有序对（x,y),其中x术语S，y属于T.在R.笛卡尔展示了如何通过实轴和它自己的笛卡尔积来描述平面的点之后，集合S*T 被称为S和T的笛卡尔积.2.52.竖直线:vertical line 水平线:horizontal line 数对:pairs of numbers 有序对:ordered pairs 纵坐标:ordinate 横坐标:abscissas 一一对应:one-to-one 对应点:corresponding points圆锥曲线:conic sections 非空图形:non vacuous graph 直立圆锥:right circular cone 定值角:constant angle 母线:generating line 双曲线:hyperbola 抛物线:parabola 椭圆:ellipse退化的:degenerate 非退化的:nondegenerate 任意的:arbitrarily 相容的:consistent 在几何上:geometrically 二次方程:quadraticequation 判别式:discriminant 行列式:determinant3.(1)In the planar rectangular coordinate system, one can set up aone-to-one correspondence between points and ordered pairs of numbers and also a one-to-one correspondence between conic sections and quadratic equation.(3)The symbol can be used to denote the set of ordered pairs （x，y）such that the ordinate is equal to the cube of the abscissa.（5）According to the values of the discriminate，the non-degenerate graph of Equation （iii）maybe known to be a parabola， a hyperbolaor an ellipse.4.（1）在例1，我们既用了图形，也用了代数的代入法解一个方程组（其中一个方程式二次的，另一个是线性的）。
反三角函数练习题及答案
反三角函数练习题及答案相关热词搜索：练习题函数答案反三角函数公式反三角函数的习题反三角函数知识点篇一：反三角函数练习周末作业2012429反三角函数练习周末作业1. arcsin（－x）x?sin（arcsinx）x?arcsin(sinx)＝，x?2. arccos（-x）x?cos（arccosx）x?arccos（cosx）x?3. arctan（-x）x?tan（arctanx）x? ??3??3?sarcs?i?n?4．arctan(?)?；co??? ?3??211 arcsin（sin20070）=________________________；arcsin?)??6??5. 用反三角函数表示sinx??,x,13??3??的角x?2??6. tanx??3，x??0,2??，则x= _____________________;7. 若3cos??1?0，当?为?ABC的一个内角时，则??8. 求函数y?9. 求函数f（x）=10. 求f(x)=2arcsin2x的反函数____________________________11. 当x?0时，arctanx?arctan?2?arcsinx,x1,1?的反函数_______________________ x?+arccos的反函数_______________________ 221恒等于。
x12. 当x??-1，1?时，arcsinx?arccosx恒等于。
113．已知y?sinx与函数y?arcsinx，下列说法正确的是（）A．互为反函数B．都是增函数C．都是奇函数D．都是周期函数14．函数y?arcsin?sinx?的定义域是（）A．??1,1? B．22,2?2 C．R D．0,???15．已知sinx?a，其中a?1,??x?3?22，那么用反三角函数表示x的四个式子中正确的是A．xarcsina B．xarcsinaC．x??2?arcsina D．x??2?arcsina16．求下列各式的值（1）cos23?4??（2）sin??3???arccos(?3)???42??（3）cos??3?arcsin1?? （4）sin??arcsin3?arccos(?5?42??513)2）（17. 求满足下列不等式的x的取值范围。
高考数学三角函数逻辑推理历年真题2024分析
高考数学三角函数逻辑推理历年真题2024分析数学是高考中最核心、最关键的科目之一，而三角函数作为数学知识中的重要组成部分，在高考中也占有很大的比重。
本文将针对2024年高考数学三角函数题目进行详细的逻辑推理分析，以帮助同学们更好地掌握这一知识点。
1. 2024年高考数学三角函数选择题第一道题目：已知四个正数A、B、C、D满足A + B = C + D，且tanA = 2，tanC = 3，那么A + B的值是多少？解析：根据题意，我们可以得到tanA = tan(A + B)和tanC = tan(C + D)。
由于tanA = 2和tanC = 3，我们可以通过查三角函数表得到A和C 的值，进而求得A + B的值。
这道题考察了三角函数的运用和对已知条件的合理利用。
第二道题目：已知sinx + cosy = 1，sinx - cosy = 1/2，则cosx + siny 的值是多少？解析：根据已知条件，我们可以得到两个方程sinx + cosy = 1和sinx - cosy = 1/2。
通过这两个方程，我们可以将cosx和siny用sinx和cosy来表示，进而求得cosx + siny的值。
这道题涉及到三角函数的运算和方程的解法，需要考生掌握相关知识点。
2. 2024年高考数学三角函数解答题第一道题目：已知正数a和b满足a^2 + b^2 = 1，求证：(a + b)^2 + (a - b)^2 = 2。
解析：首先我们可以将(a + b)^2 + (a - b)^2展开得到2a^2 + 2b^2 = 2。
而根据已知条件a^2 + b^2 = 1，可以得到2a^2 + 2b^2 = 2。
由此可见，原命题成立。
这道题考察了解答题的基本技巧和运算能力。
第二道题目：已知tanx = 2，求证：sinx * cosx = 2/√5。
解析：根据已知条件tanx = 2，我们可以利用三角函数的定义关系sinx/cosx = 2/1，进而得到sinx = 2/√5，cosx = 1/√5。
sat数学考试试题
SAT数学真题精选1. If 2 x + 3 = 9, what is the value of 4 x – 3 ?(A) 5 (B) 9 (C) 15 (D) 18 (E) 212. If 4(t + u) + 3 = 19, then t + u = ?(A) 3 (B) 4 (C) 5 (D) 6 (E) 73. In the xy-coordinate (坐标) plane above, the line contains the points (0,0) and (1,2). If line M (not shown) contains the point (0,0) and is perpendicular （垂直）to L, what is an equation of M?(A) y = -1/2 x(B) y = -1/2 x + 1(C) y = - x(D) y = - x + 2(E) y = -2x4. If K is divisible by 2,3, and 15, which of the following is also divisible by these numbers?(A) K + 5 (B) K + 15 (C) K + 20 (D) K + 30 (E) K + 455. There are 8 sections of seats in an auditorium. Each section contains at least 150 seats but not more than 200 seats. Which of the following could be the number of seats in this auditorium?(A) 800 (B) 1,000 (C) 1,100 (D) 1,300 (E) 1,7006. If rsuv = 1 and rsum = 0, which of the following must be true?(A) r < 1 (B) s < 1 (C) u= 2 (D) r = 0 (E) m = 07. The least integer of a set of consecutive integers (连续整数) is –126. if the sum of these integers is 127, how many integers are in this set?(A) 126 (B) 127 (C) 252 (D) 253 (E) 2548. A special lottery is to be held to select the student who will live in the only deluxe room in a dormitory. There are 200 seniors, 300 juniors, and 400 sophomores who applied. Each senior’s name is placed in the lottery 3 times; each junior’s name, 2 time; and each sophomore’s name, 1 times. If a student’s name is chosen at random from the names in the lottery, what is the probability that a senior’s name will be chosen?（A）1/8 (B) 2/9 (C) 2/7 (D) 3/8 (E) 1/2Question #1: 50% of US college students live on campus. Out of all students living on campus, 40% are graduate students. What percentage of US students are graduate students living on campus?(A) 90% (B) 5% (C) 40% (D) 20% (E) 25% Question #2: In the figure below, MN is parallel with BC and AM/AB = 2/3. What is the ratio between the area of triangle AMN and the area of triangle ABC?(A) 5/9 (B) 2/3 (C) 4/9 (D) 1/2 (E) 2/9Question #3: If a2 + 3 is divisible by 7, which of the following values can be a?(A)7 (B)8 (C)9 (D)11 (E)4Question #4: What is the value of b, if x = 2 is a solution of equation x2 - b · x + 1 = 0?(A)1/2 (B)-1/2 (C)5/2 (D)-5/2 (E)2Question #5: Which value of x satisfies the inequality | 2x | < x + 1 ?(A)-1/2 (B)1/2 (C)1 (D)-1 (E)2Question #6: If integers m > 2 and n > 2, how many (m, n) pairs satisfy the inequality m n < 100?(A)2 (B)3 (C)4 (D)5 (E)7Question #7: The US deer population increase is 50% every 20 years. How may times larger will the deer population be in 60 years ?(A)2.275 (B)3.250 (C)2.250 (D)3.375 (E)2.500 Question #8: Find the value of x if x + y = 13 and x - y = 5.(A)2 (B)3 (C)6 (D)9 (E)4Question #9:The number of medals won at a track and field championship is shown in the table above. What is the percentage of bronze medals won by UK out of all medals won by the 2 teams?(A)20% (B)6.66% (C)26.6% (D)33.3% (E)10% Question #10: The edges of a cube are each 4 inches long. What is the surface area, in square inches, of this cube?(A)66 (B)60 (C)76 (D)96 (E)65Question #1: The sum of the two solutions of the quadratic equation f(x) = 0 is equal to 1 and the product of the solutions is equal to -20. What are the solutions of the equation f(x) = 16 - x ?(a) x1 = 3 and x2 = -3 (b) x1 = 6 and x2 = -6(c) x1 = 5 and x2 = -4 (d) x1 = -5 and x2 = 4(e) x1 = 6 and x2 = 0Question #2: In the (x, y) coordinate plane, three lines have the equations:l1: y = ax + 1l2: y = bx + 2l3: y = cx + 3Which of the following may be values of a, b and c, if line l3 is perpendicular to both lines l1 and l2?(a) a = -2, b = -2, c = .5 (b) a = -2, b = -2, c = 2(c) a = -2, b = -2, c = -2 (d) a = -2, b = 2, c = .5(e) a = 2, b = -2, c = 2Question #3: The management team of a company has 250 men and 125 women. If 200 of the managers have a master degree, and 100 of the managers with the master degree are women, how many of the managers are men without a master degree? (a) 125 (b) 150 (c) 175 (d) 200 (e) 225Question #4: In the figure below, the area of square ABCD is equal to the sum of the areas of triangles ABE and DCE. If AB = 6, then CE =(a) 5 (b) 6 (c) 2 (d) 3 (e) 4Question #5:If α and β are the angles of the right triangle shown in the figure above, then sin2α + sin2β is equal to:(a) cos(β)(b) si n(β)(c) 1 (d) cos2(β)(e) -1 Question #6: The average of numbers (a + 9) and (a - 1) is equal to b, where a and b are integers. The product of the same two integers is equal to (b - 1)2. What is the value of a?(a) a = 9 (b) a = 1 (c) a = 0 (d) a = 5 (e) a = 11Question #1: If f(x) = x and g(x) = √x, x≥ 0, what are the solutions of f(x) = g(x)? (A) x = 1 (B)x1 = 1, x2 = -1(C)x1 = 1, x2 = 0 (D)x = 0(E)x = -1Question #2: What is the length of the arc AB in the figure below, if O is the center of the circle and triangle OAB is equilateral? The radius of the circle is 9(a) π(b) 2 ·π(c) 3 ·π(d) 4 ·π(e) π/2 Question #3: What is the probability that someone that throws 2 dice gets a 5 and a 6? Each dice has sides numbered from 1 to 6.(a)1/2 (b)1/6 (c)1/12 (d)1/18 (e)1/36 Question #4: A cyclist bikes from town A to town B and back to town A in 3 hours. He bikes from A to B at a speed of 15 miles/hour while his return speed is 10 miles/hour. What is the distance between the 2 towns?(a)11 miles (b)18 miles (c)15 miles (d)12 miles (e)10 miles Question #5: The volume of a cube-shaped glass C1 of edge a is equal to half the volume of a cylinder-shaped glass C2. The radius of C2 is equal to the edge of C1. What is the height of C2?(a)2·a /π(b)a / π(c)a / (2·π) (d)a / π(e)a + πQuestion #6: How many integers x are there such that 2x < 100, and at the same time the number 2x + 2 is an integer divisible by both 3 and 2?(a)1 (b)2 (c) 3 (d) 4 (e)5Question #7: sin(x)cos(x)(1 + tan2(x)) =(a)tan(x) + 1 (b)cos(x)(c)sin(x) (d)tan(x)(e)sin(x) + cos(x)Question #8: If 5xy = 210, and x and y are positive integers, each of the following could be the value of x + y except:(a)13 (b) 17 (c) 23 (d)15 (e)43Question #9: The average of the integers 24, 6, 12, x and y is 11. What is the value of the sum x + y?(a)11 (b)17 (c)13 (d)15 (e) 9Question #10: The inequality |2x - 1| > 5 must be true in which one of the following cases?I. x < -5 II. x > 7 III. x > 01.Three unit circles are arranged so that each touches the other two. Find the radii ofthe two circles which touch all three.2.Find all real numbers x such that x + 1 = |x + 3| - |x - 1|.3.(1) Given x = (1 + 1/n)n, y = (1 + 1/n)n+1, show that x y = y x.(2) Show that 12 - 22 + 32 - 42 + ... + (-1)n+1n2 = (-1)n+1(1 + 2 + ... + n).4.All coefficients of the polynomial p(x) are non-negative and none exceed p(0). Ifp(x) has degree n, show that the coefficient of x n+1 in p(x)2 is at most p(1)2/2.5.What is the maximum possible value for the sum of the absolute values of thedifferences between each pair of n non-negative real numbers which do not exceed 1?6.AB is a diameter of a circle. X is a point on the circle other than the midpoint of thearc AB. BX meets the tangent at A at P, and AX meets the tangent at B at Q. Show that the line PQ, the tangent at X and the line AB are concurrent.7.Four points on a circle divide it into four arcs. The four midpoints form aquadrilateral. Show that its diagonals are perpendicular.8.Find the smallest positive integer b for which 7 + 7b + 7b2 is a fourth power.9.Show that there are no positive integers m, n such that 4m(m+1) = n(n+1).10.ABCD is a convex quadrilateral with area 1. The lines AD, BC meet at X. Themidpoints of the diagonals AC and BD are Y and Z. Find the area of the triangle XYZ.11.A square has tens digit 7. What is the units digit?12.Find all ordered triples (x, y, z) of real numbers which satisfy the following systemof equations:xy = z - x - yxz = y - x - zyz = x - y - z11 / 1111 / 11。
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Angles, arc lengths, and trig functions
1.
Which of the following is the measure of the above angle in degrees?
A.180∘
B.195∘
C.210∘
D.225∘
Correct answer: B Difficulty level: 2
2.
Which of the following is the measure of the above angle in degrees?
A.120∘
B.135∘
C.150∘
D.210∘
Correct answer: C Difficulty level: 2
radians, what is the value of θ in degrees?
3.If θ=4π
9
A.20∘
B.36∘
C.80∘
D.720∘
Correct answer: C Difficulty level: 2
4.If θ=240o, what is the value of θ in radians?
A.2π
3
B.7π
6
C.4π
3
D.3π
2
Correct answer: C Difficulty level: 2
5.If θ=315o, what is the value of θ in radians?
A.5π
4
B.3π
2
C.5π
3
D.7π
4
Correct answer: D Difficulty level: 2
6.Which of the following is the value of tan(225∘)？
A.−1
B.−√2
2
C.√2
2
D.1
Correct answer: D Difficulty level: 2
7.
In the circle C shown at left, a central angle of 150o intercepts an arc 15πcentimeters (cm) in length. Which of the following best approximates the length of the radius of the circle?
A.**
B.**
5cm
C.18cm
Correct answer: D Difficulty level: 3
8.
Which of the following is the measure of the above angle in radians?
A.5π
3
B.7π
4
C.11π
6
D.23π
12
Correct answer: A Difficulty level: 3
9. A circle has a circumference of 12π feet (ft). An arc, x, in this circle has a central angle of 45∘.
What is the length of x?
ft
A.3π
2
B.3πft
C.270ft
D.540ft
Correct answer: A Difficulty level: 3
)?
10.Which of the following is the value of sin (π
2
A.-1
B.0
C.√2
2
D.1
Correct answer: D Difficulty level: 3
radians, which of the following shows the measure of θ in degrees?
11.If θ=π
2
A.45o
B.90o
C.135o
D.180o
Correct answer: B Difficulty level: 3
12.
In the diagram at left, circle C has a radius of 6 feet (ft). Which of the following best approximates the measure of the arc length s?
A.23 ft
B.46 ft
C.220 ft
D.1320 ft
Correct answer: A Difficulty level: 4
13.
In a circle with center C and radius 4 meters (m), a central angle of x∘ intercepts an arc of 20 m as shown in the diagram at left. Rounded to the nearest degree, which of the following best approximates the value of x?
A.5
B.80
C.286
D.304
Correct answer: C Difficulty level: 4
)?
14.Which of the following is the value of sin(5π
6
A.−√3
2
B.−√2
2
C.1
2
D.150
Correct answer: C Difficulty level: 4
)?
15.Which of the following is the value of sin(π
3
A.1
2
B.√3
3
C.√3
2
D.√3
Correct answer: D Difficulty level: 4
16.
Which of the following is the value of the sine of the angle shown above?
A.−√2
2
B.√2
2
C.45
D.315
Correct answer: A Difficulty level: 4
17.
Which of the following is the value of the cosine of the angle shown above?
A.1
2
B.√3
3
C.√3
2
D.√3
Correct answer: C Difficulty level: 4。