' ”*‘to'; .3 -'v‘. A--- A; ’1?" 'L't ‘fi -. Ar I 1.11:: ‘: 3}“ A- > v, . ”94‘ "til.- 1 8/! r-u-u— - “fir-f , 4}.) wall .3 u -: a u a, .. 3 ‘, ‘: '5 » e V “'1 .1 .\' :r. v.- V‘ my u“!— ' ‘1; $93. Ar." 1;." 4 ‘ a"; '44:;ng 3w. "5 A -.~ y. 15‘ {$4};- 7 ‘n F“ J . .I'.»\ I -‘~ "vi-1‘33“. ‘ ' n". 9"; ' {“Z' 5‘21, L . . 'u‘} >- - s 3‘. . . ,- 1:; d a :‘(f .‘, H: Q’fn '3 Jr j. 123‘ ‘ r ..u ¢ $.92" 5": a??? fi4¢g§h _ 1. .. 3W . u” £435“ L- llll‘l \\ \llllll‘ll‘lllllll llllll 3 1293 01028 1792 55:: LIBRARY Michigan State Unlverslty This is to certify that the dissertation entitled LEARNING THROUGH NEGOTIATION: AN ANALYSIS OF STUDENT-INITIATED DISCOURSE IN THE COLLEGE MATHEMATICS CLASS presented by Mary Anne Loewe has been accepted towards fulfillment of the requirements for Ph.D. English degree in MS U is an Affirmative Action/Equal Opportunity Institution 012771 PLACE DI RETURN aoxmmaomhmmmmm TO AVOID FINES Mum on or bdoro date duo. DATE DUE DATE DUE DATE DUE . r ‘ at»: ‘ «3.; 9, 5 m l ‘- I " '1‘ . -. _..——— __J|—]| l .T—lF—T—T MSUI-Anmmmwomlylm LEARNING THROUGH NEGOTIATION: AN ANALYSIS OF STUDENT-INITIATED DISCOURSE IN THE COLLEGE MATHEMATICS CLASS BY Mary Anne Loewe A DISSERTATION Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of English 1994 ABSTRACT LEARNING THROUGH NEGOTIATION: AN ANALYSIS OF STUDENT-INITIATED DISCOURSE IN THE COLLEGE MATHEMATICS CLASSROOM BY Mary Anne Loewe The use of international graduate students to teach American students in the university has been the subject of some debate. Lhdversities need the international teaching assistants (ITAs) to teach the rising undergraduate enrollment in freshman courses, while parents complain that their children are not receiving an adequate education. Studies have shown that~ most of these protests center in the mathematics departments, where the students complain of ITAs with poor English and an inability to adequately answer student questions. This study examines the way American students negotiate the content of the course in what is apparently a difficult speech event. It expands upon already existing research in conversational analysis to describe a particular classroom discourse pattern -- the student-initiated interaction sequence -- and provide a possible framework for future analysis of classroom discourse. One hundred twenty three student-initiated sequences were isolated from 18 hours of observation and audiotaping of six American and Chinese first year mathematics teaching assistants. These sequences were analyzed for (a) overall structure and function, and (b) amount and function of their embedded negotiation structures. The appearance of these sequences and their embedded structures was contrasted in the American and international TA classes in order to provide a description of the negotiation strategies used by students in the classes of international TAs. The findings revealed no differences in the American and international TA data in terms of the overall structure of the student-initiated sequences or the number of turns taken to successfully negotiate a resolution to the students' question. Differences did emerge, however, in terms of the functions of the student initiations and the kinds of embedded negotiation structures. The students corrected and disagreed with the international TAs less, made less use of alignment talk such as accounts and conversational repair, and made less repair involving course content in the ITA classes. These results have ramifications for both ITA and freshman orientation programs and raise questions regarding what is considered successful interaction in the college mathematics class. Copyright by MARY ANNE LOEWE 1994 ACKNOWLEDGMENTS As with any dissertation, there are many people who contributed to the success of this project. A special note of appreciation is extended to the members of dissertation committee for their consistent support throughout this process: to the chairperson of the committee, Dr. James Stalker, for his advice and encouragement on this project as well as throughout my graduate program; to Dr. Mary Bresnahan and Dr. Paul Munsell for their helpful critiques and support; to Dr. Michael Lopez for also agreeing to be a part of this process. I am also grateful to Dr. Gerald Ludden, Director of Undergraduate Studies, Barbara Miller, Administrative Assistant, and to the teaching assistants and students in the Department of Mathematics at Michigan State University for their cooperation in this and previous studies. Finally, I would like to thank my husband, Tom, for his patience, especially during the final preparations. TABLE OF CONTENTS LIST OF TABLES ................................................... ix TRANSCRIPTION NOTATION ............................................ X INTRODUCTION ...................................................... 1 CHAPTER I. Students’ Questioning as a Guide to Classroom Processes ............................................... 4 Factors Contributing to Students' Complaints ............ 4 Pedagogical Factors .................................. 4 Cultural Factors ..................................... 8 Linguistic Factors ................................... 9 Pertinent Research .................................. 11 Students' Questions as Guides to Classroom Processes. . . 13 Questions as Guides to Students’ Thought Processes. .14 Questions as Guides to Nature of Student/TA Relationship ........................................ 14 Pertinent Research .................................. 16 II . Theoretical Framework .................................. 19 Conversational Analysis ................................ l9 Question/Answer’SequencerStructure .................. l9 Alignment Talk ...................................... 20 Classroom Interaction .................................. 25 Question/Answer SequenceeStructure .................. 25 Alignment Talk ...................................... 26 III . Methods ................................................ 2 9 Subjects ............................................... 29 Observations ........................................... 30 Analysis ............................................... 33 Tabulations ............................................ 37 vi TABLE OF CONTENTS (CONT’D) IV. Findings ............................................... 40 Structure of Student-Initiated Sequences ............... 42 Demand Ticket ....................................... 42 Presequence ......................................... 45 Resolution Sequence ................................. 46 Frequency and Type of Sequence ......................... 47 Requests-to-do—Problems ............................. 47 Assertions .......................................... 51 Requests ............................................ 54 Phatic Language ..................................... 60 Use of Embedded Sequences .............................. 63 Types of Negotiation ................................ 66 Types of Repair ..................................... 71 Use of Other Alignment Talk ......................... 74 Summary ............................................. 83 V. Discussion of Findings ................................. 86 The Kind of Interaction in Mathematics Classes ......... 86 Structure of Sequences .............................. 86 Function of Sequences ............................... 88 Use of Negotiation” ................................. 88 Comparison of ATA and ITA Classes ...................... 89 Structure of Sequences .............................. 89 Function of Sequences ............................... 90 Use of Negotiation .................................. 92 Strategies Employed by Students ........................ 93 VI . Conclusion ............................................. 96 Ramifications for ITA Training ......................... 97 Areas for Further Study ................................ 99 NOTES ON CHAPTERS ............................................... 101 APPENDIX A: SAMPLE DOCUMENTS FROM FRESHMAN MATH ................ 102 APPENDIX B: ANNOTATED INTERNATIONAL TA BIBLIOGRAPHY ............ 106 vii TABLE OF CONTENTS (CONT’D) APPENDIX C: TA AND STUDENT CONSENT FORMS ....................... 114 APPENDIX D: PROFILES OF PARTICIPATING TEACHING ASSISTANTS ...... 116 APPENDIX E: SAMPLE PAGE OF NOTES ............................... 119 APPENDIX F: LIST OF REPAIR SEQUENCES ........................... 120 TRANSCRIPT ...................................................... 121 BIBLIOGRAPHY .................................................... 162 viii LIST OF TABLES ELBE BEE 1 Sections of Entry Level Math Courses .............. 5 2 Average Attendance per TA Over Three Day Observation ....................................... 41 3 Average Number of RP Sequences .................... 48 4 Average Number of Sequences by Function ........... 50 5 Types of Repair ................................... S9 6 Frequency of Repair ............................... 7O 7 Types of TA—Initiated Repair ...................... 72 8 Types of Student-Initiated Repair ................. 73 9 Use of Alignment Talk ............................. 82 ix TRANSCRIPTION NOTAT ION (...) Pauses are marked with a series of dots. The length of the pause is not strictly relevant to this study, so the measurement has not been included. Long pauses such as those for writing on the board are marked as such. [ ] Utterances that were not audible are marked with square brackets. Attempts to fill in those blanks will have the approximations within the bracket. ( ) Empty parentheses indicate the observer’s interjections or alternate descriptions such as nonverbal behavior. [ Single brackets indicate overlapping speech or interruption. St: Indicates student is speaker. TA: Indicates teaching assistant is speaker. Stzz Use of a subscript indicates that the speaker is not the same student as the one initiating the interaction. Other Transcripts follow generally accepted punctuation rules where possible. A period indicates a fall in intonation with a brief pause as would generally be considered a sentence ending. A.question mark indicates a rise in intonation at utterance end, in generally accepted question intonation. Stutters, repetitions and pause fillers such as uh and ah are represented. However, standard spelling for pronunciation is used throughout. Contractions such as can't and don't are represented in the way the students used them. The complete explanations of problems are not written out if there was no continuing discussion of the problem. Samples of transcribedother potential reasons for the students' negative reactions to international teaching assistants. IBailey (1982) found that students not sharing the 12 major field of study with the ITAs gave them a poorer end of term evaluation than those students who shared a major field with the international TA. Ainsworth (1986) and Sardokie— Mensah (1991) both discuss the ITA/student relationship in terms of understanding and responding to cultural differences in behavior and expectations in the college classroom. Finally, Bresnahan (1993) attempted to draw a relationship between the negative messages American students receive regarding international TAs and the attitudes of these students toward ITAs. She determined, however, that while positive messages regarding foreigners did have an impact on the attitudes of Americans, negative messages did not. Eisenstein (1983) and.Ryan (1983) both.correlated.the negative judgments held by native speakers regarding non-native speaker phonology, syntax, lexicon and intonation to judgments about the non-native speaker's socio-economic class or ethnic group. The other research notwithstanding, most of the concern surrounding international teaching assistants has been focused on the development of programs to help international teaching assistants become more intelligible to American undergraduate students, although a major complaint has been that second language acquisition research has not sufficiently measured the level of oral proficiency that should be attained by the ITA in order to be considered an effective teacher (Ard 1987, 1989). Mellor (1988) offered suggestions on ways for ITAs to help improve their spoken English with lttle or no extra time spent in practice. Stevens (1989) suggested using drama to 13 improve segmental and suprasegmental features of ITA pronunciation. Others are more hi-tech, offering suggestions for the use of video (Axelson 1990) and computer-assisted pronunciation work (Stenson 1992). Discipline-specific research has been conducted in the language used by teachers in chemistry and mathematics (Anderson-Hsieh 1990). Douglas and. Myers (1989, 1990) have researched. the language of chemistry lab teachers, offering an analysis of errors in the language used by chemistry ITAs as well as suggestions for improvement. This study contributes to discourse level research such as the NS-NNS research conducted by Tyler (1988 and 1992), Blum-Kulka (1989), and Gumperz (1977) who consider discourse level features such as the appropriate use of discourse markers, the perception of speech acts such as requests, and the interference of the NS system of conversational inference as factors that contribute to communication problems. Students’ Questions as Guides to Classroom Processes The situation in which the ITAs and students find themselves makes the achievement of quality education difficult. Several questions present themselves: 1. What is the nature of the interaction between the TAs and their students? 2. How does student participation in ITA classes differ from that taking place in classes taught by their American counterparts? 3. What, if any, conclusions can be drawn regarding the strategies the students employ to understand the ITAs’ presentation? 14 This study addresses these concerns by looking at the questions students ask in class. Even though studies suggest that as much as 80% of classroom interaction is initiated by the teacher (Dillon 1990), an analySis of students’ questions offers an insight into the students’ learning processes and the instructor/student relationship> by' giving' information regarding the success of the communication process. Questions as Guides to Students' Thought Processes As guides into the students’ learning processes, questions allow the teacher to determine the point at which students have arrived in their knowledge of the subject matter and to adjust the teaching accordingly -- before formal evaluation affects their grades. As Dillon (1986) notes, any student question involves an underlying set of presuppositions that gave rise to the question (p. 333). The experienced teacher attends to these presuppositions first. If the presuppositions are invalid, revealing some underlying misunderstanding of the material, the specific question goes unanswered in favor of remedying the error in. the presupposition. If the presuppositions are found valid, the question is answered, possibly giving rise to another question.. Through these question/answer sequences, the teacher follows the student’s thought processes. Questions as Guides to the Nature of the TA-Student Relationshi The teacher’s treatment of student questions can affect the teacher/student relationship because of the importance the 15 student places on the negotiation process. How well teachers answer their students' questions may be determined by the level of the teachers’ subject matter knowledge. .According to a study conducted by Carlson (1988), subject matter knowledge affected. teachers’ (a) length of responses, (b) factual content of responses, (c) relating of material to students’ personal lives, (d) use of humor, and (e) use of metatalk. Analyzing student questions effectively for indications of student misconceptions and using that knowledge in teaching is addressed by Dillon (1986), Flannery (1989), and at the college level, Aldridge (1989). In a study conducted by the ITA Program at Michigan State University, the second most cited source of trouble for the students (after the TAs’ English) was the teachers' inability to explain problems adequately (Rittenberg and Wieferich 1988). They indicated that ITAs either discouraged clarification questions, or they simply recopied the problem rather than address the student’s specific concerns. These practices created frustration, supporting the students' negative views of the process as a whole, the ITAs in particular, and possibly their fears of being able to master the material. The number of student questions may signal the level of involvement achieved by the students in class. If students lose the courage to ask questions in class because of inappropriate responses, their level of involvement with the material decreases (Marzano 1989) and the communication process breaks down. Dillon (1986) points out that if 16 students lose confidence in their ability to learn the material in the context of the class, then their desire lessens, the feeling of needing to know -- at least in that situation -- is lessened (p. 336). Students could undertake to master the material, if at all, through other means such as friends or help room, rather than attendance in class. As the students quit participating, perhaps even attending, the teacher-student relationship is damaged. The teaching assistant cannot determine the extent to which the students have mastered the material, and therefore cannot adjust class time accordingly. The communication process has been undermined. Pertinent Research Pertinent classroom interaction research focuses mostly on the teachers’ use of questions and those studies focusing on student interaction tend to be limited to classes in preschool to 12th grade. While, at the college level, emotional and. educational maturity may account for some differences in the interaction patterns, this research in elementary and high school does offer some insight into the functions of student-generated questions, their importance for the students, and their role in curriculum planning. Ralajthy (1984) holds that students can improve their retention of expository materials through self-questioning. Kenzie (1991) describes the development and improvement of study skills of handicapped students through the use of student-generated questions. Gillespie (1991) summarizes 17 other research into student-generated questions in secondary level content area reading classes. Several studies suggest that classroom atmosphere in large part determines whether the students feel comfortable in asking questions in class (Ortiz 1988). Pizzini (1991) suggests that small-group, problem solving formats might allow for more student-generated questions than teacher-directed activities do. At the college level, shyness and gender have been cited as factors. Beins (1988) describes how shy students were encouraged to ask questions in freshman psychology classes by being allowed to write them on paper. Pearson (1990, 1991) found females asked fewer questions in classes taught by males. Finally, it has been suggested that the international status of the TA has an effect on the types and number of questions asked in class. Katchen (1984) studied student- initiated question/answer sequences similar to those appearing in this study. She found that in comparison with American TAs, students asked fewer questions of the international teaching assistants but asked more Yes-No questions of them. Her conclusions imply that the students accommodated to the situation by requiring less interaction from the ITAs in class. Much more work is needed at the college level to determine the type of interaction that takes place between the faculty and students in the various disciplines. This study is a beginning look at the interaction in freshman 18 mathematics. Only after a complete examination is completed will ITA trainers know the discourse standards in these classes and really be able to assist the international teaching assistants in meeting their students' needs and expectations. In summary, as students enter a freshman mathematics course, they appear-to'be entering a novel situation, not only because it departs so much from their high school experience, but because they often face teachers whose language use and classroom expectations differ from their own. The freshman mathematics system in the Department of Mathematics is highly structured.and tightly controlled, leaving little room for the teaching assistants to learn their craft and adding fuel to the complaints the students already had about the international teaching assistants. The analysis of the students’ use of questions and negotiation strategies serves to define, in part, the kind of interaction taking place in mathematics classes. From this it is possible to determine how students negotiate content and meaning in classes where communication is considered to be difficult, i.e. in those taught by international teaching assistants. Chapter 2: Theoretical Framework Conversation Analysis QuestiongAnswer Seggence Structure This study follows along' the line of Sinclair and Coulthard (1975), McHoul (1990) and others who use conversational analysis as a basis for classroom interaction research. The question/answer sequences presented here stem firmrthe basic structural unit of the conversation, termed the adjacency pair. The adjacency pair sets up the turn-taking process by selecting the next speaker and arranging for transference of the floor. It is two utterances long, each one produced by separate successive speakers (Sacks 1972). The first utterance belongs to a group of first pair parts and requires as a response the second utterance, belonging to the group of second pair parts. Examples of adjacency pairs are Greeting-Greeting, Question-Answer, Complaint-Justification or Apology. Tsui (1989) proposes a basic conversational unit that is tripartite, in which the initiator’s second turn may be eliminated. This occurs in interaction where the two participants are especially close or where the initiator does not accept the response but does not wish to continue with a disagreement, as in the following where A does not wish to 19 20 debate politics with B. A: I hope Smith gets in. B: I voted for Jones. A: silence Demand Ticket Nofsinger (1975) expands the adjacency pair structure by postulating a "demand ticket" which can (a) get the attention of the intended partner, (b) give the speaker’s role to the person who uttered it, (c) switch the speaker’s role from one to another, and which (d) obligates the initiator to make some statement and obligates the acceptor to listen (p. 3). This demand ticket is a type of conversational "ticket," such as a comment about the weather or the conversational partner’s name, that Sacks postulated is used to open conversations. It differs somewhat from Sacks’ ticket in that it is coercive in returning the floor to the speaker: (2.1) A: Yuh know something? B: What? C: It’s time for lunch. (Nofsinger p. 2) In the above, person A’s "Yuh know something?" requires the response "What?" from person B, which in turn requires the return of the floor to person A. Alignment Talk Within these sequences other interaction sequences are often inserted, serving to expand upon the initial utterance (Schegloff 1972). For example, (2.2) P1: How do I get to your place? (1) P2: Where are you now? (2) P1: I’m at home. (3) E5: Well, take your road to . . . (4) 21 These embedded or insertion sequences can last a number of turns before an answer is finally given. Linde and Labov (1975) discuss the distribution of a proposition across more than one turn in these sequences. In the above example, the embedded sequence appears in lines 2-3. In order to give an adequate answer, P2 needs clarification before giving the answer. Jefferson (1972) proposes the misapprehension- clarification side sequence as a type of embedded sequence. (2.3) P1: If Percy goes with Nixon, (1) I’d sure like that. (2) P2: Who? (3) P1: Percy. That young fella that uh - (4) his daughter was murdered. (1.0) (5) P2: Oh ye:ah. Yeah. (6) Lines 3—5 contain the embedded sequence in this case. The misapprehension results in the question on line 3 and is clarified in the answer beginning on line 4. Both of the above are examples of alignment talk (Stokes and Hewitt 1976). When conversations do not go as planned, people attempt to determine why and how things have not worked out. They attempt to sort out exactly where in the interaction the misalignment occurred. Repair Conversational repair (under which category side sequences fall) is one type of alignment talk. Repair can be made to the proposition or to the surface structure and can be handled in a number of ways (Schegloff 1972): 22 (2.4) self-initiate/ P1: Turn right . er . (1) self-repair I mean left at the (2) next corner. (3) (2.5) self-initiate/ P1: Turn left . uh . (1) other-repair I mean (2) P2: right? (3) P1: Yea. At the next (4) corner. (5) (2.6) other-initiate/ P1: Turn right at the (1) self—repair next corner (2) P2: Right? You sure? (3) P1: I mean left at the (4) next corner. Yea. (5) (2.7) other-initiate/ P1: Turn right at the (1) other-repair next corner. (2) P2: You mean turn left (3) at the next corner. (4) P1: Yea. That's right. (5) (2.8) third party P1: Turn right at the (1) repair Shoprite. (2) P2: Is that after the (3) Sunoco? (4) P1: What? (5) P3: He wants to know (6) where the Shoprite (7) is. (8) P1: Oh, yea. It's right (9) there after the (10) Sunoco Station. (11) In (2.4), the speaker realizes a misstatement and corrects it. In (2.5), the bearer fills in the correct word after the speaker has initiated the repair. In (2.6) and (2.7), the repair is initiated by the second person in the conversation and carried out either by the original speaker (2.6) or the originator of the repair (2.7). The last is an example of a third party offering assistance to the other two principle participants and was discussed by Schegloff as a type of other-initiate/other-repair. Self-repair was considered to be preferred over other-repair because of the greater frequency 23 of occurrence. Other-initiate repair generally yields self- repair. Accounts and Metatalk In addition to repair, other types of alignment talk pertinent to this study include the use of accounts and metatalk such as formulations and framing devices. Accounts include such structures as¢U M- Ema/mug.) 1 . )1qu M __ _____,".LEL_-..._ .4. ‘7 399a "D .2166" £41.17» ~--—Jg\m¢ (A {IVA-— L [TL‘IL‘ LL U ,‘i.’ ,1. ‘9- Z/‘ ‘7" {J M. _.._...-..—.. ~— 06 ___-_ - .— — ~—— .——_.__.~ ._._. ..___._ - ._ _” ____. qfi —— APPENDIX F LIST OF REPAIR SEQUENCES TA Initiated TA Repaired TA Initiated Student Repaired APPENDIX F List of Repair Sequences Student Initiated 7 Student Repaired ublblbL/JH (DWHHO) .21 Student Initiated .27 TA Repair pppbppppwwwwwwwwwwHHHr—AH H \o 120 mmpwwwwwwwwwmmpt—IHH mmtbobole-‘l-‘HH St: TA: St: TA: 1.2 St: TA: St: TA: St: TA: St: TA: St: TA: TRANSCRIPT INTERACTION SEQUENCES PER TA FATA 1 What’re we supposed to put down for six? On six? It was pi over three plus k pi. Cuz if the...the tangent has a period of uh pi. You put down two pi? No, I didn’t. Yea, cuz I...You got credit for putting down the right answer and then there was four points. The half of it was for getting the right period of pi of tangent so...If there are questions on the [ ] I just wanted to say that uh he’s going kinda quick through this stuff in lecture you know and we’re trying to keep up [writing it down] and not catching a lot of this stuff so I hope I learn a lot in here. (laugh) SERIOUSLY! I hear everything in terms of sine and cosine. You see in this figure this tangent of u plus v formula? I that’ll show up in [ So the object of a lot of this stuff [ Pardon? The object of a lot of this stuff is [finding the easiest thing to do?] Yea, that's all you do. You pretty much have to memorize Well, if you know like two formulas and the sine and cosine of addition ones, you’re all set. Really? Yea. So it should work out. 121 St: TA: St: TA: St: TA: St: TA: St: TA: 1.6 St: TA: St: 122 Is there a couple of different ways to go about doing this? Cuz I didn’t do it that way. Yea. I’m I’m sure there is. What did you do? Convert this in terms of tangent? One over tangent [ Yea, one over tangent x, then I got [ Ok, hang on a second. You have one plus tan of x over one plus tan of x and ah you clear. You combined this, right? Yea, I got the common denominator. One plus tan of x ah excuse me, tan x plus one over tan of x and this is my division line. Yea, it’ll work the same way cuz you get one plus tan of x over one times tan of x over one plus tangent of x and they cancel. Can you just give us the answer for number four? Number four. It’s, uh, cotangent. So, this is number four. Let me just..The cotangent of theta plus two pi equals cotangent of theta and you can use this as...let me just set it up. This is one over tan theta plus two pi. That works out, or have cosine theta plus two pi over sine theta plus two pi. Both of those methods work out. Would the hypotenuse be the square root of two? Yea. Thank you. (completes problem) You don't have to take it any farther than that do you? Cuz [ ] Actually, I do. Because I have to find all answers between zero and two pi. Oh, ok. St: TA: St: TA: St: TA: St: TA: St: TA: 123 How do you know that? Cosine theta equals cosine two minus theta? Ok that comes from [ Why? Hmmm? Why’d you do that? Ok. The...I might as well do both of them. Do you do you know that sine of x equals sine of pi minus x? Yes. Ok. The same the same concept holds when I’m looking at cosine. Write this a little bigger than usual. Well, how come you didn't do it on number five [ ]? Know what I mean? Ok. Uhm. When when I draw my line, ok, I draw my line through to find my points. Now I if I can [find] original cosine original sine curve and I draw my line, you know it only intersects once, so this is really the same as uh one eighty minus pi over two. .And.I still get still ninety degrees the same way. Now in this case, uhm, this is pi, this is two pi. Now I sort of draw some lines in through here, choose some points. NOW“With this point and this point they're the same they’re the same function. Because the distance from here to here is the same distance as from here to here. And all I do, in this case, is I'm just gonna [ ]. If that was sine half I would have I would've done it. Uh. But when...that’s why I went to the original cosine curve the original sine curve. It only intersected once so I was ok. Uhm. You can give decimal representations to these. It’s not really inportant, cuz you you sort of um you just add pi, so it you want to figure out you know you can add one, three point one four to get the answer but it’s not it’s not essential. Ok. Thirteen. St: TA: St: TA: St: 1.10 St: TA: 124 Explain where you get the three pi over two again? Please. Ok. Here’s uhm, I want to know when it’s zero. Ok. It's zero. This is ninety degrees. Ok. Cuz if there’s a zero here so ninety plus two plus one eighty, cuz it's also zero down here, gives me two seventy. But that's represented as three pi over two cuz this is uh this here is pi over two. This is pi, and this angle is pi plus pi over two, so I get three pi over two. And so it’s uhm, it’s easy to think of these in terms of breaking this up into four parts on this graph. This middle is pi, this is pi over two this is three pi over two. Ok. That's general. Now see this I think is a much easier method.of looking at it. Just breaking it up into four cuz you can get sine the same type of a thing cuz it’s zero at pi, the high point is pi over two the low point is at three pi over two, and then it's at two pi so these are the two types of methods. That that will help on this. Rm:me I can see this a lot easier using this type of a of a diagram, but it you look at the unit circle and the [ ] the same. If anything is zero and ones this is this is nice sort of. Ya gotta get used to it depending on in high school you got it one or the other ways, and I did it this way in high school. That’s why I'm very familiar with this. Some people use the other one, so different methods [ ] Can I go on? Where’d you get the five pi over three again? Uhm. This equals two pi minus pi over three and that’s six pi over three. It's always nice to connect this in terms of three. So I have six pi over three. And it’s really straightforward. Oh. Ok. What mode should we be in for our calculator? We're gonna be in degree mode. It’s gonna be mode four. .11 St: TA: St: TA: .12 St: TA: .13 St: TA: 125 Isn’t that the other way around? Yea. B and c? You’re right. Sorry about that. See I want to put a c here. Gamma. I just get confused with gamma. I'm sorry. Uhm. So you don’t want an exact answer? Well, see when you trace, you’re gonna you’re gonna if you’re close enough on tracing, because this is such a small interval, that usually you’re fairly close on stuff. You could find something like one point eight six and maybe the correct answer is one point eight one you know and you're gonna run out of if you spend a lot of time on.a problem like this. On the test you’re going to uh, you’re not gonna really have time to find out precision in here and I’ve been fairly lenient on, I mean, if it’s close, I've been giving credit because if you can get this far, you know how to do it. And the question’s really do you know how to solve this problem is really what (names supervising professor) is looking for. And so with a scale like this in radiens, you should be able to come extremely close uh at just. You know if you want to find out when there is negative and when it’s positive but...I think I think I’d just graph it. Because the times when you can't when you substitute. You have something equal to a number, like equal to a constant, but if you have a this equal x, I mean I know how to substitute to do stuff, I’m going to graph it. So unless it’s uh like something that we know like sine of two x cosine of two x, stuff like that, uh, I’m just not going to spend the time trying to do it algebraically. Because if I have a cosine to the first, anything to the first power means I can usually get a quadratic or a cubic of that term without sines or cosines or other things around and I can't do that here. So I can’t do a nice simple substitution, so I'm going to kind of quickly graph it. Try to get as close as I can. You use radiens mode? Yep. St: TA: St: TA: St: TA: St: TA: 126 Why’d you choose radiens? Because this is, uh, these numbers are more accurate because they’re smaller numbers, and.generally the answer you want...This is going to be periodic and you’re going to have...I’m not sure what the graph looks like. But if I graph these two what’s going to happen is I [ ]. Out here’s the first line curve, the second one, the third one. This answer this sine cosine [ ]. I'm going to have a period I'm gonna have a periodic with pi and so just radiens is a lot simpler. Do you graph, um, the top equation you have or do you graph sine three x equal y? It’s your choice. I mean if [ Which one’d be simpler? Hmm? Which one'd be more simple? This is this is going to tell you where it crosses the axis, and this is going to give you intersecting points. So this one is simpler cuz what's going to happen, uh let me call this f of x. Ok what’s going to happen is that f of x at one point. I mean this is the, uh, the shift x y button...Give me y guys. Ok, None of these is going to be less than zero. The next [ ] is greater than zero and I know I can just take a number between them. So that this is easier cuz I can tell when it's zero. Ok. But if you want to find intersection points, that's ok, but you can’t go wrong finding out when is it positive when it's negative and it’s zero someplace in between. And so, that’s that’s another way. But there’s anything like, I'll just write something up, three sine two x minus x squared equals zero, you have an x squared in there and.you can’t do anything with that just graph it. You know graph either three sine two x equals x squared, you can do that, or go with the positive negative. But, this is actually pretty [ ], so it’s [ ] fairly straightforward. .16 St: TA: St: TA: .17 St: TA: .18 St: TA: St: TA: 127 Wait. When you graph that what’d you how’d you plug it into the calculator? You want to graph this one or this one? The top one. Ok. I'm going to do y equals sine parenthesis three x close parenthesis plus cosine x, and it should work out like that. Like that. Cuz you got to put the [parenthesis around it], or you could do y equals sine three x colon y equals cosine x. Uhm. Let me make one other quick note. We’ve got some time. I have no clue what this one looks like. This is, uh, give an example. Sine three x sine squared of three ijlus cosine x equals zero. You do the same thing. You graph it on on your calculator as y equal parenthesis sine three x, um, hit the x y button, here, ok, um, ok. If I have the square here, put parenthesis around the sine or‘whatever, square it, that’s yea. That's how they [ ] on t 11 e calculator, and so that’s how you treat those. So, that's the simplest way of doing it. [ ] and then everything works out. (pause) Other questions on that? The ten was given? No, I just chose it. I arbitrarily chose it. I didn’t I didn’t know anything, so I just chose a number. Such as ten. Nice simple numbers and so when I cross multiply...(finishes problem). Wouldn't the reason why eight, why it has to be less than eight is cuz if it was longer than eight and had two values it would [could] be on this side, and then if it were longer than that it would be on this side, and you can't have it? So you’re [going into] the definition, whole theory behind the triangle [ I'm just saying, you didn’t know where it came from. Isn’t that why? So I have something like this [ St: TA: St: TA: St: TA: 1.20 St: TA: St: TA: 1.21 St: TA: St: St: 128 Cuz if it were longer than that one side you couldn’t put in two spots. Cuz it would bring it back to the other side of that line [ Here and. I can’t bring it this way. Right. It has, that’s right. You’re swinging your lines but I'm not. I really am not sure how to explain this thoroughly. (looks at student, both nod) I’m lost. What're you trying to figure out? Ok. I’m trying to find out what c is. Why don't you just use law of sines to find angle a then use the law of cosines? Uhn. Yea. That would work. very very easily. The chapter was on law of cosine, so I tried to apply everything to law of cosine first. If you just used the law of sines would it be wrong? Uh. No, no. I didn’t even see it like that, (person’s name). Can you [ Well, I just used the law of sines to find angle a. It just happens that the top just doesn't work anyway. Hang on. It just so happens that alpha equals sine negative one point three four. This can't happen. See, I never did learn this the easy way. What’s 5? S is the, half of the perimeter. This is when you know only three sides. Oh. What number was that? TA: 1 .23 St: TA: St: TA: 129 Uhm. Eight five number nineteen. It’s talking about distances east of north and west of north. There’s a nice diagram of it. That’s important. When we [do the test] will he give us a diagram do you think? Maybe not. Maybe not? Just remember that angle of declination is measured down from the horizontal and angle of inclination is measured up from the horizontal. St: TA: St: TA: St: TA: 2.3 St: TA: St: TA: St: TA: St: TA: 130 MATA 2 This is my question. So, if no one else wants to hear then. Can you just use eight then? Right. Ok. Still, do it exactly. Well, first of all figure out how many feet per rotation. Can we have that is writing? After the exam. How about thirteen? [no answer..TA goes on to next section] I just didn’t know where you got it from. Where I got this thing from? Yea. Pulled it out of the book. So we (writes on board) and that’s what we had to come up with. So on the test if we just wrote down cotangent of theta and then [ If you had a satisfactory explanation, such as simply the cotangent of theta.had period pi plus shifting everything horizontal to left two pi shifts everything down exactly two cycles two period and.you’re right back on top of where you started, something like that, yea. So we have (continues problem) St: TA: St: TA: St: TA: St: TA: St: TA: St: TA: 131 So when we solve these we going to use uhm equations, or you want graphics? Uhm. I’m saying, you could probably do either way. You could, I mean this is absolutely correct way to think of it. Uhm. In the book I think they’re really trying to make you use the identityu Use the equations. Uhnn But this, but this graphic approach helps you think of it, helps you come up with an answer cuz in the other book, I mean of a test, it’s something that might say simplify, but you don’t know what you're supposed to be aiming at. Here we have an idea graphically of what we're supposed to be aiming at. Can you always do that? Yep, you cross multiply. Ok, the the one plus tan of x is equal to secant two over four. It doesn’t apply when everything's squared cuz I [ Y o u mean you want this as secant? I started out that one plus tangent x and I made that secant x. You know how that's one plus tangent x squared equals secant squared x. Right. I tried it like that. I don't know if that's right or not, but I still come up with tangent x. To get secant of x you kind of took the square root. It’d be it would not be one plus tangent of x. What I’m.saying'is that I just assumed since you could do it with squared you could do it with not being squared, so I said one on plus tangent x equals secant x. Well, this is the identity that we have. That's the identity in the book, true know solved true everything. But the secant of x which would be the square root of this number, so this is not an identity. This is not true. St: TA: St: TA: St: TA: St: TA: St: TA: St: TA: St: TA: St: Stz: St: TA: St: TA: 132 Ok. Actually, a good way to do this (completes problem). Do number six? Number six? (does problem) I lost you over one point sine x one plus sine x over sine x. There right there. Ok. This piece right here sine x over sine squared x. There's a sine of x in each of them. How many sine x’s are in this one? One. How many sine of x’s are in this one? How many what’s? Sine x’s. Oh. Sine x. But how'd you Right. But how'd you get rid of the one from the from the one before that? Cuz we have a one plus sine x and we have a one plus sine down here. These two things are multiplied. This whole [quantity] is multiplied. Up. The one above. The one plus sine x minus. Yea that one right there. How’d you get rid of the one plus sine x? Oh. I really, I really didn't get rid of this one. I combined this one and this one. One minus cosine squared. [ 01 1b get the sine. [ And this one came along for the ride. St: TA: St: TA: St: TA: St: TA: St: TA: 2.13 St: TA: St: TA: St: TA: 133 Oh, Ok. Now I see. All right. So I should be a little more explicit? Can you do seventeen? Seventeen. Ah, we’ll do fifteen, which is a fast one. They wanted us to do it by graphics. Oh they did? Ok, in that case, stick it into your calculator. That’s all I had to do it? Yea. Graph the first one sine t minus cosine t over cosine t plus one. That’s your first function colon. Graph y equals tangent of x. All right. [We] did it this way. You’re not going to do forty three? Oh. I need to do forty three? Thank you. Oh that's a good one. Can I erase this one? Forty three. We have sine of the inverse tangent of x. So [ Um. Are you on eight point three? Ha. Yes I know'what I'm doing . Sine squared of x minus one. So how we going to solve this? They have two pi. What they have is actually equivalent to this and uh after class if anyone wants to know how they got that I’ll go over it. We just have like five or so minutes before I have to give you the quiz and uh sixty two to sixty six. See how many of those we can do. St: TA: St: TA: St: TA: St: St: TA: St: TA: St: TA: St: 134 Is that the square root? Oh, sorry. So this equals forty five sine of theta. It looks like you’re adding two pi to me. I’m adding. No I’m just adding pi. Ok. You started from the first point [ ] to the third point. Isn’t there two pi between there? No. This, see here’s the period is pi so from here from here to here is moving down one pi. And this is pi. And that’s going to two pi and that’s zero it's going to pi. So everything moving exactly onejperiod” Exactly one pi. I see what you’re saying. That’s all we have to write is pi k? That’s, yea. Because that's correct. You see why it’s correct? Cuz if we put two in for k we get two pi. We put one in for k would be this one. Zero in for k to get this one. This is just in the first. This is just on the first cycle. Already have a graph over here. Here's pi two pi zero. Keep going. Three pi, four pi, come back this way. That’s where zero, cuz we're solving this thing right here. Ok. How do you know that, that one angle is forty five degrees? That, that’s a good question. Uhm. Just basically because we have to use a little bit of geometry of a baseball diamond that the pitching mound is is, uh, directly (draws) in line (draws) between uh [ second and home TA: St: TA: St: TA: 2.21 St: TA: St: TA: St: TA: St: TA: 135 Yea. Second and home. Thank you. And uh we know this. See this is a square. It's given in the beginning sixty degrees sixty degrees sixty degrees sixty degrees. And so this line bisects it. So it bisects these angles. Yea. I did say that. Thank you for pointing it out. Wouldn't that be a right triangle? That's the thing. We’re not sure. This is a right triangle. It may just be the way I’ve drawn it. I mean forty feet away from home base could be like this. I mean I’m not sure that forty foot mark is dead center. I mean we have enough information using law of cosine. That’s not on there. Not on there? Ok. You could've just took half of eighty nine degrees to get that angle. Again, I’m not sure that thing bisects jig Maybe it does. Maybe it does cut it in half. Uhm. like this one. I, uh, ok. Maybe it does but I don’t remember my geometry well enough to know that this cuts it. It probably does, but we don’t really need it for this problem. Does anybody know for sure does it bisect it? To satisfy our own curiosity? I tried it and came up with the same answer. Ok. Let's let’s do it this way. We [ ] do it your way. So for this problem again then we just use the law of...All right a and b law of cosine c squared equals eighteen squared plus twenty six cosine of one forty one. One hundred forty one degrees. .And now it's just use the calculator to figure out [ Oh. It's wrong. I did it wrong. Oh, ok. And you get forty one point fifty six. Do you think the test’ll have uh proofs of uh [ identities St: TA: St: TA: St: TA: St: TA: St: TA: 136 Yea. Yea. That'll be like the first thing on there. MATA 3 Is it curved? I don’t know the answer to that question” Ah: I believe the highest grade (interrupted by 3.2). What’s the average? Ninety eight. I don't know if this is going to be curved any differently. I do not know that. So, ah, check the adding, ok? Check the adding. Make sure I added yp the points. Look closely at each problem. Make sure you agree with. my grading. I’m not going to be too cooperative changing grades cuz I did spend.a lot of time going over each. problem to see if there was some semblance of understanding. Ok. I know there are certain certain things everyone had problems on. Number three? Can you do that one? Sure. (Answers 3.4 first.) Any questions of that problem? A lot of people messed up on what the angle of depression was. When you give partial credit is there some specific thing that you [ I’ll tell you, [name], when I grade these when I grade these papers, uh yea, he gives me a piece of paper that tells me specifically what partial credit is. Now, I will usually give more partial credit and I’m being totally serious. I give more partial credit than I’m allowed to on paper. That’s because I can see lot of times what you’re trying to do and I know what we’ve covered in class. That’s why you want to come to me if you disagree with the grading of a problem. You know. come to me first and I will look at it and I will show St: TA: St: TA: St: TA: St: TA: 137 you exactly what kind of credit I gave you. Ok. Sometimes I’ll give you a couple extra points if it looks like you’re doing ok. This...All right problem three. Could you do number six please? Ok. The first thing I did was draw a picture. I wanted to see what the sine the sine of x looked like (completes problem). One last stepu Ok. 'You you were asking about partial credit. This is worth something, this is worth something, and then writing down this and this is worth something, and then writing down this and.tw01pi is worth something. This is worth something. And these two are worth somethingu Ok: That’s hOW'he tells me what to do. Then I look at the whole problem and say, well they know what they’re doing and give them a few extra points so take some time to memorize on page five forty eight figure eight point two point four. This triangle and another triangle to memorize. Yea. Do number seven please? All right. When you see the words "within point zero one accuracy" that should tell you you can use your calculator. (completes problem) That’s six point two eight minus one point eight eight. So it'll come out four point four one. Wait. How’d you get the five point two four? How did I get this number? Oh, ok. I'll [explain it] better. This is uh this is two pi times the number of times the radius in inches. So this is going to be sixty two inches. Ok. That’s how big around it is. And then what I did was I said, I should do this.. I said, how many feet is that? That’s in feet? Cuz this is in feet. Yea. That's all I did. Be careful with that kind of stuff. St: TA: St: TA: St: TA: St: TA: St: TA: St: TA: 138 Can you do number nine? Number nine? Ok. You do eight? Nine and eight? Let me make a choice between those two. For number nine, uh, the sine...Here’s how you do it. In the middle one, wouldn’t x be squared, too, then? Because [ No. No. When I write this, actually, all I’m doing is taking...Ok, this bit of notation here means take the cosine of something and raise it to the fourth power. Ok, in other words, ok, I’m taking, all right, I’m taking the cosine of x to the fourth and I’m just rewriting it as this. Cosine t the fourth of x. Ok? [ Uh, ok. I’m not touching the x’s. They're [locked inside the sine and cosine] ok. Then there I just expanded it out using that first rule. Ok. Now... Can you multiply one plus sine x over cosine x by [ ]? Cuz...Oh, it'll cancel out. Oh, yea. It's good that you’re thinking. Uhm. I have this written in my notes. So, I will finish this way. But you’re right. In fact that's what I told the afternoon class. Go home and see if you can find an easier way to do this. Nobody who did it, everybody pretty much did it this way, which surprised me cuz I thought I was [ I did. it a different way. Ok. Good. Wait on it. And if it works, then that's good. Ok, my next step would be. .13 St: TA: .14 St: TA: .15 St: TA: .16 St: TA: .17 St: TA: St: TA: 139 Is that the answer then? Ah. I would, this looks pretty darn correct to me. (laughter) I mean the sine squared of anything plus the cosine squared of anything is equal to one. That's a verification that that identity is true. So, yea. yea that’s an even better way; This is what I'm looking for, the shortest way possible. The first way is never the shortest almost. Ok. Ahm. What was two? Two was the tangent of x. You mean that’s forty five degrees. Uh. Yea, sorry. (changes board) I [ ] where you came up with sixty over four. I can understand.howryou go the fifteen after that. You got pi over three times one fourth. Ok. (pause) It came from here. Watch. Um. I'm just rewriting this. I didn’t start with pi over three. I started with pi over twelve. And that’s, uh, that's pi over three times one fourth, ok? So, pi over three is the sixty. So it’s sixty over four and then that's the fifteen. And I wouldn't've, you know, even thought to use this triangles if I hadn’t looked at the examples. So it's a special type problem that 1H1 is cooked Lu) especially for an exercise. Any other questions? Would it work if yo just put the top sixty minus forty- five where could just do one half minus one over square root of two and just solve that? Instead of doing that long [ Instead of splitting it apart? Yea. I'm sorry. Wait. Co...I’m sorry. St: TA: St: TA: St: TA: St: TA: St: TA: St: TA: St: TA: St: 140 Well, the cosine of sixty is one half. Right? Yea. The cosine of forty—five is one over square root of two. The cosine of fifteen [ Are are you doing this? Are you using like a distributive property? Are you saying that this is equal to this? Right. No. That’s not true. Why not? Because this , um, uh, no it's not even true in this special case. Basically it's just...ok, if I could use distributive then I could say cosine theta plus beta is equal to cosine theta plus cosine beta. Ok. That would also be true [ Thmfis ...we have the other identities. And not I got to I can't do that. I have to memorize this long thing. They’re they’re just not equal that’s all. Um yea. It’s just a rule. What I thought you were going to say, and this is another alternative, you could've also used, uh, I think thirty and forty—five and then you would’ve gotten nice nice angles that you can deal with. I got it, but where’d.you get negative the square root of [ l? I got it. Ok that's that's important to know because I hate it when I disagree with the back of the book but I double and triple checked it. I got positive [ ]. Well, well the thing is is that they say you can can tell by looking at the at the preamble of the problem. Let's say if the sine of x is two thirds we know'what that does for a unit circle. It puts the sine right about here and the angle here and.here. .And then they say x is between pi over two and pi. Ok. Oh. 141 TA: Yea. See what I'm saying? if x is in this quadrant there aint no way you can have a positive cosine. I’m pretty sure that it’s supposed to be minus. Ok. St: Ok. 3.19 St: Can you do thirty? TA: Do which? St: Thirty seven. (several people). TA: Thirty seven. Sure. (pause) Oh yea that is a long one. 3.20 St: I don’t understand how you got...yea that right there. That’s the part I don't understand. How can you [ TA: Wait. I haven’t written this part down yet. St: Yea. [ ] Right over there on the left. How [ TA: Here? St: Yea. Where’d you get that? TA: The expansion. Ok, well let’s see sine of two theta, sine of two x, sine of two of anything is equal to (pause) sine x cosine x. Yea. I'm pretty sure this is correct. Cosine x. St: All right. TA: Sine of x, and then I have x plus y. So I replace this one with.y and this one with.y so I have cosine of this, sine of this. St: Oh, Ok. TA: Sine of this cosine of this. St: Ok. I was doing it differently. TA: That's ok. Grill me. St: I was using the distributive property. [ TA: Oh god. I was wondering. St: TA: St: TA: St: TA: St: TA: St: TA: St: Stz: TA: St: TA: 142 Sine x two theta and then taking the two theta and then expanding instead of doing that. Oh. Ok. Um. Alls I can say is very very loudly, don’t distribute these things. I mean that's part of the part of the pain of trig is you have to literally memorize how to expand this. Ok. Uh, so for instance when you first learned algebra they said that this (pause) was equal to this. And they didn’t tell you why. They just said it is. Well, now they're telling you that this is equal to this and they’re not telling you why. Just telling you it is. Yea. Can you finish? Huh? Can you finish that one? Can I? Yea. Ok. Ah. Sure. This is a (completes problem). Would you start off with cosine two, you know. Well, I take this [ Right now yea I understand that. But would you start off with cosine two theta of sine two theta there? Where you 're at where you have cosine two theta, should it should it say that or...cuz before I think we had sine two theta there first. Yea. It doesn’t matter. I mean, as long as I only have them once. Ok. I mean you got your commutative and and your associative for your commutative you can rewrite it in any order. Just made sure you don't have two of anything. Ok. You know there’s no set way to write it down. Do you still need me to finish it or or was that your question? I I get the next step. After that I have problems. [ Mm Hmm. St: TA: St: TA: .23 St: TA: St: TA: St: TA: St: TA: .24 St: 143 combining things. combining them and getting it to look like the back of the book? I got the next step where you break down the sine of theta or cosine. I get that part. Ok. Let me show you...Yea I was able to get it into uh a form similar to this and then I got bored with the problem. Ok. So cosine two theta... How do you know when you’re when you’ve made it to the last step? When it matches the back of the book. (laughter) Yea. But like on a test. [ Literally. Oh, uh, you mean on a test. Yea. On the test there's no reason to even worry about that because they will always tell you when to stop. In this Oh. problem, well this problem wouldn't be assigned because you have to know the answer before you knowaou’re done. Literally. Otherwise you could do this forever. You could go backwards and rebuild that and say wow look how I've simplified. You know? (laugh) So, so the point is don’t worry about when you’re done. Ok. Uh if they say simplify any expressions involving only sine theta that means get anything that looks like a two theta out of there. Ok. 'Uh. So actually, see what I should say is if this if this problem said simplify this in terms of sine theta cosine theta, I would be done after this step. Literally. I mean that would be the end of the test problem. Because I would have no I would have no cosine two theta sine two theta I would be done. I, uh, I wasn’t done in this case becauseeI didn’t match the book. Ok. Why? TA: .25 St: TA: .26 St: TA: St: TA: 144 Uhm: The simplest answer I can give is because it's just the difinition of the arctangent. For an arctangent we plus in a number and it spits out an angle. Ok. So it’s just the backwards. We started with the angle and took the tangent. It’s just I’m working backwards. Ok? On number fifty nine, how come in the back of the book it says it's the arctangent of fifty over i. I just seem to get tangent of fifty over i. Ahm, because what you end up with is this. Well, let me ask you if this is what you ended up with. Ok, when you write the triangle down, I’m just doing this for the people that have trouble doing the problem, the reason the answer looked like it did was because it said "express theta as a function of 1." So you want theta by itself. That’s the reason. So they just write theta as equal to the arctangent of fifty over 1. Theta is now a function of 1. Plug in different 1's and get theta. Other than that it's just like the previous section. Just real quick. You can use a quadratic on trig functions but you can’t distribute them? That doesn’t make sense. Oh, ok. Let me see if I can explain what’s going on [ I know what you did. I just don’t understand [ Ok. I can I can shOW'you the diffrence. Uhm. When I. . .that’s a good question. There’s an answer forming in the depths of my mind. I thought about that quite a bit because somebody in the afternoon asked me the exact same question. The same thing. Ok. This is just a god-given law of trig. Ok. It's just the way it is. Ahm. If you write them out, if you write the cosine as angle, if you write it like this (pause) basically a plus b is some angle. So I take an angle b from here to here and this is a plus b. It’s the angle a plus b. Not, I haven’t overlapped them I've written a then b. Now if I take the cosine of the a plus b, that’s this number, this length. Now for reasons I don’t know off the top of my head, this number this cosine of a plus b is not equal. Yea it kind of makes sense too because here's angle a and here’s the cosine of the angel a. Ok. Pretty big number; And here's b. IHere I overlapped b. I added them here I’m just going to written what b was from a standard position. Its cosine is this line St: TA: St: TA: St: TA: St: 145 two when I add those cosine together. I don't understand. Isn’t that a? This is this is a and this angle from here to here approximately. Ok don’t let my drawing confuse you. Taking the two [ You said b didn't overlap a. Huh? You said b didn’t overlap a. In this picture in this picture they do not. In this picture I am rotating it. In this picture I’ve very badly drawn an a and a bigger b. Ok. The point is is here’s your cosine a and here’s your cosine b. When you add them together like this you get some huge number bigger than one. By the definition of cosine. If but this can be greater than one so that’s just a reason why it can be greater than one so that's just a reason why it wouldn’t work. Now back to your question. I am using no that that’s an intrinsic. That's an inherent property of the cosine. That I can't do that. What I was doing before was just taking taking that, and notice all I’m doing is I’m saying instead of writing cosine )L I’m going to write the letter c. Ok. Now I’m just I’m going just back to this then, so I’ll substitute back. ‘You see I’m not really using any property of the cosine. I'm just throwing it around. I’m not taking it apart and distributing it and placing it in places. Does that help? That's a good question. Uhm. I'm not sure how to answer more clearly than this. I’m not really using property of the cosine. I'm just writing the cosine as one thing. Ok? Here here’s an example. What you might be thinking of is this (explains a related example on board). See what I'm saying? There’s a distinction between messing with the rules of trig and using them algebraically. Ok. St: TA: St: TA: St: TA: St: TA: St: TA: St: TA: St: TA: St: TA: 146 FITA 4 Um. For my graph on number fifteen, I have two places greater than five hundred and you only put one point down. You mean the part B number fifteen? Uh huh. Number fifteen part A. Most of people got part A. Most of people miss part B What was the range you used? Range? The range of the graph or of this? Yea. The graph? Yea. The graph. The graph [ ] so we will continue goes down adn here we continue goes up. I don’t understand how you used you used ten not twelve. Here? Yea. This thirteen. Oh. Thirteen. How’d you know how to use ten instead of thirteen? Ok. We need twenty six minus two x greater than zero. But we have twenty six greater than [this two x] divided by two divide by two who is thirteen. X less than thirteen. Oh we have to check this condition and this condition. Ok? The domain should satisfy this three condition. And so this [why] we get the solution from zero to ten or we can draw the graph. We can draw graph here (completes graph)- Ok? Any other questions? St: TA: St: TA: Stz: St: TA: St: TA: Stz: TA: St: TA: St: TA: 147 Can you just factor it though? If...You cannot factor it. You can. the formula. Quadratic formula. Two times two, negative one, one [ ] If you don’t know how to factor it, you can use the formula. Yea? Isn’t there an easier way to do it than using u? Pardon me? Can you do it a different way than using u? isn’t there a different way than letting us equal x squared? This is the easiest way. ‘You think there is another easy way? Got to be. Pardon me? It it’s just confusing. Confusing. Ok. Or the other way is [ is u] you can [ ] by this x square over square x square x square. Ok minus fifteen. Then you have to think about...This one is for for one unit. For unit. So this become a quadratic equation. Then you can do the factoring faster. This is to the power four and we don't have any formula. Can't you just use two x squared minus five and x squared plus three? Yea. 'You can do that but, why I show guarantee everybody can solve for u but, if you do two x square minus five, x squared plus three, not everybody can do that. Ok. You are very familiar with the factoring you can do this one. But a lot of you are just beginning. [Haven’t] learn how to do the factoring right, ok? But we will get we will find that u...this is one plus...If you are very good you can do this way. If you are beginner you can follow this way. This guarantee you can solve it. St: TA: St: TA: St: TA: St: TA: St: TA: St: TA: St: TA: 148 I don’t understand what you’re doing over here. Here? Yea. The root of this equation will be this one. Right? Mmm Hmm. Understand this one? Ok. And from there we know the root of complex: It’s not real. Since it's complex, the the means the inside the [discriminamy] is negative. This is the only criterion. IflflL So this one, we require this one. Actually it’s four cubed less than zero. So we can write our equation. Our polynomial has this form. And the only criteria is p squred minus four cubed. Like if I pick up p is one then q is one. It’ll be ok. Ok? P is one now pick up q is two. It'll be ok. It’ll satisfy this [discriminant] is less than zero. So you do that with all the complex factors? Yea. This one here you, uh, do the factoring, you find our [it's true] it’s complex. So there are infinite many solutions. So are you allowed to do it like the first part or no? This two. Number one. Is that right also? I think. Yea this one also in your book. Only give you one solution. Mmm hmm. But there are more ways to do it? Actually there for this one for real there is only one case. For this one there are infinite many. Yea. St: TA: St: TA: St: TA: St: TA: TA: St: TA: St: TA: 4.11 St: TA: St: TA: 149 So if that was like on a test we could put x plus one cubed and get it right? Which one? The first one. Yea. And it’d be right? It’d be right. There’s nothing to it then. Pardon me? This is for more complex. [ Why go through step two when you can just scratch that off right there and be done? Yea. But I just want to explain, uh, since I don’t think your book give you give you this solution: The book give you one of the solutions conditions. The second case, that’s why I explain, and sometimes they will uh tell you the only one real [true] complex then you have to know how to find out the complex one if they give you more condition. Ok? You know the k minus one? Where’d you get the minus one right there? Is it one minus one? K times one» Then thirty seven ok (points and underlines answer on board - student nods and writes something down). You can also take the sixth root, can't you? Pardon me? You can also take the sixth root? YOU mean St: TA: St: TA: St: TA: St: St: TA: St: TA: St2: TA: St: TA: St: 4.14 St: TA: St: TA: 150 [ That’s how they did it in another class and that’s how Yea. You can do that. Comes out the same. Yea. You can do that. X minus one, will be...Yea you can do that. You have answer? Ok? That's good idea. We didn’t have to do that one. Pardon me? 3 This wasn’t on our homework. (louder) We didn't have to do thirty nine. We didn’t have to? [ It only went up to thirty eight. Do I have to do? It wasn't on our homework. No? Ok. Then we go on to something else. five. Twenty five. Twenty five? Twenty five. You do seventeen before that? Ok. What section are we on? Four five page two eighty three. We go to four St: TA: St: TA: St: TA: St: TA: 4.18 St: TA: St: TA: St: TA: St: TA: St: TA: St: 151 Does it matter that you multiply those two together as opposed to like uh two plus i squared [ ]? Here? Yea. I will multiply by two plus i squared and this two together. I feel this will be easier. Oh. All right. Ok? How’s there two complex? (along with a student) Since we have three three zeroes. We have three zeroes. Ok. I got it. Ok? How’d you know when it’ll be a double double root? Double root? Double root. Ok. If this is a double root, they must [ ] triple root. They must what? Ok. If this is double root? Mmm hmm. Ok. This means there two zeroes here. Uh two real here. Then I have another one. Must have three zero. It's impossible. Since in in this case we can factor in its [ ] form but we cannot. And this if you have a [ ] here. Actually the complex happen here. Right. Ok. Here and here. So here is a. Here is...just is conjugate. Ok? Ok. St: TA: St: TA: St: TA: St: TA: St: TA: St: TA: St: TA: St: TA: 152 What? What? Complex zeroes. So we have one integer, two complex non-real. Ok. That’s on problem twenty nine. Couldn’t you, um, couldn’t you just like graph it? Find the one, and then you know there’s going to be three [ ] for the other two. Uh. Yea. You can do that. From the graph you know only one [cross] here. Only one. Real. And since uh the [ ] cross x axis once so you have to you will know one for real, two for complex. Ok. What if it was degree three? Degree three? Yea. If...in this case degree three. Ok. We have this two complex, ok. We have two two complex then the third one will be either real or complex. Do you think it will be complex? No. Should be two. It’s a pair so the other one will be real. So how do you find that you multiply those two together and then and then and then you factor it or [ Yea. factor it. Multiply those two together then what are they? Yes, well. The other way is...You know this way is real. Mmm hmm. If you know this one then you can.uh.have...Since we have no any other information, we assume this c ok. So we have another x-c. Ok. And can be any number any real number. Ok? St: TA: St: TA: St: TA: St: TA: 153 Number seven? Number seven. Ok. Fifteen. Fifteen? Ok. Why’d you use the five times? Five times thirty five? [ Yea. I'm just checking you. Ok. Ok? (laugh) Thank you. It’s five times the initial. St: TA: St: TA: St: TA: 5. St: TA: St: TA: St: TA: St: TA: St: TA: 154 MITA 5 Thirteen? (lists on board) Eighteen? Eighteen? Eighteen. (lists on board) I’ve got a problem way back in eight point two, number 11? Eleven. (lists on board) Ok. Uh Afterward we covered this section we’ll do the other the other thing. Say that again? No solution. Because you see b should be bigger than a. From this graph b should be bigger than a. Right? Right. All right. So we [see] there no solution. You have to check formally, uh, to to do that to do that thing right. Now. How to do it formally. I’m just confused because in the book it says b is less than h which is a sine beta. Then there is no solution. B is less than h. H is [ ] here. Right. But even b is less than h, even b is bigger than h but less than a it still gives you no solution: What in this situation even if b bigger than.h but less than a. Still no solution, right? If b is bigger than h but is is less than a, if b is here but the angle has to be here. This one. The reason is b is less than a not less than h. I St: TA: St: TA: St: TA: St: TA: St: TA: St: TA: 155 guess b is less than a. The reason is b is less than a, not b is less than h. Let’s compute uh h. Can tell you more reason. Let’s computed h yea. H equals a sine one sixteen degree. That's one sixteen degree that eleven on. (reading’ in Zbook) angle, sine of, one is ten, that’s bigger than that's less than ten. Less than b. Less than b. B is bigger than h. That’s why I’m confused. Because it says that b has to be smaller than h for there to be no solution. Oh. No no. Forget about this. The reason is not this. Ok. Yea. Different from the textbook. The criterion says compare b with h but here is different. Is this on the test? Hm? About the test. Is it like going to have story problems on it? Like these things? The previous test you shouldn’t have such a kind of problem on.i:n. The problems are usually [ ] by formula, ok? This one right here? When b is less than a then there’s no x right? B less than a. 'You can tell (reads in book) when beta is acute b less than a you have two solutions. So these things right here only so for acute angles. What was the average? Forty eight. Forty eight? (nods) We had some people did very good, uh, excellent. Ninety six or uh ninety five. So some students did very good. St: TA: St: TA: St: TA: St: TA: St: TA: 156 How did we how’re we supposed to know'we were supposed to put that down. I mean it didn’t really say you know put both solutions down. You compare b with uh, first you compute h. Fine. This is h. You compare b even with h. You compare h you can compute h compare with b. You know h is less than b. Ok. If h is less than b and here b is less than a right. B is less than a. So you have you have two solutions you have two solutions. There is there is a criterion in the textbook. If this condition satisfied we have two solutions. Yes, but it it just said solve the triangle. I mean be more specific. You want two or you just want one. [ Yea, solve the triangle. But but in this problem [ IfIIgotcme answer that was right, would we almost get the next one, the second solution right? Uh. Why take off so many points? Just from the law of sine, right, you compute sine alpha but you have no no reason to say sine alpha a is this. Just this. It it can be this. Right. Just from this law of sine you can’t you can’t say that only this one [ ]. So if from this you only get one answer it mean you lose the other one. The other one. Right. So just from here you get two alphas, and uh yea there is a criterion in the but there is also you can do this. You get two alpha [ ] like this. So there is alpha, there is.also alpha. Anyway this section expecially there are there are criterions to understand how how many solutions you have. So according to the grading policy you give one answer you just get sixty six points. To get side c, uh, c you use alpha I mean a squared plus b squared equals c squared? No. It’s not a right triangle. Instead of right triangle have the formula Pythagorean law. ZUL. Formula right? I think and if and, uh, in the triangle this triangle is not right. So instead you use the law of cosine. Right. That’s the uh uh case of the Pythagorean law. Not a right triangle. It’s the laws of cosine. St: TA: St: TA: St: TA: 5.12 St: TA: St: TA: St: TA: 5.14 St: TA: St: TA: 157 The law of cosine special case of the law of cosine is when alpha is ninety degree so so uh cosine alpha is zero so you get just uh the right hand side. You get a squared. A squared the left left hand side. You get a squared the right hand side. A squared b squared plus c squared right.? Minus something minus something is zero so you get Pythagorean law. It’s a special case of the law of cosine. And next I want some of you to put your solutions. Do five? Pardon? Five? (lists on board) He curving any [of these]? Yea. I guess should be curved. What’s he gonna drop a test? (laugh) What was the high? What? What was the high grade? High score is what six ninety sixg What'd I say? Ninety six. In our section or overall? Pardon? In our section or overall? In our section. [That seem] pretty high. St: TA: St: TA: St: TA: St: TA: St: TA: St: 158 Number eleven. (lists on board). Twenty one. Ok. (lists on board). Thirty seven? Ok. (lists on board) Could we uhm just take r r one times r two and take the absolute value of each one because the absolute value of 2 one [ Right. Right. That's another way to do this to do this is to to compute to compute 2 one absolute value from this formula squared and compute 2 two 2 two absolute value from this similar formula. Right. And multiply that and.you know that theorem one cuz theorem one tells you that to get uh absolute value of z one times 2 two just multiply their their absolute values. Multiply their ablsolute values so that's another method. (writes on board) It’s this formula so you can compute the these two these two absolute values first. On number twenty nine, I got the right answer, but I don't understand why the second one is square root of thirteen and the cosine of what angle equals 2 k pi. Oh. Yea. There are two problems in this group. The first is writing the trig formula for theta between zero and 2 pi. Right. The next is general algebra for every angle. [ ] the first you got angle theta in, uh, zero. Three sixty in this interval. The next [ ] all the possible answers so that uh so uh twenty nine. What would you do if it said find exactly theta? TA: .21 St: TA: .22 St: TA: 5 .23 St: TA: St: TA: 159 No. No this time you can’t find you can’t find exact solution. Why? cuz, uh, there is not a special angle. Right. So you cannot find exact solution. You just have theta over here. You do twenty one? (lists on board) Ok. Now graph the now graph the root. Ok. Now root, uh, so this x this number and you know that z is uh this number two cuz this is one of the cube root of three. Right. So three is just cubed. Ok. So x is this CIS cubed. And you just [ theorem] by [ theorem] two cubed and CIS. Here you just multiply this angle by three» By this exponent multiply this by three. So that’s (pause to write) this is just eight. Right. CIS this angle is pi (writes) Ok. Trig formula of 2. Is the eighth root of fifty the same as the square root of fifty to the one fourth? Yea. That’s the same thing. You can...this this is equal to eight. You just multiply and have many forms. The the inside just the inside is different. Suppose have one half this whole thing is one half the fourth root is supposed one fourth. You can multiply these two things. You get many forms you can multiply. This is, uh, there is probably for this formula right. There's just power four that’s right. But how about if you have square root of fifty to one fourth? Is that the same? Yea. Just the same. Just different forms. You can put a square up here or you can put a squre right here. 6. St: TA: St: TA: St: TA: St: TA: 160 MITA 6 How did you get the one third? Huh? How’d you get the one third? Just you can graph this function because the graph for the function is through this point through this point. Because is hard to for you to decide. Because you gotta find x is equal to one third. Is hard to find but you can use the the graph for this function to find the to find the x intercept. X equals one third. Just graph this function. Wouldn’t it be from negative one to zero instead of from negative infinity? Cuz how would you have negative infinity if its square [ You.mean.negative square here but [I mean] x is very close to two but less than two. Suppose x [ Hmm squared equals to negative three equals to two (writes) two (writes) minus...This is (writes) in this case ok. In this case 2 squared equals to two minus one third. Right. That means x squared minus two equals negative one third. Right. So the reciprocal z squared minus two equals to one negative one third. So is equals to negative three. If this is negative one n, this here is negative one n over :1. There is n or it would be negative infinity, ok? So this interval, because this x squared could be very near close to could be very close to two x squared. Very close to two in the axis. This means here is two x squared would be very close to two very close to two. And the reciprocal. So x squared minus two very close to zero. So the reciprocal one over x squared minus two very, very, uh, is really negative infinity. Any other problem? Just notice that notice this one because 2 squared could be very close to two from right side or left side. If it is from left side that means the reciprocal would be go to positive infinity; Here 2 squared.go to two goes to two left side goes to two then the range would be from from uh left side to to negative infinity. Go zero 9“) to negative infinity. St: TA: St: TA: St: TA: St: TA: St: TA: 161 Are you going to give us an example? Yea. I will give you an example in the exercise. Can you twenty three, too, please? Oh sure. (lists on boand) X power five minus two x squared plus four [ That’s twenty one. Oh: X power four minus three x square minus four x minus two over x ndnus three is greater than equal to two. Right? (completes problem) Now like on twenty one where you got the negative one point seven one, you said to use the [trace key]. 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