Monday 29 July 2013

summary for July 30 's lesson

TRIGO













1.Definition of trigonometry, identify tan, cot, sin, cos, cosec, sec in a given triangle.



2.Unit circle and "CAST" rule , memorize the symbol and values of trigonometry in different angles.


3.Different trigonometry rules: sine rule, cosine rule, area formula, special relationship  and how to transfer in different  trigo values.


4.Memorize values of special angles.

Thursday 25 July 2013

Trigonometry Graph - Transformation

GRAPH SKETCHING
ANALYSIS OF SINE GRAPHS

in the following activities, post your responses as a comment.

Activity 1:

Define the following terms, where possible identify the context when it is applied.
A. Amplitude
B. Wavelength
C. Frequency
D. Phase
E. Characteristics of the Wave

Activity 2:
Study the following graphs with reference to y = sinx and (1) identify the transformation (s) of the graphs eg. enlargement, shift in x-direction. (2) State whether the new graph is similar to y = sinx. (3) suggest possible the 'transformed' equation. Note that x is in radian.

original
y = sinx


transformation 1



 transformation 2


transformation 3



transformation 4






Wednesday 24 July 2013

TI84 TRIG graphing

Summary of the lesson





Oh.
Here are the conditions for the ambiguous case:

  • The angle A is cute (i.e., A < 90°).
  • The side a is shorter than the side c (i.e., a < c).
  • The side a is longer than the altitude h from angle B, where h = c sin A (i.e., a > h).

Tuesday 23 July 2013

Friday 19 July 2013

19th July Math Lesson Summary

Bearing 
- It is the angle measured from the true north .
- It is measured in the clockwise direction
- It is basically the reference from true north measured in a clockwise angle . 

Application .
- Able to use bearing for navigation . Bearing gives the reference of one object to another , leading to the desired direction one should proceed .
- Map making as we can quickly determine a location from a reference point .

Heuristic problem solving 
- Read and find the keywords to better understand the question .
- Analyze what the question is asking for .
- Apply previous knowledge to the question .
- Draw a model if necessary to simplify the question and understand it better .


Thursday 18 July 2013

Trigonometry (Bearing - Application) (GROUP TASK 3)


ACTIVITY 3

Objective:
To provide students to apply the concept of Trigonometry Bearing using a real world context.
To enable students to apply the procedural skills as well as the Mathematics Problem solving Heuristics.
To create a Collaborative working scenario in an authentic learning environment

Task:
Each group will be assigned a Real World problem.
Your product should have the following:
A) Assumptions made
B) Visual representation of problem
C) Solution (showing explicitly the Heuristics used)
D) Analysis of suitability of solution (why do you think the solution is feasible/logical ?)

Posting:
Submit your e-solution in the Group Page and Label it as 
TRIGONOMETRY - BEARING

GROUP'S PROBLEM SCENARIOS
=================================================================
GROUP 1

=================================================================
GROUP 2

=================================================================
GROUP 3

=================================================================
GROUP 4




Trigonometry (Bearing - Application) (GROUP TASK 1)

ACTIVITY 2

Objective:
To enable students to apply the procedural skills as well as the Mathematics Problem solving Heuristics.

Task:
Study the problem given and analyse the solution attached.
Use the Mathematical Heuristics techniques done in previous lessons as a guide.
Post as comment:
  What do you think is the heuristics used in solving the problem? Justify your answer




Guess this equipment


18 July Math Scribe summary

What we did today:
- Groups presented Cosine and Sine rules, and asked questions

Cos rule:
1. Included angle - The angle connected by two sides. etc A connected by b and c in the diagram below
2. Relation between pythagoras theorem and cosine rule:
c^2 = a^2 + b^2 - 2abcos90
As cosine 90 is 0,
c^2 = a^2 + b^2

- Derivation of area formula: 1/2 ab sinC







- Went through a few questions for trigonometry

Wednesday 17 July 2013

Trigonometry (Bearing - Introduction) (INDIVIDUAL TASK)

Bearing (introduction)

ACTIVITY 1
Objective
To have a better appreciation of the application of trigonometry in the real world by using a collaborative tool (Linoit).

Task 1
Individually review the resources included in the linoit and answer the questions included. Post your findings/responses/resources in the Linoit.





Monday 15 July 2013

Trigonometry Activity 2: SINE and COSINE Rules

The following videos show the derivation of (i) SINE rule and (ii) COSINE rule.
The visual representations are not supported by auditory explanation.

Task
Your task, as a group, is to incorporate the explanation from Mathematical perspective using either Viva Voce technique or clear and concise text explanation.

SINE RULE





COSINE RULE


Sunday 14 July 2013

Heuristic Problem solving in Mathematics (Activity 1)



Problem
Two men facing a tall building notice the angle of elevation to the top of the building to be 30o and 60o respectively. If the height of the building is known to be h =120 m, find the distance (in meters) between the two men.

Heuristic Problem Solving
Stage 1
Question any assumptions
Key in your assumption/s in the linoit individually.





Stage 2
Solve the Problem Individually (my perspective)
Once your assumptions and hypothesis have been clarified, start working on the problem individually. No discussion with anyone for this stage.

Stage 3
Collaboration - seeking multiple perspectives
In the big group of 5-6 students, present and discuss your solution with the group. 
Focus on the assumptions/hypothesis, Process and the validity of the final solution(s).
Post your Group's solution as a New Post with the following criteria:
Post Title: Heuristic Solution (Group number)
Present the following:
- Assumptions Made
- Process (how you solved the problem) - what method/approach you used eg diagrams, on-line resources
- Final Solution

Stage 4
Peer Assessment
Review the solution(s) from other groups. 
To facilitate process, follow the following:
Group 1 to comment on Group 2
Group 2 to comment on Group 3
Group 3 to comment on Group 4
Group 4 to comment on Group 1

Stage 5
Self Assessment
Review the comments made by the other group.
Be prepared to defend or justify your POV.
However, if an error has been committed - learn from it.




Tuesday 9 July 2013

Lesson Summary 10/7/13- Unit Circle

Unit Circle

•Why is the unit circle used in the derivation of Trig ratios?

•What is the underlying concept behind unit circle? What is unit circle?

•How is unit circle linked to ASTC?


Class Q's
π=180˚?
Uses?
How˚ and radian are related?
What are ˚ and radian?
Why use radian mode?
Why does radian mode produce diff values from ˚mode
Radian functions?




Trigonometry : Unit Circle 2

An activity on Unit Circle

ACTIVITY 1:

http://www.mathsisfun.com/algebra/trig-interactive-unit-circle.html


ACTIVITY 2

http://www.sonom.gr/upload/B/trig1/en_trig1.html

Trigonometry - Unit Circle

A lesson on Trigonometric Ratios Using Unit Circle concept:

Please Click here to download lesson.
Unit Circle Lesson Plan - Home - Welcome to...


 powered by GoBookee.net

Monday 8 July 2013

Lesson summary 9th July

Odd numbers
-Trigonometry is the study of triangles using functions like sine, cosine, tangent, cosecant, secant and cotangent
-It was theorised officialy by Hipparchus, a greek mathematician, but it has first been used during the egyptian and babylonian period.
-Used in physical engineering, architecture, navigation, astronomy (To calculate distances and height)
-Also used in urban planning which is evident in rome, with the presence of aqueducts and many grand buildings such as the pantheon, colosseum etc.
-Trigonometry is actually ratios that has been converted into angles

Q: Why do we need to come up with sin, cos and tan when we already know the sum of angles in a triangle and we already have Pythagoras theorem.
A: It is used in metrology, to minimise error

Important notes:
-All museums in England is free
-Some free walking tours are conducted by experts (win-win situation)

Lesson summary

Linear Law 

1) Linear law graph question will not provide you a scale, you must state your own scale on the graph paper

eg
x-axis, 1cm --- 2 units
y-axis, 1cm --- 5 units

2) Accuracy must be within half a small square

3) Plotted points must be smaller than 1 square

4) Plot using a X

5) Best fit line ---> linear regression

6) Label x axis,  y axis and equation of the graph

7) For E math the graph can exceed the last point but it cannot exceed for A math

8) No need to waste time to draw a nice table

TRIGONOMETRY REVIEW

DIAGNOSTIC QUIZ

Complete the following QUIZ within 7 minutes. Record your score as a comment
(index number: SCORE)

Alternative link to Linoit

Saturday 6 July 2013

Semester 2 Update


1. Scheme of Work (Syllabus for Semester 2)

Term 3
Wk 1       (AM)  LINEAR LAW
Wk 2-3    (EM) TRIGONOMETRY
                        - Sine Rule, Cosine Rule, bearings, Angle of Elevation, 3D problems (EM)
Wk 4-6    (AM) TRIGONOMETRY (AM)
Wk 7   (EM) PROPERTIES OF CIRCLES 
Wk 8       (AM) CIRCULAR MEASURE
Wk 9-10  (AM) BINOMIAL THEOREM 
Term 4
Wk 1 Revision
Wk 2 EOY Exam

Self Directed Learning (AM) URVES & CIRCLES

2. Assessment 

Level Test 2 (10%) 
Wk 7-8
format:   1 hour
Marks:    40 marks
Topics
EM
     Coordinate Geometry
     Trigonometry
AM
     Coordinate Geometry
     Trigonometry
     Linear Law

Paper 3 - AM (10%) 
PT2       - EM (10%)

Tuesday 2 July 2013

2 July 2013 Lesson Summary - Wee Chang Han, (20)

Linear Law


Purpose
To convert an equation to linear form, namely Y = mX + c.

Concept

Replacing the value of the y-axis and x-axis with the Y and the X value will result in a linear graph. This would simplify the graph and allows us to discover the relationship between the more variables more quickly and efficiently.


EG: y = ab^x

convert to Y = mX + c format

lg both sides:

lg y = lg a + x lg b

Convert to form Y = mX + c

lg y (Y) = x (X) lg b(m) + (lg a)


Full explanation inclusive of graph and conversion to Y = mX + c.
Mr Johari's Workings on the Whiteboard


Example 3 Linear Law - Owen and Ryan

yx^n = c
y = c / (x^n)
lg y = lg c - lg x^n
lg y = -n lg x + lg c
Y=mX + c
CONVERTING TO LINEAR LAW :
Y = mX + c where
Y = lg y
m = -n
X = lg x
C = lg c

AFTER DRAWING GRAPH where y-axis - lg y and x-axis - lg x :
Value of n : Gradient of graph * -1
Value of c : 10^ y-intercept

Homework : Coordinate Geometry

Submission of Coordinate Geometry:

1 CARISSA LIEW EN HUI
2 DESIREE LEE RU YI
3 LEONG HOI MUN, JEMIMA                Y        Y
4 YAO PEIMING                                               Y
5 ZHONG XINTONG
6 CHEN ZIXIN
7 CHIN WAI KIT
8 FARRELL NAH JUN HAO
9 GAN WAN CHENG ISAAC                   Y        Y
10 LEE KAI EN
11 LIM YONG XIANG JUSTIN
12 MASON SIM                                       Y        Y
13 MIKOWICZ KAELAN THADDEUS
14 NG YUE HAO, SHAUN
15 OWEN ONG CHAU SIONG
16 POON JIA QI
17 RYAN TAN ZHENG NING                            Y
18 SEBASTIAN DENZEL SUPRIYADI
19 TEH HOWE WEE
20 WEE CHANG HAN                              Y       Y

Linear Law worksheet Example 6 By Crystal, Jemima and Justin

a) The linear equation is Y=mX+c

∵ the Y intercept is (0,6)

∴ c=6

Substitute to get m

∵m*7+6=0

∴ m=-6/7

∴ X^2= -6/7 ln y+ b

ln y=-7/6 (x^2-6)

Then we can express y in terms of x

y=e^(-7/6 (x^2-6))

b) Substitute 1 into the equation

ln y=-7/6 (x^2-6)

∴ln (1)= -7/6 (k^2-6)

0=-7/6 (k^2-6)

k1=√6   k2=-√6 (rejected ∵it can't be negative)

Linear Law worksheet Example 7 (By: Mason, Kai En, Shaun)

Firstly, we find the gradient and y-intercept, to be able to substitute in the values once we have made the equation in linear law form, Y = mX+c

y-intercept = 0.7 (as question states straight line cuts vertical axis at 0.70)

gradient = y2-y1/x2-x1 = 0.7-0 / 0-(-0.233) = 3.00 (3sf)

Now, we will make the equation y = 2-px^q into linear form.

px^q = 2-y
Apply lg to both sides
lg(px^q) = lg(2-y)
expand lg(px^q), and bring the power q to the front
lg(p) + qlg(x) = lg(2-y)

Now, we have an equation in the form Y=mX+c
lg(2-y) = qlg(x) + lg(p)
where lg(2-y) is Y
q is m,
lg(x) is X
and lg(p) is c

Hence, we can find the value of q by subbing in the gradient, and the value of lg(p) by subbing in the y-intercept.

q = 3

lg(p) = 0.7
p = 10^0.7
p = 5.01 (3sf)

Hence, p = 5.01, q=3