TRIGONOMETRY REVIEW

DIAGNOSTIC QUIZ

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

Alternative link to Linoit

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%)

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