Download trig_show - Gortnor Abbey

2008 SLSS
LEARNING OUTCOMES
 Outline and emphasise best practice in teaching
trigonometry
 Teaching for understanding as basic principle in teaching
trigonometry
 Using ICT in the classroom to enhance understanding
 Using the calculator as a teaching resource
 Provide a summary of trigonometry which can be used in the
classroom for revision of topic
 Assessment: Current strengths and weaknesses in trigonometry
Why Trig ?
Connections between
Trigonometry
and the other parts of the course.
 Coordinate Geometry of the Circle. Q1. Paper 2
Parametric form of a circle
x  4  3cos  ,
y  2  3sin 
 Vectors. Q2. Paper 2
Dot product
 




a . b  | a |  | b | cos  ( where  is the abgle between a and b )
 Coordinate Geometry of the Line. Q3. Paper 2
Slope of a line: Angle between two lines.
Slope = tan ratio :
m1  m2
tan =
1  m1m2
 Complex Numbers. Q3. Paper 1
Polar Form of a complex number
x  yi  r  cos  i sin   .
 Differentiation. Q6 and Q7. Paper 1
The differential of trigonometrical functions
 Integration. Q8. Paper 1
Integral of trigonometrical functions: Using trigonometrical identities.
 2cos4 cos2 d
=  (cos6 + cos2 ) d
 Further Calculus Option. Q8. Paper 1
Maclaurin Series for sin, cos  and tan-1  : Max and Min Modelling:
Integration by parts.
Syllabus
Revise Meaning of Main Facts
and
Theorems from
Junior Cert. Geometry
LEAVING CERT TRIGONOMETRY
SUMMARY OF THE SECTIONS IN L.C. TRIG.
The Unit Circle.
Simple identities with proofs.
Radian Measure ( Length of arc and Area of sector)
Angles on page 9 of tables.
Solving Right Angled Triangles
TRIG
Sin Formula
Cos Formula
Solving Non-Right Angled Triangles
Area of triangle
More Identities from page 9 of tables
Solving Trig equations.
Limits of Trig functions.
Proofs required.
Unit Circle: Page 9
UNIT CIRCLE
(x,y)
1
A
x
y
y  SinA
x  CosA
NOW THE TRIANGLE IS
1
sinA
A
cosA
1
A
CosA
SinA
(cosA,sinA)
1
SinA
(CosA,SinA)
(x,y)
A
CosA
x  CosA
y  SinA
SinA
1
(0,0)
A
CosA
Using Pythagoras Sin2 A  Cos 2 A  1
opp Sin A
Tan A 

adj Cos A
Tan A 
Sin A
Cos A
These are our first two identities on page 9 in tables.
Sin2 A  Cos 2 A  1
Sin2 A  1  Cos 2 A
Cos 2 A  1  Sin2 A
Sin A
Tan A 
Cos A
Sin A
CosA 
Tan A
SinA  TanA .CosA
2
Sin
A
Tan 2 A 
Cos 2 A
1
Sec A  1  Tan A 
Cos 2 A
2
2
1
Ex .1 Show that 1  Tan A 
Cos 2 A
2
LHS = 1  tan 2 A
2
Sin A
= 1
2
Cos A
Cos 2 A  Sin 2 A
=
Cos 2 A
1

 RHS
2
Cos A
Use the CD on this
Ex .2 Show that Cos 4 A  Sin4 A  Cos 2 A  Sin2 A
LHS = Cos 4 A  Sin4 A
= (Cos2 A  Sin2 A)(Cos2 A  Sin2 A)
= (Cos2 A  Sin2 A)(1)
= Cos A  Sin A = RHS
2
2
BASIC GUIDES IN PROVING IDENTITIES.
1. Start with one side (usually the more complicated).
2. Convert unfamiliar terms to sin and cos, if possible.
3. Keep an eye on the target.
4. At each step do what seems sensible.
5. Try another approach if things become too complicated.
Note: Emphasise algebraic procedures.
Ex .3 Show that Sec - Tan .Sin  Cos
LHS = Sec -Tan .Sin
1
Sin

.Sin
Cos Cos
1  Sin2

Cos
Cos 2

Cos
 Cos = RHS
Radian Measure
o
Radius = r
a
Radius = r
o
2 rads
1 rad
b
a
b
Length of arc = r
Measure of the angle in radians =
Length of arc ab =2r
Length of arc
Length of radius
Semi-Circle
Radius = r
Length of semi-circle = r
= (r)
= r units
Measure of the angle in radians =

Length of Arc(semi-circle)
Radius
r
radians
r
 180 =  radians
ANGLES ON PAGE 9 IN TABLES
NOTE :  radians  180 
0
180
90
60
45
30
Link to Java Script.
ANGLES > 900
90o
(-270o )
SIN + ALL +
180o
(-180o )
180o - 

180o + 
360o - 
0o
360o
TAN + COS +
270o
(-90o )
Link to Java Script
UNIT CIRCLE
(0,1)
(1,0)
(-1,0)
(0,-1)
Find the Sin 120
Step 1. Draw a circle with the four quadrants
Step 2. Mark in the 120 line
Step 3. Take 120 from the BASE LINE angle 180  60
 Sin120  Sin60 
As it is in the 2
nd
3
2
quadrant it is a  answer.
120
90o
0o
360o
180o
270o
Other examples : (i) Sin 123 (ii) Cos 300 (iii) Tan 290
Answer s (i) 0.8387
(ii) 0.5
(iii) - 2.7475
Use your calculator to check answers
making sure you understand why you get a
positive or negative answer.
1
Ex .1 If 0  A  360 , find the values of A for which Sin A  2
S
wT
A
Cw
Forget about the sign at first.
1
If Sin A 
2
 A  30
As Sin is Negative the angle can only be in the 3rd or 4 th quadrant.
Add 30 to 180 and subtract 30 from 360
So A  {210 and 330 }
3
Ex .2 If Cos A  , 0  A  360 , find two values of A.
2
wS
A
wT
C
Forget about the sign at first.
If Cos A 
3
2
 A  30 
As Cos is Negative the angle can only be in the 2 nd or 3rd quadrant.
So A  {150 and 210 }
SOLVING RIGHT ANGLED TRIANGLES
a
Ex .1 Find ab in the following triangle
3
c
ab  3 2  4 2
2
ab  9  16  25
2
b
4
ab  25  5
Some common triplets used
5
3
4
3 ,4 , 5
10
6
13
5
8
12
6, 8, 10
5, 12, 13
USE OF SIN , COS, TAN FORMULAE.
A
ADJACENT
900
A
900
ADJACENT
Opp
Sin 
Hyp
Adj
Cos 
Hyp
Opp
Tan 
Adj
8
Ex .1 Find the length of the side x in the diagram.
30
x
x
cos 30 
8
3 x

2 8
 3
x  8 
  4 3
2


Ex .2 Find the size of the angle x in the diagram.
5
tan x 
4
tan x  1.25
x  tan1 1.25
x
4
5
5
Ex.3 If Cos  , find Sin and Tan2 , 0    90
13
Note: If given ratio always draw right angled triangle
x

Adj = 5
Adj 5
Cos 

Hyp 13
By Pythagoras 13 2  x 2  5 2
 x  12 (Note triplet)
Opp 12
Sin 

Hyp 13
2
144
 Opp   12 
2
2 

   
Tan   (Tan )
25
 Adj   5 
2
Ex3. If SinA 
1
, for 0  A  90 Find (i) 2 SinA
2
(i )
1
2SinA  2(
)
2
Multiply above and below by 2
2
2

.
2 2
( ii ) SinA 
(ii) Sin2A
2

2

2 2
2
1
2
From the handy angles (page 9) A  45
Sin 2 A  Sin90  0
 2A  90

2
SOLVING NON RIGHT ANGLED TRIANGLES
We use three things here.
1. Area of triangle.
2. Sine Rule
1
a .b .SinC
2
a
b
c


SinA SinB SinC
3. Cosine Rule a 2  b 2  c 2  2b.c.CosA
B
a
c
A
A  B  C  180
C
b
1. Area of triangle.
1
a .b .SinC
2
It is important to emphasise that in order to use the
area of triangle formula that you need two sides and the
included angle.
Example 1. Find the area of the triangle ,
correct to one decimal place
8cm
37o
10cm
1
Area  8.10.Sin 37
2
 24.1cm 2
Example 2. The area of the triangle shown is 16.2 cm 3
Find the measure of the angle A. 5
A
8
1
Area of   ( one side)(other side)(Sin of Included Angle)  16.2
2
1
Area of  ( 8 )( 5 )SinA  16.2
2
 20 SinA  16.2
16.2
 SinA 
20
 SinA  0.81
 A  54 06'
USE OF SIN AND COS RULE.
MOST IMPORTANT RULE.
You use the COS RULE if given
(a) Three sides
(b) Two sides and the included angle.
You use the SIN RULE in all other cases.
Ex .1 Find the size of the largest angle in the triangle
7
9
8
Given three sides so you have to use the Cos Rule
The largest side is always across from the largest angle.
You always start with the side across from the angle
you are dealing with.
7
X0
8
9
9 2  7 2  8 2  2( 7 )( 8 )Cos x
81  47  64  112Cos x
81  111  112Cos x
81  111  112Cos x
 30  112Cos x
Cos x  .2679
x  7428'
30
112
x  Cos 1 .2679
Cos x 
Ex .2 Find the length of the side x in the following diagram.
9
x
320
10
Given two sides and the included angle Cos Rule.
You always start with the side across from the angle
you are dealing with.
x 2  9 2  10 2  2(9)(10)Cos 32
x 2  81  100  180(.8480 )
x 2  181  152.64
x 2  28.36
x  28.36
x  5.3254
a
Ex 3. Given triangle abc find bc .
120o
5cm.
30o
b
bc
5

Sin120 Sin 30
bc 
5 Sin120
Sin30
Using page 9
bc 
3
2
5
1
2
5
3
2
 21  5
3
2
. 21  5 3
c
a
Ex 4. Given triangle abc find abc .
120o
b
8
5

Sin120 Sinabc
8
5

3 Sinabc
2
5 3
Sinabc 
16
Sinabc  .54127
abc  Sin 1.54127  32.77
8cm
5cm.
c
MORE DIFFICULT PROBLEMS ON SIN AND COS RULES.
Ex .1 A and B are the goalposts at one end of a football pitch and F
is the corner flag. (as shown in the diagram), A, Band F lie in
a straight line with AB  7.4m and BF  28m.
A player is at position P, such that PBF  85 and PFB  25
A
7.4 m B
28 m
F
P
Calculate (i) PF
(ii) PA (iii) APB
7.4 m
A
B
950
28 m
850
F
250
700
P
This is the Master Diagram. Fill in all the angles you know.
Now split up the diagram into the individual triangles.
A
7.4 m
B
B
28 m
850
950
F
A
35.4 m
250
250
700
P
P
F
P
B
28 m
850
F
250
700
P
To find PF use the Sin rule
PF
28

Sin85 Sin70
28 Sin85
PF 
Sin70
PF 
28(.9962 )
.9397
PF  29.68 m
35.4 m
A
F
250
29.69 m
P
To find AP use the Cos rule.
AP  35.4 2  29.69 2  2( 35.4 )( 29.69 )Cos 25
2
AP  1253.16  881.5  2102.1Cos 25
2
AP  2134.5  1905.15
2
AP  229.35  15.14
A
7.4 m
B
950
15.14 m
P
To find APB use the Sin rule.
15.14
7.4

Sin95 SinAPB
7.4 Sin95
SinAPB 
15.14
SinAPB  .4896
APB  Sin1 .4896  29.31
Ex .2. In a triangle abc, ab  7cm , bc  8 cm and the area of the triangle
abc is 14cm 2 . Find two values of the angle abc.
Make a sketch of the resulting triangles
c
8 cm
a
7 cm
b
Area of abc  21 ( 7 )( 8 )Sinabc  14 cm 2
28 Sinabc  14
Sinabc 
A
wS
w
T
C
14
28
Sinabc 
The angle is in the first and second quadrent.
abc  150
abc  30
1
2
Sketch of the two triangles
c
8 cm
300
a
a
7 cm
c
b
1500
7 cm
b
8 cm
THREE D PROBLEMS
Ex .1 pqr are points on level ground. rw is a TV mast which
is held in place by two cables pw and qw. If rw is 5m and
the wpr  wpr is  45 and wqr  120 , show that pq  5 3
w
5m
r
1200
450
p
450
q
Isolate the various triangles NOTE the right angles.
w
5m
w
5m
r
5m
r
450
1200
5m
r
450
p
Using the Cos rule we
q
p
5
Tan45 
pr
5
1
pr
5
Tan45 
qr
5
1
qr
pr  5 m
qr  5 m
q
can now find pq
pq  5 2  5 2  2( 5 )( 5 )Cos120
2
pq  25  25  50( .5 )
2
pq  75
2
pq  75
pq  5 3
2007 Paper 2, Q5 (c).
s
s
h
h
r
r
60
p
s
c
r
60
30
q
h
r
30
p
p
q
c
q
s
s
h
3
p
h
r
h 3
q
c
h
To find |prq|, use the cosine formula
r
r
2
60
30
p
q
tan 60 
h
pr
h
pr
h
| pr |
3
3
tan 30 
h
rq
1
h

3 rq




2
 h 
 h 
c 

h
3

2


 h 3 cos prq
 3
 3
h2
2
c   3h2  2h2 cos prq
3
h2
2
c   3h2  2h2 cos prq
3
3c 2  h2  9h2  6h2 cos prq
2
But given 3c2  13h2
 13h2  h2  9h2  6h2 cos prq
3h2  6h2 cos prq
1
cos prq  
2
prq  120
LENGTH OF ARC OF CIRCLE
Length of arc  r (  is in radians)
Ex .1 Find the length of the arc ab in the diagram below.
o
0
7 30
b
a
METHOD 1
Length of arc  r (  is in radians)
Length of arc  7

6
3.142
Length of arc  7
6
Length of arc  3.665 cm
METHOD 2
Length of circle  2R
We have 30 parts of a full circle
30
Length of arc 
of a circle
360
1
Length of arc  of 2R
12
1
Length of arc  2 7  3.665
12
AREA OF SECTOR OF CIRCLE
1
Area of sector  r 2 (  is in radians)
2
Ex .1 Find the area of the sector oab in the diagram below.
o
0
7 45
METHOD 1
1 2
Area  r  (  is in radians)
2
1 2
Area  (7)
2
4
Area  19.24 sq .units
b
a
METHOD 2
Area of circle   R 2
We have 45 parts of a full circle
45
Area of sector 
of a circle
360
1 2
Area of sector  R
8
1
Area of sector   ( 7 )2  19.24
8
Ex .2 2000 Q.4 (a) Paper 2. The area of a sector of a circle
is 27cm 2 . The radius is 6cm. Find in radians the measure
of the angle in the sector.
1 2
Area  r  (  is in radians)  27
2
1
Area  ( 6 )2   27
2
18  27

6cm
27 3

 radians
18 2
NOTE You can convert radians to degrees as follows.
 radians  180 1 radian 
180

3
180 3
radians 
.  85.94
2
 2
COMPOUND ANGLE FORMULAE
Sin( A  B )  SinACosB  CosASinB
Sin( A  B )  SinACosB  CosASinB
NOTE Sign changes
Cos( A  B )  CosACosB  SinASinB
Cos( A  B )  CosACosB  SinASinB
NOTE Sign changes
TanA  TanB
Tan( A  B ) 
1  TanATanB
TanA  TanB
Tan( A  B ) 
1  TanATanB
NOTE Sign changes
COFUNCTION FORMULAE
Sin(90  A)  CosA
Ex 1.
Ex 2.
1
Sin(90  45)  Cos 45 
2
3
Sin(90  30)  Cos60 
2
Cos(90  A)  SinA
Ex 1.
1
Cos(90  45)  Sin45 
2
Ex 1.
3
Cos(90  30)  Sin60 
2
EXAMPLES ON COMPOUND ANGLE FORMULAE
Ex .1 Find an expression in surd form for Sin75
Sin75  Sin( 45  30 )  Sin45Cos 30  Cos 45 Sin30
1
3 1 1

.

.
2 2
2 2
31

2 2


31 2
.
2 2
2
6 2
4
Ex .2 If Tan(x  45 )  a find Tanx in terms of a.
Tanx  Tan45
Tan( x  45 ) 
a
1  TanxTan45
Tanx  1

a
1  Tanx .1
Tanx  1  a  aTanx
Tanx  aTanx  a  1
Tanx( 1  a )  a  1
Tanx 
a1
1 a
Sin(A  B)
Ex .3 Prove that
 TanA  TanB
CosACosB
Sin(A  B) SinACosB  CosASinB

CosACosB
CosACosB
Divide each term by CosACosB
SinACosB CosASinB

 CosACosB CosACosB
CosACosB
CosACosB
SinA SinB

 CosA CosB
1
 TanA  TanB
DOUBLE ANGLE FORMULAE
Cos 2 A  Cos A  Sin A
2
2
This formula is easy to derive.
Cos 2 A  Cos( A  A )
Tan2 A  Tan( A  A )
 CosACosA  SinASinA
 Cos A  Sin A
2
2
Sin2 A  2 SinACosA
This formula is easy to derive.
Sin2 A  Sin( A  A )
 SinACosA  CosASinA
 2 SinACosA
2TanA
1  Tan2 A
This formula is easy to derive.
Tan2 A 

TanA  TanA
1  TanATanA
2TanA

1  Tan2 A
EXAMPLES ON DOUBLE ANGLE FORMULAE
Ex .1 Express Sin3A in terms of SinA
Sin3 A  Sin( 2 A  A )
 Sin2 ACosA  Cos 2 A .SinA
 2 SinACosACosA  ( 1  Sin2 A )( SinA )
 2 SinA( Cos 2 A )  SinA  2 Sin3 A
 2 SinA( 1  Sin2 A )  SinA  2 Sin2 A
 2 SinA  2 Sin3 A  SinA  2 Sin3 A
 3 SinA  4 Sin3 A
3
, wher 0  A  90 , find the value of
5
(i) Sin2A
(ii) Cos2A
Ex 2 If SinA 
From the right angled triangle.
5
3
A
4
CosA 
5
4
( i ) Sin2A  2SinACosA
3 4
 2. .
5 5
24

25
( ii ) Cos2A  Cos 2 A  Sin2 A
2
 4  3
    
5 5
16 9


25 25
7

25
2
DOUBLE ANGLE FORMULAE IN TAN
1  Tan2 A
Cos 2 A 
1  Tan2 A
Pr oof of this
Cos 2 A  Cos 2 A  Sin2 A
 Cos 2 A  Cos 2 ATan2 A
 Cos 2 A( 1  Tan2 A )

1
2
(
1

Tan
A)
2
Sec A
1  Tan2 A

1  Tan2 A
2TanA
Sin2 A 
1  Tan2 A
Pr oof of this
Sin2 A  2 SinACosA
2 SinA

.Cos 2 A
CosA
 2TanA .
1
Sec 2 A
2TanA

1  Tan2 A
HALF ANGLE FORMULAE
1
Cos 2 A  ( 1  Cos 2 A )
2
This formula is easy to derive.
1
Sin A  ( 1  Cos 2 A )
2
2
This formula is easy to derive.
Cos 2 A  Cos 2 A  Sin2 A
Cos 2 A  Cos 2 A  Sin2 A
Cos 2 A  Cos 2 A  ( 1  Cos 2 A )
Cos 2 A  ( 1  Sin2 A )  Sin2 A
Cos 2 A  2Cos 2 A  1
Cos 2 A  1  2 Sin2 A
2Cos 2 A  Cos 2 A  1
2 Sin2 A  1  Cos 2 A
1
Cos A  ( 1  Cos 2 A )
2
1
Sin A  ( 1  Cos 2 A )
2
2
2
CHANGING PRODUCTS INTO SUMS AND DIFFERENCE S
THE PROOFS OF THESE ARE NOT REQUIRED
2CosACosB  Cos( A  B )  Cos( A  B )
2 SinACosB  Sin( A  B )  Sin( A  B )
2 SinASinB  Cos( A  B )  Cos( A  B )
2CosASinB  Sin( A  B )  Sin( A  B )
SUMS AND DIFFERENCE S INTO PRODUCTS
THE PROOFS OF THESE ARE NOT REQUIRED
A B
A B
CosA  CosB  2Cos
Cos
2
2
CosA  CosB  2 Sin
SinA  SinB  2 Sin
A B
A B
Sin
2
2
A B
A B
Cos
2
2
SinA  SinB  2Cos
A B
A B
Sin
2
2
These last two sets of identities are
used mainly for solving Trig Equations
TRIG EQUATIONS
Type 1. Quadratic
Ex .1 Find all the solutions of Cos 2 x  Cosx  Sin2 x  0
in the domain 0  x  360
Cos 2 x  Cosx  Sin2 x  0
Change Sin2 x into 1 - Cos 2 x
Cos 2 x  Cosx  ( 1  Cos 2 x )  0
Cos 2 x  Cosx  1  Cos 2 x  0
2Cos 2 x  Cosx  1  0
Factorise
( 2Cosx  1 )( Cosx  1 )  0
2Cosx  1  0
Cosx  1  0
1
Cosx 
2
Cosx  1
We now have two simple trig equations.
The reference angle is 60
It is a negative answer so it is
in the quadrants shown.
wS
A
wT
C
x  120
x  240
x  0
x  360
SOLUTION ON AUTOGRAPH
4
y
3
2
1
x
–30
30
–1
–2
–3
–4
60
90
120
150
180
210
240
270
300
330
360
390
Ex .2 Find all the solutions of 2Sin2 2 x  Cos 2 x  1
in the domain 0  x  360
2 Sin2 2 x  Cos 2 x  1
Change Sin2 2 x into 1 - Cos 2 2 x
2( 1  Cos 2 2 x )  Cos 2 x  1
2  2Cos 2 2 x  Cos 2 x  1
2Cos 2 2 x  Cos 2 x  1  0
Factorise
( 2Cos 2 x  1 )( Cos 2 x  1 )  0
( 2Cos 2 x  1 )( Cos 2 x  1 )  0
2Cos 2 x  1  0
1
Cos 2 x 
2
The reference angle is 60
It is a negative answer so it is
in the quadrents shown.
wS
A
wT
C
2x  120 + n 360
2x  240 + n 360
n = 0 2x  120°  x = 60°
n = 0 2x  240°  x = 120°
n = 1 2x  480°  x = 240°
n = 1 2x  600°  x = 300°
n = 2 2x  840°  x = 420°
n = 2 2x  960°  x = 480°
Outside the domain 0  x  360
Cos 2 x  1  0
Cos 2 x  1
2x  0 + n 360
n = 0 2x  0°  x = 0°
n = 1 2x  360°  x = 180°
n = 2 2x  720°  x = 360°
n = 3 2x  1080°  x = 540°
Outside the domain 0  x  360
All the values of x are
60, 120, 240,300, 0, 180 and 360
Ex .3 x  0 and x  60 are two solutions of aSin2 2 x  Cos 2 x  b
Find the value of a and the value of b, where a , b  N.
Using these values of a and b, find all the solutions of
the equation where 0  x  360.
aSin2 2 x  Cos 2 x  b  0
At x  0
aSin2 0  Cos0  b  0
a( 0 )  1  b  0
b1
At x  60
aSin2 60  Cos60  b  0
2
 3 1
a
  b0
 2  2
3 1
a  b0
4 2
3a 3
a2
 0
4 2
Now the equation is 2 Sin2 2 x  Cos 2 x  1  0 which can
be solved in the same way as the previous example.
Type 2. Double Angles
Ex .1 Write Cos2x in terms of Sinx. Hence, find all the solutions
of the equation Cos2x - Sinx  1, in the domain 0  x  360.
Cos 2 x  Cos 2 x  Sin2 x
Cos 2 x  ( 1  Sin2 x )  Sin2 x
Cos 2 x  1  2 Sin2 x
Now we have to solve the equation Cos2x - Sinx  1
Replacing Cos2x with 1 - Sin2 x we have.
1  2 Sin2 x  Sinx  1
2 Sin2 x  Sinx  0
Factorise
Sinx( 2 Sinx  1 )  0
Sinx  0
x  0
x  180
x  360
2 Sinx  1  0
1
Sinx 
2
x  210
x  330
Type 3. Sums and difference into products
Ex .1 (i) Express Sin5x  Sin3x as a product of Sin and Cos.
(ii) Find all the solutions to the equation Sin5x  Sin3x  0
in the domain 0  x  180.
(i)
5x  3x
5x  3x
Sin5 x  Sin3 x  2 Sin
Cos
2
2
 2 Sin4 xCosx
( ii )
Sin5x  Sin3x  0
2 Sin4 xCosx  0
Factorise
NOTE : This is already factorised
(2Sin4 x)( xCosx)  0
2 Sin4 x  0
Sin4 x  0
4x  0 + n 360
4x  180 + n 360
n = 0 4x  0°  x = 0°
n = 0 4x  180°  x = 45°
n = 1 4x  360°  x = 90°
n = 1 4x  540°  x = 135°
n = 2 4x  720°  x = 180°
n = 2 4x  900°  x = 225°
n = 3 4x  1080°  x = 270°
Outside the domain 0  x  180
Outside the domain 0  x  180
All the values of x are
0, 90, 180, 45 and 135
Cosx  0
x  90
SOLUTION ON AUTOGRAPH
Limits 1
1. The first thing normally tried in evaluating any limit is to evaluate
f(a)
1
2
3
4
f (a) =
 Lim f ( x) =
k R
k
xa
k  R, k  0


k
kR

k

kR
0
k
0
Limits 2
If f (a) does not give one of the previous answers, we need to try something else.
0 
i.e. if we got , , or  , - (indeterminate forms)
0 
2. Find a common factor and cancel
sin
3. Lim
= 1 Special Limit
x 0

Examples: Evaluate the following limits
2
2
x
 x-6
x  4 x  3 4. Lim
1. Lim
x2
x2
x-2
x3
sin 5
sin 4
5. Lim
2. Lim
 0
 0
2
4
3. Lim cos 4
6. Lim 4 cos 2 9
 0
 0
7. Lim 4 tan 2 2 x
x 0
sin 2 x
8. Lim
 0
3x 2
sin 4 x - sin 2 x
9. Lim
 0
3x
d
Area of  abc  Area of sector abc  Area of  adc
b

r
a
1 2
r sin
2
 sin

r
c

1 2
r
2



sin
sin 
sin 


As   0
1
sin 



Limit

Limit
 0
 0
sin 

sin 
sin 

tan



 1
1 2
r tan
2


tan
sin 

1
cos 

1

1

1
Limits
Link to Excel
Example 1.
Find limit
 0
Limit
 0
sin5θ
sin4θ
5 Sin5 4
4 5  Sin4

5
 1 1
4

5
4
2007 Q4
(cosA  sinA)2  cos 2 A + sin2 A + 2cosAsinA
 1 + 2cosAsinA
 1 + sin2A
6cos 2 x  sinx - 5  0
6(1- sin2 x)  sinx - 5  0
6sin 2 x  sinx - 1  0
(2sinx  1)(3sinx + 1)  0
2sinx  1  0
1
or
sinx 
2
3sinx + 1  0
1
sinx  
3
x  30
x  199
x  150
x  341
(i)
c
What if ?
90o
a

o
ac
cos 
ab
ab cos  ac
c
b
ab  diaameter = 2r
 ac  2rcos
a

o
b
(ii)
c
a
1 2
Area of semicicle =  r
2

o
b
Area of region abc = Area of aoc + Area of sector obc.
1
1
Area of region abc = ao ac sin + (r)(r)(2 ).
2
2
1
Area of region abc = r(2rcos )sin + r 2 .
2
1 2
Area of region abc = r (2cos sin ) + r 2 .
2
1 2
1 2
2
Area of region abc = r sin2α + r α = πr
2
4
1
2
 sin2α + 2α = π ( multiplying each term by 2 )
2
r
2006 Q 4
2006 Q 5
2006 Q 5
2002 Q.4
(a) T he length of an arc of a circle is 10cm. The radius of the circle
is 4cm. The measure of the angle at the centre of the circle is  .
(i) Find  in radians.
(ii) Find  in degrees, correct to the nearest degree.
( b ) (i) Write Cos2x in terns of Sinx.
(ii) Hence find all the solution of the equation
Cos 2x - Sinx  1
in the domain 0  x  360.
( c ) A triangle has side a, b and c.
The angles opposite a, b and c are A, B and C respective ly.
(i) Prove that a 2  b 2  c 2  2 bcCosA .
(ii) Show that c(bCosA - aCosB)  b 2  a 2
c
C
A
a
B
c
(a)
(i) Length of arc  r (  in Radians)
 10  4 r
10
r 
 2.5 radians
4
( ii )  radians  180
180
1 radian 

2.5 radians 
180

2.5  140
(b) Identity
( c ) (i) This is the proof of the Cos rule.
( ii ) To show that c(bCosA - aCosB)  b 2  a 2
2
2
2
a

b

c
As a 2  b 2  c 2  2 bcCosA  CosA 
 2 bc
b2  a2  c 2
2
2
2
As b  a  c  2 acCosB  CosB 
 2 ac
Sub in for CosA and CosB into c(bCosA - aCosB)  b 2  a 2
a2  b2  c 2
b2  a2  c 2
c(b
-a
)  b2  a2
- 2bc
 2 ac
Simplify this and we get b 2  a 2  b 2  a 2
2002 Q.5
Sin7
 0 Sin 2
(b) xyz is a triangle where xy  8cm. and yz  6cm.
(a) Evaluate Limit
Given that the area of the triangle xyz is 12cm 2 , find
(i) the two possible values of xyz
(ii) the two possible values of xz , correct to one decimal place.
(c) A is an obtuse angle such that

 4 3


Sin  A    Sin  A   
6
6
5


(i) Find SinA and TanA.
1
(ii) Given that Tan(A  B)  , find TanB and express you answer
2
p
in the form
where p,q  Z and q  0
q
(a)
Limit
 0
Sin7
Sin2
Sin7
7
7

= Limit
 0
Sin2
2
2
7 7
=

2 2
1
( b ) Area  a .b .SinC
2
1
Area  8.6.Sin  12
2
 24 Sin  12
1
Sin 
2
  30 and   150
z
x
6
8

y
Case 1
Case 2
z
x
x
8
30
z
6
8
130
y
xz  8 2  6 2  2( 8 )( 6 )Cos 30
2
3
 64  36  72
2
 37.646
xz  37.646
xz  6.136
xz  6.1
6
y
xz  8 2  6 2  2( 8 )( 6 )Cos130
2
 3
 64  36  72

 2 
 162.353
xz  162.353
xz  12.741
xz  12.7
( c ) (i)

 4 3


Sin A    Sin A   
6
6
5


Using the formula SinA  SinB


  A  6    A  6 
  A  6    A  6 


Sin A    Sin A    2 Sin
Cos 


6
6
2
2





 A  6  A  6 
 A  6  A  6 
 2 Sin
Cos 


2
2



 2 6 
 2 A
 
 2 Sin Cos    2 Sin ACos  
 2 
6 
2
3 4 3
4
 2 SinA

 SinA 
2
5
5
Drawing a right angled triangle
5
4
A
3
4
 TanA 
3
( ii )
1
Tan(A  B) 
2
TanA  TanB
Tan(A  B) 
1 - TanA.TanB
4
1
3  TanB
 4
2 1 - 3 .TanB
8
4
 2TanB  1  TanB
3
3
4
8
2TanB  TanB  1 
3
3
6TanB  4TanB 3  8

3
3
10TanB  5
1
 TanB  2
Trigonometric Formulae: Proofs Required.
© SLSS 2007