1

QUESTION 1 (PICTURE SUMS)

1.1 ( )
In the diagram below are A 2 ; − 2 3 and AR ⊥ OR . Reflex angle
ˆ = β.
ROA

Calculate:

1.1.1 cos ( −β ) (3)

1.1.2 sin ( 30 − β ) (5)

1.1.3 sin 2β (3)

1.2 If 2sinα = 4m; cosα = −m and 0  α  180 , calculate without the use
of a calculator and using a sketch, the value of m. (5)

1.3 If 4tanθ = −3 and θ  (180 ; 360 ) , without the use of a calculator,

(5)
determine the value of 5sinθ − sec 2θ . (IEB ONLY)

1.4 If sin40 = r , determine the following in terms of r :

1.4.1 cos50 (3)

1.4.2 sin80 (3)

[27]
 2

QUESTION 2 (EXPRESSIONS)

Without the use of a calculator, simplify the following:

1 − cos 2 110
2.1 (6)
tan35.sin 2 55

cos10.cos340 − sin190.sin( −20)

2.2 (8)
sin80.cos 20 + cos100.cos70

2.3 ( sin + cos  ) 2 + ( sin − cos  ) 2 (4)

cos 2 45.sin120
2.4 (7)
cos120.sin270

sin(180 − n ) + tan(90 − n ) + cot(360 − n )

2.5 (IEB ONLY) (6)
cos(180 + n )

2.6 cos70.cos10 + cos20.cos80 (5)

[36]
 3

QUESTION 3 (IDENTITIES)

Prove the following identities:

sinβ sinβ 2 tanβ

3.1 + = (5)
1 − sinβ 1 + sinβ cosβ

sin x + cos x
3.2 sin( x + 45) = (4)
2

1 − cos 2α
3.3 = tan α (3)
sin 2α

2 tanθ
3.4 sin2θ = (5)
1 + tan 2 θ

sin2 − cos 
3.5 = cot  (IEB ONLY) (6)
1 − cos 2 − sin 

1 cos a
3.6 − = tan a (5)
cos a 1 + sin a

1 (3)
= sec 2  (IEB ONLY)
3.7
(1 − sinθ )(1 + sinθ )
[31]
 4

QUESTION 4 (TRIGONOMETRIC EQUATIONS)

4.1 Given: cos35 = sin( x + 35)

Determine the general solution of the equation without the use of a
calculator. (5)

4.2 Determine the general solution of the equation: tan3θ.cot 24 − 1 = 0 (5)
(IEB ONLY)

4.3 Determine the general solution of 3cos x − 2sin x + 2sin2x = 6cos 2 x (8)

4.4 Determine the general solution of sin2β = cos (β + 60 ) (6)

4.5 Given: f ( x ) = sin( x + y ) − sin( x − y )

4.5.1 Prove that f ( x ) = 2cos x sin y (2)

4.5.2 Hence, or otherwise, solve sin3x − sin x = 0 for x  [ −180;0] . (6)

4.6 Solve cos y = sin 2 y for y  [0;270] . (5)

4.7 If p sin x = −3 and p cos x = 3, p  0 , determine the value of x for

0  x  360 . (4)

[50]
 5

QUESTION 5 (TRIGONOMETRIC GRAPHS)

5.1 The sketch graph below, shows the curves of the trigonometric
functions f and g for the domain θ  0 ; 360 .

5.1.1 Write down the defining equations of the curves by completing

the following:

(i) f (θ) (2)

(ii) g (θ) (2)

5.1.2 Write down:

(i) the period of g (1)

(ii) the range of f . (1)

5.1.3 Complete:

(i) f ( 90 ) = ... (1)

(ii) g (180 ) = ... (1)

 6

5.1.4 Solve for θ in the interval 0  θ  360 :

(i) f (θ) = 0 (1)

(ii) g ( θ ) = −1 (1)

5.1.5 For which values of θ is f decreasing? (2)

5.1.6 If A ( 21,5 ; 0,73 ) is given, write down the coordinates of B. (2)

5.1.7 For what values of  is f ( θ )  g ( θ ) ? (2)

5.2 The sketch below shows the curves of the two functions f ( x ) = −2 sin x
and g ( x ) = tan x for 0  x  270 .

5.2.1 Write down the coordinates of T, the turning point of the curve (2)
f.

5.2.2 Write down the period of g . (1)

5.2.3 Write down the range of f . (1)

5.2.4 Write down the coordinates of D if C is C (120 ; − 1,73 ) . (2)

5.2.5 Calculate the length of PQ if PQ is parallel to the y − axis and

passes through the point ( 45 ; 0 ) . Give the answer correctly
(3)
up to 2 decimal places.
 7

5.3

Source: ongtecksheng.blogspot.com

A tsunami or tidal wave is a sea wave caused by an earthquake. The

water moves downwards to the sea bottom and then rises back to the
same height as in the beginning. The water mass then rises the same
distance upwards as it has settled downwards. Before a tsunami, the
normal depth of water is 10 meters above the sea floor (the distance
from sea surface to the bottom). The diagram below shows the first
tsunami wave modelled by comparison y = a sin bx + q (ignoring the
scientific impact that the sea wave will have, if it passes below the
ground surface, the x − axis).

The x − axis represents the time in minutes for the formation of the wave
and the y − axis represents the height of the wave.
 8

5.3.1 Determine the values of a, b and q of the functions. (3)

5.3.2 Give the maximum height that the tsunami wave reached
between 0 minutes and 20 minutes. (1)

5.3.3 By first determining the general solution, determine how long the
tsunami wave was below ground ( x − axis) for the time interval (5)
0 minutes to 20 minutes.

5.4

Source: https://webstockreview.net/explore/bedtime-clipart-sleeping/

During deep sleep, one's breathing is very normal. The volume of air in
the lungs follows a cosine-shaped ratio ranging between 1 000 ml and
5 000 ml over a 5-second interval. The graph below represents the
volume of air in a person's lungs versus the time during deep sleep. The
volume of air in a person's lungs is 3 000 ml before deep sleep.

5.4.1 Determine an equation that relates the volume of air in the lungs
to the time in seconds in the format V ( t ) = acosbt + q . (4)

5.4.2 Give the volume of air in the person's lungs during 6 seconds of
deep sleep. (2)
 9

[40]
QUESTION 6 (2D AND 3D TRIGONOMETRY)

6.1 ABCD is a parallelogram with:

ˆ = 120
• BCD
• DB = 14 cm
• BC = 12 cm
• AE ⊥ DB

6.1.1 ˆ , rounded off to one decimal digit.

Calculate the size of BDC (3)

6.1.2 Calculate the area of BCD . (3)

6.1.3 Determine the length of AE. (3)

6.2 In the figure, AB represents a light pole that leans 5 from the vertical
PB (the original position of the light pole), directly in the direction of
Kevin standing at point C.

• B and C are on the same horizontal plane.

• Ĉ = 20
• AB = 10m

Calculate the distance from C to B, rounded off to one decimal digit. (4)
 10

6.3 In the diagram below, PQR is an equilateral triangle with sides x units
long. PQ, QR and PR are extended to A, B and C respectively such that
AQ = RQ, RB = PR and CP = PQ. The triangle thus formed is also
equilateral.

Determine, in terms of x , in the simplest root form:

6.3.1 the area of PQR . (3)

6.3.2 the area of ABC . (6)

 11

6.4 The diagram below shows the boundary lines of a school sports field
QVWR. QV RW and VW ⊥ RW . PQ is a vertical pylon for a floodlight.
ˆ = x and RVW
The angle of elevation of P from R is y . QRV ˆ =y.

6.4.1 ˆ in terms of x and y .

Express RQV (3)

PQ cos ( x − y )
6.4.2 Prove that VR = . (6)
sin y

6.4.3 Calculate VR, without using a calculator, if the height of the pylon
is 25 m, x = 75 and y = 30 . Leave your answer in surd form. (4)

6.4.4 The site supervisor at the school wants to determine the area of
QRV. Help the terrain supervisor by calculating the area of this
triangle to the nearest integer. (5)
 12

6.5 Refer to Diagram 1 below:

A rectangular prism box with SR = 12 cm , RQ = 5 cm and BQ = 6 cm
is given.

The lid of the box, ABCD is now opened at a 30 angle to position
DCMN, as shown in Diagram 2 below.

6.5.1 Write down the length of NC. (1)

6.5.2 Calculate NE, the perpendicular height of P above the base of (3)
the box.

6.5.3 ˆ : sin NDE

Calculate the value of sin DEN ˆ . (3)
 13

6.6 The elevation angle of the Carlton Centre DE, as measured from three
different points on the same horizontal plane A, B and C is α, β and θ
respectively. ABC is a straight line. It is also given that:
1
• tan α = ;
4
1
• tanβ = ;
2
1
• tanθ = ;
3
• AB = 900 m;
• BC = 300 m;
• DE = h and
• DBC ˆ =φ

6.6.1 Show that DC = 3h . (2)

6.6.2 Hence, write down DB and DA in terms of h . (3)

6.6.3 Show that 1 200h cosφ = 90 000 − 5h 2 . (3)

12h 2 − 810 000 (3)

6.6.4 Show that cosφ = .
3 600h

6.6.5 Hence, determine the value of h. (3)

[61]
 14

QUESTION 7 (MORE QUESTIONS)

7.1 In the diagram below A ( − 4;7 ) , OAC ˆ = m.

ˆ = 90 and BOA

Calculate (rounded to ONE decimal digit):

7.1.1 OA (2)

7.1.2 m (2)

7.1.3 the coordinates of C. (4)

1 1
7.2 If tan x = m + ; 90  x  270 and m 2 + 2 = 1 , calculate the
m m
value(s) of x without the use of a calculator. (6)

7.3 Show, without the use of a calculator, that:

2− 2
sin22,5 = (5)
2

7.4 Solve for m and n if m + n  0;180 and m − 2n  0;180, and it is

given further that:

3
cos ( m + n ) = − and cos ( m − 2n ) = 1
2 (5)
 15

7.5 3 B 1
If sin A = and cos = − for B  0;360 and A  90;180 ,
2 2 2
calculated without the use of a calculator:

B 
7.5.1 sin  + 30  (5)
2 

B 
7.5.2 sin 2  + A  (4)
2 
A
7.5.3 2 sin 2 B − 3 cos (3)
2

cos α 1− x 2 (8)
7.6 If x = , prove that = − sin
1 − sin α 1+ x 2

7.7 Express cos 2 θ in terms of cos2θ and then prove that

2+ 3 (6)
cos15 = .
2

7.8 The roots of the quadratic equation 2x 2 − 2 = 3x are given by sin A

and secB respectively. Calculate the values of A and B when it is given
that A and B are angles in the same quadrant for the interval 0;360
(7)
(IEB only)

[57]
 16

MEMO GRADE 12 TRIGONOMETRY

QUESTION 1

1.1.1 OA 2 = OR 2 + AR 2 Pythagoras ✓ 4 units

✓ cosβ
( 2) 2 + ( −2 )
2
 OA = 3 1
✓ (3)
2
 OA = 16
 OA = 4 units
2 1
 cos ( −β ) = cosβ = =
4 2
1.1.2 sin ( 30 − β ) ✓ compound angle
✓ special angles
= sin30 cosβ − cos30 sinβ ✓ ratios
 1   1   3   −2 3  ✓
1 3
+
=    −   
 2   2   2   4  4 4
✓ answer (5)
1 3
= +
4 4
=1
1.1.3 sin2β ✓ double angle
= 2sinβcosβ ✓ ratios
✓ answer (3)
 −2 3   2 
= 2
 4   4 
 
3
=−
2
1.2 2sinα = 4m ✓ sinα = 2m
 sinα = 2m ✓ sketch
✓ Pythagoras
cosα = −m 1
✓ m2 =
5
✓ answer (5)

 ( 2m ) + ( −m ) = 1 Pythagoras
2 2

 4m 2 + m 2 = 1
 5m 2 = 1
1
m2 =
5
1 5
m = = ( p  0)
5 5
 17

1.3 −3 ✓ r =5
tanθ = and θ  (180 ; 360 ) ✓ substitute
4
✓ double angles
✓ simplified
✓ answer (5)
r 2 = 25 Pythagoras
r = 5
5 sinθ − sec 2θ
1
= 5 sinθ −
cos 2θ
1
= 5 sinθ −
1 − 2sin 2 θ
 −3  1
= 5  −
 5   −3 
2
1− 2  
 5 
25
= −3 −
7
46
=−
7
1.4.1 sin40 = r ✓ x = 1+ r 2
✓ co-function
x = 1+ r 2 ✓ answer (3)

cos50
= cos(90 − 40)
= sin 40
=r
1.4.2 sin80 ✓ double angle
= sin2 ( 40 ) ✓ substitute
✓ answer (3)
= 2sin 40 cos 40
= 2(r ) ( 1+ r )
2

= 2r 1 + r 2
[27]
 18

QUESTION 2

2.1
1 − cos 2 110 ✓ sin 2 110
sin35
tan35.sin 2 55 ✓
cos35
sin 2 110
= ✓ cos 2 35
sin35 ✓ sin70
.sin 2 (90 − 35)
cos35 ✓ 2sin35.cos35
sin110 ✓ 2 (6)
=
sin35
.cos 2 35
cos35
sin(180 − 70)
=
sin35.cos35
sin70
=
sin35.cos35
sin2 ( 35 )
=
sin35.cos35
2 sin35.cos35
=
sin35.cos35
=2
2.2 cos10.cos340 − sin190.sin( −20) ✓ cos20
sin80.cos 20 + cos100.cos70 ✓ − sin10
cos10.cos(360 − 20) − sin(180 + 10). − sin(20) ✓ − sin(20)
= ✓ cos10
sin(90 − 10).cos 20 + cos(90 + 10).cos(90 − 20)
✓ − sin10
cos10 cos 20 − sin10 sin20 ✓
= sin20
cos10 cos 20 − sin10 sin20 ✓ compound angles
cos (10 + 20 ) ✓ answer (8)
=
cos (10 + 20 )
cos30
=
cos30
=1
2.3 ( sin + cos  ) 2 + ( sin − cos  ) 2 ✓ simplify
✓ 2sin 2  + 2cos 2 
= sin 2  + 2 sin  cos  + cos 2  + sin 2  − 2 sin  cos  ✓ common factor
+ cos 2  ✓ answer (4)
= 2sin 2  + 2cos 2 
(
= 2 sin 2  + cos 2  )
=2
 19

2.4 cos 2 45.sin120 ✓ cos60

✓ sin60
cos120.sin 270
2
cos 2 45.sin(180 − 60) ✓
= 2
cos(180 − 60).sin 270
3
cos 2 45.sin 60 ✓
= 2
cos 60.sin 270 1
 2  3  ✓
2
  
 2   2  ✓ −1
= 3
 1 ✓ −
 2  ( −1)
(7)
2
 
6
= 4
1
−
2
3
=  −2
2
= −3
2.5 sin(180 − n ) + tan(90 − n ) + cot(360 − n ) ✓ sinn
cos(180 + n ) ✓
sin n
sin(90 − n ) cos n
sin n + − cot n ✓ − cot n
cos(90 − n )
= ✓ − cosn
− cos n ✓ cot n
cos n ✓ − tann (6)
sin n + − cot n
= sin n
− cos n
sin n + cot n − cot n
=
− cos n
sin n
=
− cos n
= − tan n
2.6 cos70.cos10 + cos 20.cos80 ✓ co-functions
= cos ( 90 − 20 ) .cos ( 90 − 80 ) + cos 20.cos80 ✓ co-functions
✓ compound angle
= sin20.sin80 + cos 20.cos80 ✓ negative angle
= cos 20.cos80 + sin20.sin80 ✓ special angle (5)
= cos ( 20 − 80 )
= cos ( −60 )
= cos 60
1
=
2
[36]
 20

QUESTION 3

3.1 sinβ sinβ 2 tanβ ✓ LCD

+ = ✓ simplify
1 − sinβ 1 + sinβ cosβ
sinβ sinβ ✓ 1 − sin 2 β
LHS = +
1 − sinβ 1 + sinβ ✓ cos 2 β
sinβ (1 + sinβ ) + sinβ (1 − sinβ ) ✓ 2 sinβ (5)
 LHS =
(1 + sinβ )(1 − sinβ )
sinβ + sin 2 β + sinβ − sin 2 β
 LHS =
1 − sin 2 β
2 sinβ
 LHS =
cos 2 β
2 tanβ
 LHS =
cosβ
 LHS = RHS
3.2 sin x + cos x ✓ compound angle
sin( x + 45) = ✓ special angles
2
✓ simplify
 LHS = sin( x + 45)
✓ rationalize (4)
 LHS = sin x cos 45 + cos x sin 45
 2  2
 LHS = sin x   + cos x 
 2   2 
   
2
 LHS = (sin x + cos x )
2
2(sin x + cos x )
 LHS =
2
sin x + cos x
RHS =
2
sin x + cos x 2
 RHS = 
2 2
2(sin x + cos x )
 RHS =
2
 LHS = RHS
 21

3.3 1 − cos 2α ✓ double angle

= tanα
sin2α ✓ double angle
1 − cos 2α ✓ simplify (3)
 LHS =
sin2α

 LHS =
(
1 − 1 − 2sin 2 α )
2sin αcos α
2sin 2 α
 LHS =
2sinαcos α
sinα
 LHS =
cos α
 LHS = RHS
3.4 2 tanθ ✓ 2sinθcosθ
sin 2θ =
1 + tan 2 θ sin θ
✓
LHS = sin 2θ cos θ
 LHS = 2 sinθ cosθ sin 2 θ
✓
2 tanθ cos 2 θ
RHS = ✓ identity
1 + tan 2 θ
✓ simplify (5)
2 sinθ
 RHS = cosθ2
sin θ
1+
cos 2 θ
2 sinθ cos 2 θ + sin 2 θ
 RHS = 
cosθ cos 2 θ
2 sinθ cos 2 θ
 RHS = 
cosθ 1
 RHS = 2 sinθ cosθ
 LHS = RHS
3.5 sin2 − cos  ✓ double angle
= cot 
1 − cos 2 − sin  ✓ common factor
✓ double angle
sin2 − cos 
 LHS = ✓ simplify
1 − cos 2 − sin  ✓ common factor
2sin  cos  − cos  ✓ simplify (6)
 LHS =
( )
1 − 1 − 2sin 2  − sin 
cos  ( 2sin  − 1)
 LHS =
1 − 1 + 2sin 2  − sin 
cos  ( 2sin  − 1)
 LHS =
sin  ( 2sin  − 1)
 LHS = cot 
 LHS = RHS
 22

3.6 1 cos a ✓ LCD

− = tan a ✓
cos a 1 + sin a simplify
1 cos a ✓ identity
 LHS = − ✓ common factor
cos a 1 + sin a ✓ simplify (5)
1 + sin a − cos a
2
 LHS =
cos a (1 + sin a )

 LHS =
(
1 + sin a − 1 − sin 2 a )
cos a (1 + sin a )
sin a + sin 2 a
 LHS =
cos a (1 + sin a )
sin a (1 + sin a )
 LHS =
cosa (1 + sin a )
 LHS = tan a
 LHS = RHS
3.7 1 ✓ simplify denominator
= sec 2 θ
(1 − sinθ )(1 + sinθ ) ✓ identity
✓ reciprocal (3)
1
 LHS =
(1 − sinθ )(1 + sinθ )
1
 LHS =
1 − sin 2 θ
1
 LHS =
cos 2 θ
 LHS = sec 2 θ
 LHS = RHS
[31]

QUESTION 4

4.1 cos35 = sin( x + 35) ✓ 55

 sin( x + 35) = cos35 ✓ nZ
GRAPHICAL METHOD:
✓ x = 20
✓ 125
 sin( x + 35) = cos35
✓ 90 (5)
 x + 35 = 55 + n.360 nZ
 x = 20 + n.360
or
 x + 35 = 125 + n.360
 x = 90 + n.360 nZ

OR
 23

REFERENCE ANGLE METHOD:

ref  = 55
I: x + 35 = 55 + n.360 nZ
 x = 20 + n.360
or
II: x + 35 = 125 + n.360 nZ
 x = 90 + n.360
4.2 tan3θ.cot 24 − 1 = 0 ✓ =1
 tan3θ.cot 24 = 1 ✓ tan24
✓ 24 + n.180
1 ✓ nZ
 tan3θ =
cot 24 ✓ 8 + n.60 (5)
 tan3θ = tan24
 3θ = 24 + n.180 nZ
 θ = 8 + n.60
4.3 3cos x − 2sin x + 2sin2 x = 6cos 2 x ✓ double angle
✓ common factor
 −2sin x + 2sin2 x = 6cos 2 x − 3cos x ✓ common factor
 −2sin x + 4 sin x cos x = 3cos x ( 2cos x − 1) ✓ equate
3
 −2sin x (1 − 2cos x ) = 3cos x ( 2cos x − 1) ✓ tan x =
2
 2sin x ( 2cos x − 1) − 3cos x ( 2cos x − 1) = 0 1
✓ cos x =
 ( 2cos x − 1)( 2sin x − 3cos x ) = 0 2
✓✓ solutions (8)
GRAPHICAL METHOD
1
 cos x = or 2sin x = 3cos x
2
3
 x = 60 + n.360 tan x =
2
nZ x = 56,31 + n.180
nZ

OR

REFERENCE ANGLE METHOD:

1
 cos x = or 2sin x = 3cos x
2
3
ref  = 60 tan x =
2
I: x = 60 + n.360 ref  = 56,31
nZ I: x = 56,31 + n.180
IV: x = 300 + n.360 nZ
 24

4.4 sin2β = cos ( β + 60 ) ✓ co-functions

✓ simplify
 sin2β = sin ( 90 − β + 60) ✓ 2β = 30 − β + n.360
 sin2β = sin ( 30 − β ) ✓ β = 10 + n.120
✓ 2β = 150 + β + n.360
GRAPHICAL METHOD: ✓ β = 150 + n.360 (6)
 2β = 30 − β + n.360 nZ
 3β = 30 + n.360
 β = 10 + n.120
or
 2β = 180 − ( 30 − β ) + n.360 nZ
 2β = 150 + β + n.360
 β = 150 + n.360

OR

REFERENCE ANGLE METHOD:

ref  = 30 − β
I: 2β = β + 60 + n.360 nZ
 β = 60 + n.180
or
II: 2β = 180 − ( β + 60 ) + n.360 nZ
 2β = 180 − β − 60 + n.360
 3β = 120 + n.360
 β = 40 + n.120
4.5.1 f ( x ) = sin( x + y ) − sin( x − y ) ✓ compound angle
 f ( x ) = sin x cos y + cos x sin y ✓ compound angle (2)
− (sin x cos y − cos x sin y )
 f ( x ) = 2cos x sin y
 25

4.5.2 HENCE: ✓ sin3x = sin x

sin3 x − sin x = 0 ✓ x = 0 + n.180
 sin ( 2 x + x ) − sin ( 2 x − x ) = 0 ✓ x = 45 + n.90
✓ nZ
 2cos 2 x sin x = 0 ✓ first 2 solutions
 cos 2 x sin x = 0 ✓ last 2 solutions (6)
 cos 2 x = 0 or sin x = 0

GRAPHICAL METHOD:
cos 2 x = 0 or sin x = 0
 2 x = 90 + n.360 nZ
 x = 45 + n.180
or
0 
x =   + n.360 nZ
180
 x  −180; −135; −45;0
OR

REFERENCE ANGLE METHOD:

cos 2 x = 0
ref  = 90
I: 2 x = 90 + n.360 nZ
 x = 45 + n.180
IV: 2 x = 270 + n.360 nZ
 x = 135 + n.180
or
sin x = 0
ref  = 0
I: x = 0 + n.360 nZ
II: x = 180 + n.360 nZ
 x  −180; −135; −45;0

OR
 26

OTHERWISE:
GRAPHIC METHOD:
sin3 x − sin x = 0
 sin3 x = sin x
 3 x = x + n.360 nZ
 2 x = 0 + n.360
 x = 0 + n.180
or
 3 x = 180 − x + n.360 nZ
 4 x = 180 + n.360
 x = 45 + n.90

 x  −180; −135; −45;0

OR

REFERENCE ANGLE METHOD:

sin3 x − sin x = 0
 sin3 x = sin x
ref  = x
I: 3x = x + n.360 nZ
x 2 x = 0 + n.360
 x = 0 + n.360
or
II: 3 x = 180 − x + n.360 nZ
 4 x = 180 + n.360
 x = 45 + n.90

 x  −180; −135; −45;0

 27

4.6 GRAPHICAL METHOD: ✓ co-function

cos y = sin2y ✓ equation 1
 cos y = cos(90 − 2y ) + n.360 ✓ equation 2
✓ first 2 solutions
 y = (90 − 2y ) + n.360 nZ ✓ last 2 solutions (5)
 3 y = 90 + n.360
 y = 30 + n.120
or
 y = −(90 − 2y ) + n.360 nZ
 − y = −90 + n.360
 y = 90 − n.360

 y  30; 90; 150; 270

OR

REFERENCE ANGLE METHOD:

cos y = sin2y
 cos y = cos(90 − 2y ) + n.360
ref  = 90 − 2y
I: y = (90 − 2y ) + n.360 nZ
 3 y = 90 + n.360
 y = 30 + n.120
or
IV: y = 360 − (90 − 2y ) + n.360 nZ
 − y = 360 − 90 + n.360
 y = −270 − n.360

 y  30; 90; 150; 270

 28

4.7 p sin x = −3 ✓ sketch

−3 ✓ Pythagoras
 sin x = ✓ p 2 = 18
p
p cos x = 3 ✓ p = 3 2 (4)
3
 cos x =
p

p 2 = ( 3 ) + ( −3 )
2 2
Pythagoras
 p 2 = 18
p = 3 2 p0
[41]

QUESTION 5

5.1.1 (i) 1 ✓ a=2

a=
2
2 − (−2) = 2 ✓ equation (2)
 f ( θ ) = 2sinθ
(ii) 1 ✓ b=2
a=
2
1 − ( −1) = 1 ✓ equation (2)
360
b= =2
180
 g ( θ ) = cos 2θ
5.1.2 (i) periode = 180 ✓ answer (1)
(ii) −2  y  2 ✓ answer (1)

OR

y   −2;2
5.1.3 (i) f ( 90) = 2 ✓ answer (1)
(ii) g (180 ) = 1 ✓ answer (1)
5.1.4 (i) θ  0;180;360 ✓ answer (1)
(ii) θ  90;270 ✓ answer (1)
 29

5.1.5 f decrease: ✓ 90  x

x  ( 90;270 ) ✓ x  270 (2)

OR

f decrease:
90  x  270
5.1.6 B (158,5;0,73 ) ✓ 158,5
✓ 0,73 (2)
5.1.7 21,5  θ  158,5 ✓ 21,5  θ
✓ θ  158,5 (2)
5.2.1 T ( 90; −2) ✓ 90
✓ −2 (2)
5.2.2 period = 180 ✓ answer (1)
5.2.3 −2  y  2 ✓ answer (1)

OR

y   −2;2
5.2.4 D ( 240;1,73 ) ✓ 40
✓ 1,73 (2)
5.2.5 g (45) = tan ( 45 ) = 1 ✓ 1
f (45) = −2 sin ( 45 ) = − 2 ✓ − 2
✓ 2,41 (3)
( )
PQ = 1− − 2  2,41 units
5.3.1 y = a sin bx + q ✓ q = 10
q = 10 ✓ b = 18
✓ a = −12 (3)
 y = a sin bx + 10
360
b= = 18
20
 y = a sin18 x + 10
1
a = − 22 − ( −2 )  = 12
2
 y = −12sin18 x + 10
5.3.2 max = 22 ✓ answer (1)
5.3.3 y = −12 sin18 x + 10 5
✓ sin18 x =
 0 = −12 sin18 x + 10 6
10 ✓ x = 3,14
 sin18 x = ✓ x = 6,86
12
✓ difference
5
 sin18 x = ✓ answer (5)
6
 30

GRAPHICAL METHOD:
18 x = 56,44 + n.360 nZ
 x = 3,14 + n.20
or
18 x = 123,56 + n.360 nZ

x x = 6,86 + n.20
 Total time = 6,86min − 3,15min
 Total time = 3,71 or  4 min

OR

REFERENCE ANGLE METHOD:

ref  = 56,44
I: 18 x = 56,44 + n.360 nZ
 x = 3,14 + n.20
II: 18 x = 123,56 + n.360 nZ
x  x = 6,86 + n.20
 Total time = 6,86min − 3,15min
 Total time = 3,71 or  4 min
5.4.1 V ( t ) = acosbt + q ✓ 2 000
✓ 28,8
1
a=
2
5 000 − 1 000 = 2 000 ✓ 3 000
✓ equation (4)
360
b= = 28,8
12,5
1
q = 5 000 + 1 000 = 3 000
2
 V ( t ) = 2 000cos ( 28,8t ) + 3 000
5.4.2 V ( 6 ) = 2 000cos ( 28,8 ( 6 ) ) + 3 000 ✓ substitute
✓ answer (2)
 V ( 6 ) = 1 015,77 ml
[40]

QUESTION 6

6.1.1 sinD ˆ sin120 ✓ sine rule

2
= ✓ simplify
12 14
✓ answer (3)
 sinD ˆ = 12sin120
2
14
 D̂ 2 = 47,9
 BDCˆ = 47,9
 31

6.1.2 B̂ 4 = 180 − 120 − 47,9 ✓ B̂ 4 = 12,1

 B̂ 4 = 12,1 ✓ area rule
✓ answer (3)
1
area BCD = .DB.BC.sinBˆ 2
2
1
 area BCD = (14 )(12) sin (12,1 )
2
 area BCD = 12,61 cm 2
6.1.3 area BCD = area BAD ✓ 17,61
1 ✓ substitute
17,61 = (14 )( AE ) ✓ answer (3)
2
 AE = 2,52 cm
6.2 Â = 180 − 20 − 85 ✓ Â = 75
 Â = 75 ✓ sine rule
✓ simplify
BC 10 ✓ answer (4)
=
sin75 sin 20
10 sin75
 BC =
sin 20
 BC  28,2 m
6.3

6.3.1 1 ✓ substitute
area PQR = x.x.sinBˆ 2 ✓ 60
2
1 ✓ answer (3)
 area PQR = x 2 sin ( 60 )
2
3 2
 area PQR = x
4
 32

6.3.2  P̂ = 180 − 60 = 120 ✓ 120

✓ substitute
BC 2 = x 2 + ( 2 x ) − 2 x(2 x )cos120
2
✓ simplify
 BC 2 = 5 x 2 + 4 x 2 cos 60 ✓ answer
✓ area rule
 1 ✓ answer (6)
 BC 2 = 5 x 2 + 4 x 2  
2
 BC 2 = 5 x 2 + 2 x 2
 BC 2 = 7x 2
 BC = 7 x

 area ABC =
1
2
7x ( )(
7 x sin60 )
7 3 7 3x 2
 area ABC = x 2 =
2 2 4
6.4.1 ˆ = 90 − y
VRW ✓ 90 − y
ˆ = VRWˆ ✓ 180 − (90 − y ) − x
QVR corr. s; QV RW
✓ answer (3)
ˆ = 90 − y
 QVR
ˆ = 180 − (90 − y ) − x
 RQV s of 
 RQVˆ = 90 + y − x
6.4.2 In PQR: PQ
✓ tan y =
PQ QR
tan y =
QR PQ
✓ QR =
PQ tan y
 QR =
tan y ✓ sine rule
✓ substitute QR
In QRV:
sin y
VR QR ✓
= cos y
sin ( 90 + y − x ) sin ( 90 − y ) ✓ simplify (6)
QR sin ( 90 − ( x − y ) )
 VR =
sin ( 90 − y )
PQ
cos ( x − y )
tan y
 VR =
cos y
PQcos ( x − y )
 VR =
tan y cos y
PQcos ( x − y )
 VR =
sin y
cos y
cos y
PQcos ( x − y )
 VR =
sin y
 33

6.4.3 25cos ( 75 − 30 ) ✓ substitute

VR = ✓ special angles
sin30
✓ simplify
25 cos 45
 VR = ✓ answer (4)
1
2
 2 1
 VR = 25  
 2  2
 
 2
 VR = 25    2
 2 
 VR = 25 2 units
6.4.4 In PQR: ✓ simplify QR
PQ 25 ✓ area rule
QR = = ✓ 25 2
tan y tan30
✓ sin75
25
 QR = = 25 3 ✓ answer (5)
3
3
In QRV:
1
( )(
area QRV = 25 3 25 2 sin75
2
)
 area QRV = 739 m 2
6.5.1 NC = BC = RQ = 5 cm ✓ answer (1)
6.5.2 NO ✓ trig ratio
sin30 = ✓ NO = 2,5
5
 5 sin30 = NO ✓ answer (3)

 1 5
 NO = 5   =  2,5 cm
2 2
17
 NE = 2,5 + 6 = 8,5 cm of cm
2
 34

6.5.3 ✓ DN = 13 cm
✓ sine rule
✓ answer (3)

DN 2 = ( 5 ) + (12 )
2 2
( ˆ = 90
Pythagoras NCD )
 DN 2 = 169
 DN = 13 cm
In NDE:
ˆ
sinDEN ˆ
sinNDE
=
DN NE
ˆ
sinDEN DN 13 26
 = = =
ˆ
sinNDE NE 8,5 17
 sinDENˆ : sinNDE
ˆ = 26 : 17
6.6.1 In EDC: ✓ substitute
h ✓ answer (2)
tanθ =
DC
1 h
 =
3 DC
 DC = 3h
6.6.2 ED ✓ Method
tan α = ✓ DB = 2h
DA
1 h ✓ DA = 4h (3)
 =
4 DA
 DA = 4h
ED
tanβ =
DB
1 h
 =
2 DB
 DB = 2h
 35

6.6.3 In BDC: ✓ cosine rule

✓ substitute
DC = BD + BC − 2BD.BCcosφ
2 2 2
✓ Simplify (3)
 ( 3h ) = ( 2h ) + ( 300 ) − 2 ( 2h )( 300 ) cos φ
2 2 2

 9h 2 = 4h 2 + 90 000 − 1 200h cos φ

1 200h cosφ = 90 000 − 5h 2
6.6.4 In ABD: ✓ cosine rule
cos(180 − φ) = − cosφ ✓ substitute
✓ simplify (3)
DA 2 = AB 2 + DB 2 − 2AB.DBcos(180 − φ)
 ( 4h ) = ( 900 ) + ( 2h ) + 2 ( 900 )( 2h ) cos φ
2 2 2

16h 2 = 810 000 + 4h 2 + 3 600h cos φ

12h 2 − 810 000
 cosφ =
3 600h
6.6.5  12h 2 − 810 000  ✓ equate
 = 90 000 − 5h
2
1 200h  ✓ simplify
 3 600h  ✓ answer (3)
12h − 810 000
2
 = 90 000 − 5h 2
3
12h 2 − 810 000 = 270 000 − 15h 2
 27h 2 = 1 080 000
 h 2 = 40 000
 h = 200 m
[61]
 36

QUESTION 7

7.1.1 ✓ Pythagoras
OA = ( −4 ) 2 + ( 7 ) 2 Pythagoras
✓ answer (2)
 OA = 65
 OA  8,1 units
7.1.2 7 ✓ trig ratio
tan m = ✓ answer (2)
−4
 m = 180 − 60,2551187
 m = 119,7
7.1.3 OA  8,1 ✓ trig ratio
ˆ = 29,7 ✓ simplify
AOC
✓ OC = 9,3 units
In AOC:
✓ C ( 0; 9,3 ) (4)
8,1
cos 29,7 =
OC
8,1
 OC =
cos 29,7
 OC = 9,3 units
 C ( 0; 9,3 )
7.2 1 ✓ square both sides
tan x = m + ✓ simplify
m
2 ✓ substitute
 1 ✓
 tan x =  m + 
2 simplify
 m ✓ 120
1 ✓ 240 (6)
 tan 2 x = m 2 + 2 +
m2
1
 tan 2 x = m 2 + +2
m2
 tan 2 x = 1 + 2 = 3
 tan x =  3

GRAPHICAL METHOD:
 x = 60 + n.180 nZ
 x  120;240

OR

REFERENCE ANGLE METHOD:

ref  =60
I: x = 60 + n.180 nZ
II: x = 120 + n.180 nZ
x  120;240
 37

7.3 cos 2(22,5) ✓ double angle

= cos 45 ✓ equation
✓ double angle
2 1 ✓ simplify
= of
2 2 ✓ simplify surd (5)
2
 cos 2(22,5) =
2
2
 1 − 2 sin 2 (22,5) =
2
2− 2
 sin 2 (22,5) =
4
2− 2
 sin22,5 =
4
2− 2
 sin22,5 =
2
7.4 GRAPHICAL METHOD: ✓ equation A
3 ✓ equation B
cos ( m + n ) = − ✓ solving
2
simultaneously
 m + n = 150 + n.360 n  Z_ ✓ n = 50
 m + n = 150 --- ( A ) ✓ m = 100 (5)
cos ( m − 2n ) = 1
 m − 2n = 0 + n.360 nZ
 m − 2n = 0
 m = 2n --- (B )
substitute ( A ) in (B ) :
2n + n = 150
 3n = 150
 n = 50
 m = 100

OR
 38

REFERENCE ANGLE METHOD:

3
cos ( m + n ) = −
2
ref  = 30
II: m + n = 150 + n.360 n  Z_
 m + n = 150
III: m + n = 210 + n.360 n  Z_
 m + n = 210

 m + n = 150 (only) --- ( A )

cos ( m − 2n ) = 1
ref  = 0
I: m − 2n = 0 + n.360 nZ
IV: m − 2n = 360 + n.360 n  Z
m − 2n = 360

 m − 2n = 0 (only)
 m = 2n --- (B )
substitute ( A ) in (B ) :
2n + n = 150
 3n = 150
 n = 50
 m = 100
7.5.1 GRAPHICAL METHOD: B
✓ = 120
B 1 2
cos = −
2 2 ✓ B = 240
B ✓ substitute
= 120 + n.360 nZ ✓ simplify
2
 B = 240 + n.720 ✓ special angle (5)
 B = 240
 240 
 sin  + 30 
 2 
= sin (150 )
= sin ( 30 )
1
=
2

OR
 39

REFERENCE ANGLE METHOD:

B 1
cos = −
2 2
ref  = 60
B
II: = 120 + n.360 nZ
2
 B = 240 + n.720
B
III: = 240 + n.360 nZ
2
 B = 480 + n.720
 B = 240 (only)

 240 
 sin  + 30 
 2 
= sin (150 )
= sin ( 30 )
1
=
2
7.5.2 GRAPHICAL METHOD: ✓ A = 120
3 ✓ substitute
sin A = ✓ special angle
2
✓ answer (4)
 A = 60 + n.360 n  Z
or
 A = 120 + n.360 n  Z
 A = 120
B 
sin 2  + A 
2 
 240 
= sin 2  + 120 
 2 
= sin 2 ( 240 )
= sin 2 ( 60 )
2
 3
=
 2 
 
3
=
4

OR
 40

REFERENCE ANGLE METHOD:

3
sin A =
2
ref  = 60
I: A = 60 + n.360 n  Z
or
II: A = 120 + n.360 n  Z
 A = 120
B 
sin 2  + A 
2 
 240 
= sin 2  + 120 
 2 
= sin 2 ( 240 )
= sin 2 ( 60 )
2
 3
=  
 2 
3
=
4
7.5.3 A ✓ substitute
2sin 2 B − 3cos ✓ special angles
2
120 ✓ answer (3)
= 2sin 2 ( 240 ) − 3cos
2
= 2sin 60 − 3cos 60
2

3  1
= 2  − 3 
4 2
3 3
= −
2 2
=0
 41

7.6 1− x 2 ✓ substitute
LHS = ✓ LCD
1+ x 2
✓ simplify numerator
2
 cos α  ✓ simplify
1−  
 LHS =  1 − sinα  denominator
2 ✓ identity
 cos α 
1+   ✓ common factor
 1 − sin α  ✓ common factor
 cos 2 α  ✓ answer (8)
1−  2 
 LHS =  1 − 2sin α + sin α 
 cos 2 α 
1+  2 
 1 − 2sin α + sin α 
 1 − 2sin α + sin 2 α − cos 2 α 
 
 1 − 2sin α + sin 2 α 
 LHS =
 1 − 2sin α + sin α + cos α 
2 2
 
 1 − 2sin α + sin 2 α 
sin 2 α + cos 2 α − 2sinα + sin 2 α − cos 2 α
 LHS =
1 − 2sin α + sin 2 α + cos 2 α
2sin 2 α − 2sinα
 LHS =
2 − 2sin α
2sin α ( sinα − 1)
 LHS =
2 (1 − sin α )
−2sin α (1 − sinα )
 LHS =
2 (1 − sinα )
 LHS = − sin α
 42

7.7 cos 2θ = 2cos 2 θ − 1 ✓ equate

cos 2θ + 1
 cos 2θ + 1 = 2cos 2 θ ✓ cos 2 θ =
2
cos 2θ + 1 ✓ substitute
 cos 2 θ =
2 ✓ special angle
cos 2θ + 1 ✓ simplify
 cosθ = ✓ answer (6)
2
cos 2 (15 ) + 1
 cos15 =
2
cos30 + 1
 cos15 =
2
3
+1
2 3 +2 1
 cos15 = = 
2 2 2
3 +2
 cos15 =
4
2+ 3
 cos15 =
2
7.8 2x 2 − 3 x + 2 = 0 ✓ factors
1
 ( 2 x + 1)( x − 2 ) = 0 ✓ x=−
2
x = −
1
or x = 2 ✓ x=2
2 ✓ substitute
1 ✓ general solution
 sin A = − or sec B = 2 ✓ only A = 330
2
✓ only B = 300 (7)
1
cosB =
2
GRAPHICAL METHOD:
1
sin A = −
2
 A = − 30 + n.360 n  Z
or
 A = 210 + n.360 nZ
 A = 210 or A = 330
n.a
1
cosB =
2
 B =  60 + n.360 n  Z
 B = 60 or B = 300
n.a
 A = 330 and B = 300

OR
 43

REFERENCE ANGLE METHOD

1
sin A = −
2
ref  = 30
II: A = 150 + n.360 nZ
III: A = 210 + n.360 nZ
 A = 210 or A = 330
n.a
1
cosB =
2
ref  = 60
I: B = 60 + n.360 nZ
IV:B = 60 or B = 300
n.a
 A = 330 and B = 300
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