Trigonometry

Radian
c °
π =180
°
c 180
1=
π
c
° π
1=
180

Arc Length and Area of Sector

The length of the arc AB ,l=rθ

1 2
The area of the sector AOB, A= r θ
2

Example 1
2π
An arc PQ subtends an angle of 5 radians at the centre O of a circle with
radius 10 cm. Find
a) The length of the arc PQ
b) The area of the sector POQ
Leave your answers in terms of π

a)
The length of the arc PQ ,l=rθ
¿ 10 ( 25π ) cm
¿ 4 πcm

b)

1 2
The area of the sector POQ , A= r θ
2

1
¿ ( 10 )2
2 ( )
2π
5
c m2

2
¿ 20 π c m

Pythagorean Relationships
Consider the right-angled triangle
 opp a
sin θ= =
hyp c
adj b
cos θ= =
hyp c
opp a
tanθ= =
adj b

Now:
a
sin θ c a
tanθ= = = ∎
cos θ b b
c

2
a 2 a
If sin θ= ⇒ sin θ= 2
c c
2
b 2 b
If cos θ= ⇒ cos θ= 2
c c

2 2
∴ sin θ+cos θ
2 2 2
a b c
¿ 2
+ 2 = 2 =1
c c c
2 2
∴ sin θ+cos θ=1

Proving Identities
 i) Always hold one side constant
ii) Using trigonometric identities, simplify or factorize to show the
left-hand side is equal to the right-hand side
iii) Reference all identities used

Example 2
Prove cos 4 A−sin4 A +1 ≡2 cos 2 A

cos 4 A−sin4 A +1 ≡ ( cos 2 A +sin2 A ) ( cos 2 A−sin2 A ) +1

2 2
¿ cos A−sin A +1
2
¿ cos A+ cos A
2
[From 1−sin2 A=cos 2 A ]
2
¿ 2 cos A ∎

Solving Trigonometric Equations

i) sin θ=k sin θ=−k
 ° ° ° °
θ=α , ( 180−α ) θ=( 180+α ) , ( 360−α )
c c c c
¿ α , ( π−α ) ¿ ( x +α ) , ( 2 π −α )

ii) cos θ=k cos θ=−k

° ° ° °
θ=α , ( 360−α ) θ=( 180−α ) , ( 180+α )

¿ α c , ( 2 π−α c )
c c
¿ ( π−α ) , ( π + α )

iii) tanθ=k tanθ=−k

° ° ° °
θ=α , ( 180+ α ) θ=( 180−α ) , ( 360−α )
c c c c
¿ α , ( π +α ) ¿ ( π−α ) , ( 2 π−α )

Note:
i) αis always the acute angle and is always positive
ii) When working, we don’t enter the negative k value, in calculator

Example 4
Solve for 0° ≤ x ≤ 360° , 0° ≤ y ≤ 360°
a) cos 2 y −cos y=2
b) sin y=4 tan y
c) sin 2 x=0.866
d) tan (2 x−50 )=−0.7

a) 2
cos y −cos y−2=0

( cos y−2 ) ( cos y +1 )=0

cos y=2 cos y=−1

Inadmissible −1
α =cos (1)=0
° °
x=( 180−0 ) , ( 180+0 )
°
x=180

b) sin y=4 ( cos y)

sin y

sin y cos y−4 sin y=0

sin y ( cos y−4 )=0
sin y=0 cos y=4

°
y=0 ,180 ,360
° °
Inadmissible
c) −1
α =sin ( 0.866 )=60
°

0 ≤ x ≤ 360 0 ≤ 2 x ≤720

° ° ° °
2 x=60 , ( 180−60 ) , ( 360+ 60 ) , (540−60 )
° ° ° °
2 x=60 , 120 , 420 , 480
° ° ° °
x=30 , 60 , 210 , 240
 d) −1
α =tan ( 0.7 )=35
°

0 ≤ x ≤ 360 0 ≤ 2 x ≤720 −50 ≤ 2 x−50 ≤ 670

2 x−50=−35
2 x=15
°
x=7.5

2 x−50=145 2 x−50=325 2 x−50=505

° ° °
x=97.5 x=187.5 x=277.5
° ° ° °
x=7.5 , 97.5 , 187.5 , 277.5

Trigonometry 2
Addition Identities
sin ( A+ B )=sin A cos B+ cos A sin B
sin ( A−B )=sin A cos B−cos A sin B
cos ( A+ B )=cos A cos B−sin A sin B
cos ( A−B )=cos A cos B+sin A sin B
tan A+ tan B
tan ( A+ B )=
1−tan A tan B
tan A−tan B
tan ( A−B )=
1+tan A tan B

Factor Formulae
If sin ( A+ B )=sin A cos B+ cos A sin B
If A=B
sin ( A+ A )=sin A cos A+ cos A sin A
sin 2 A=2 sin A cos A

If cos ( A+ B )=cos A cos B−sin A sin B

If B= A
cos ( A+ A )=cos A cos A−sin A sin A
2 2 2
cos 2 A=cos A−sin A cos 2 A=2 cos A−1
2
cos 2 A=1−2 sin A

tan A+ tan B
If tan ( A+ B )= 1−tan A tan B

If B= A
tan A + tan A
tan (2 A )=
1−tan A tan A
2 tan A
tan (2 A )= 2
1−tan A

Example 5
Express sin 3 x in terms of sin x

sin 3 x=sin ( 2 x+ x )=sin 2 x cos x +cos 2 x sin x

¿ 2 sin x cos 2 x + ( 1−2sin 2 x ) ( sin x )

¿ 2 sin x ( 1−2sin 2 x ) +sin x−2 sin3 x

3 3
¿ 2 sin x−2 sin x +sin x−2 sin x
3
¿ 3 sin x−4 sin x

Example 6
Express cos 3 x in terms of cos x

cos 3 x=cos ( 2 x + x )=cos 2 x cos x−sin 2 x sin x

¿ ( 2 cos 2 x−1 ) cos x−2sin 2 x cos x

¿ 2 cos3 x−cos x −2 ( 1−cos 2 x ) cos x

3 3
¿ 2 cos x−cos x −2cos x+2 cos x
3
¿ 4 cos x−3 cos x

Example 7
Express tan3 x in terms of tan x

tan2 x +tan x
tan3 x=tan ( 2 x + x )=
1−tan 2 x tan x

¿
[ 2 tan x
1−tan x
2 ]
+ tan x

1−
[ 2 tan x
1−tan x
2 ]
tan x

3
2 tan x + tan x−tan x
2
1−tan x
¿ 2 2
1−tan x−2 tan x
2
1−tan x
3
3 tan x−tan x
¿ 2
1−3 tan x

Example 8
Expand sin ( A+ B+C )

sin ( [ A+ B ] +C )=sin ( A+ B ) cos C+ cos ( A+ B ) sin C

¿ ( sin A cos B+cos A sin B ) cos C + ( cos A cos B−sin A sin B ) sin C
¿ sin A cos B cos C+ sin B cos A cos C+cos A cos B sinC−sin A sin B sin C

Example 9
3 5
If sin x= 5 , cos y = 13 where x and y are acute. Find without the use of a
calculator
i) sin ( x + y )
ii) cos ( x− y )
iii) tan ( x + y )

3 12
sin x= sin x=
5 13

4 5
cos x= cos x=
5 13

3 12
tan x= tan x=
4 5

sin ( x + y )=sin x cos y +cos x sin y

¿ ( ) ( )
3 5
+
4 12
5 13 5 13
 63
¿
65

cos ( x− y )=cos x cos y +sin x sin y

¿ ( ) ( )
4 5
+
3 12
5 13 5 13
56
¿
65

tan x+ tan y
tan ( x + y ) =
1−tan x tan y
3 12
+
4 5
¿
1−
3 12
4 5 ( )
63
¿−
16

Example 10
° 1
Given that tan30 = √3 , without the use of a calculator find tan15
°

° °
tan15 =tan ( 45−30 )
° °
tan 45 −tan30
¿ ° °
1−tan 45 tan 30
1
1−
¿ √3 =2−√ 3
1+ 1
1
√3 ( )
 °
2 tan15
tan2 ( 15 ) = 2 °
1−tan 15
°
2 tan 15
° 1
tan30 = =
1−tan 15 √ 3
2 °

2 √ 3 tan 15 =1−tan 15
° 2 °

tan 15 + 2 √ 3 tan 15 −1=0

2 ° °

tan15 =2− √3
°

Example 11
sin 3 θ+sin θ
Prove cos 3 θ+cos θ
≡ tan 2θ

2 sin2 θ cos θ
2cos 2θ cos θ
sin 2θ
¿ =tan 2 θ ∎
cos 2 θ

Example 12
Solve for 0 ≤ θ ≤180, cos 5 θ−cos θ=sin 3 θ

−2 sin 3 θ sin2 θ=sin 3 θ

sin 3 θ+2 sin 3 θ sin 2θ=0
sin 3 θ ( 1+2 sin2 θ ) =0
sin 3 θ=0 1+2 sin 2θ=0

−1
α =0 sin 2 θ=
2
°
α =30

0 ≤ θ ≤180
0 ≤ 2θ ≤ 360

° ° ° °
3 θ=0 ,180 ,360 ,540
° ° ° °
θ=0 , 60 , 120 ,180

° °
2 θ=210 ,330
° °
θ=105 ,165

° ° ° ° ° °
θ=0 , 60 , 105 ,120 ,165 , 180
Past Paper Questions

2012
1)
a) The diagram shows a sector of a circle center O with an adjoining
square. The radius of the circle is 4m

π
If the sector AOC subtends an angle 3 at O, calculate giving your answer
in terms of π
e) The area of the shape OACMN
ii) The perimeter of the shape OACMN
b) Given that sin 3 = √2 and sin 4 =cos 4 = √2 , evaluate without
π 3 π 1 π π 2
, cos =
3 2
7π
using calculators, the exact value of cos 12
c) Prove the identity
1 1−sin θ
≡
sec θ+ tan θ cos θ
2013
2)
a)

The diagram shows a sector cut from a circle of center O. The angle at O
π 5
is 6 . If the perimeter of the sector is 6
( 12+ π ) cm, what is its area?

b) Solve the equation 2 cos2 θ+3 sin θ=0 for 0 ≤ θ ≤360°

1
c) Given that tan (θ−α )= 2 and that tanθ=3, use the appropriate
compound angle formula to find the value of the acute angle α

2014
3)
a) The figure below shows the sector OAB of a circle with center O,
radius 9 cm and angle 0.7 radians

i) Find the area of the sector OAB

ii) Hence, find the area of the shaded region, H
 b) Given that sin 6 = 2 and cos 6 = √2 , show that
π 1 π 3

( π6 )= 12 ( √ 3 cos x−sin x ), where x is acute

cos x +

c) Prove the identity [ 1−cos θ ]

tan θ sinθ
≡ 1+
1
cos θ

2015
4)
a) The following diagram shows a circle of radius r =¿ 4 cm, with
π
center O and sector AOB which subtends an angle, θ= 6 radians at
the center

1 2
If the area of the triangle AOB= r sin θ ,
2 then calculate the area of the
shaded region
b) Solve the following equation, giving your answer correct to one
decimal place
8 sin θ=5−10 cos θ , where 0° ≤ θ ≤360 °
2

c) Prove the identity

sinθ +sin 2θ
≡ tanθ
1+ cos θ+cos 2θ
2016
5)
a) The diagram below shows two sectors AOB and DOC . OB and OC are
x cm and ( x +2 ) cm respectively and angle AOB=θ

2π
If θ= 9 radians, calculate the area of the shaded region in terms of x

b) Given that cos 30 °= √2 and sin 45 °= √2 , without the use of a calculator,

3 2

evaluate cos 105 °, in surd form, giving your answer in the simplest
terms
sin ( θ+α )
c) Prove that the identity cos θ cos α
≡ tan θ+ tan α

2017
6)
a) The figure below shows a plot of land ABCD. Section ABC is used
for building and the remainder for farming. The radius BC is 10m
and angle BCD is a right angle
 50 π 2
i) If the building space is 3
m, calculate the angle ACB in radians
ii) Working in radians, calculate the area used for farming
b) Given that sin 3 = √2 and sin 4 =cos 4 = √2 , show without using
π 3 π 1 π π 2
, cos =
3 2
a calculator that
cos
[
π π
−
4 3
≡
]
√ 2+ √ 6
2π 2 √3
sin
3

c) Prove the identity

2
cos θ
1− ≡ sin θ
1+sin θ

2018
7)
a) A wire in the form of a circle with radius 4cm is reshaped in the
form of a sector of a circle with radius 10cm. Determine, in
radians, the angle of the sector, giving your answer in terms of π
b) Solve the equation sin2 θ+3 cos 2 θ=2 for 0 ≤ θ ≤ π . Give your answer(s) to
1 decimal place
c) Prove the identity
1 1 2 tan x
− ≡
1−sin x 1+sin x cos x

2019
8)
a) A compass is used to draw a sector of radius 6cm and area 11.32 c m2
i) Determine the angle of the sector in radians
ii) Calculate the perimeter of the sector
 b) A right-angled triangle XYZ has an angle θ where sin θ= √5 . Without
5

evaluating θ, calculate the exact value (in surd form if applicable)

of
i) cos θ
ii) sin 2 θ
2 1
c) Show that tan θ+1 ≡ 2
cos θ

2021
24
9) Given that cos M = 25 and that angle M is acute, determine the value
for tan2 M

2022
10)
a) The diagram below, shows a chord AB which subtends an angle of
0.5 at the center O of a circle of radius 10cm. Given that the area of
c

1 2
triangle AOB= r sin θ ,
2 calculate the area of the shaded region

b)
i) Show that cos 2 θ ≡2 cos 2 θ−1
ii) Hence, solve the equation cos 2 θ+cos θ+1=0 for 0<θ <2 π
Past Paper Solutions
2012
1)
a)
i)
1
Area of the sector, OAC= 2 ( 4 )
2
( π3 )= 83π m 2

Area of the square, ONMC=4 ( 4 ) =16 m2

Hence, the area of the shape, OACMN = 16 + ( 8π 2

3
m)

ii) Perimeter¿ OA + AC +CM + MN + NO

¿4+ 4
( ( ))π
3
+ 4+ 4+ 4

(
¿ 16+
4π
3)m

( 712π )=cos ( π3 + π4 )
b) cos

¿ cos ( )cos ( )−sin ( )sin ( )

π π π π
3 4 3 4

¿
2 2 ( ) ( )
1 √ 2 √ 3 √2
−
2 2

¿
√2−√ 6
4

c)
1 sin θ
Recall: sec θ= cos θ and tanθ= cos θ
 1
sec θ+ tan θ
=
1
1
+
( )
cos θ
sin θ cos θ
cos θ cos θ

¿ (
cos θ 1−sin θ
1+ sinθ 1−sin θ )
cos θ ( 1−sinθ )
¿
( 1+ sin θ ) (1−sin θ )
cos θ ( 1−sin θ )
¿ 2
1−sin θ

Recall: 1−sin2 θ=cos 2 θ

cos θ ( 1−sin θ )
¿ 2
cos θ
1−sin θ
¿ ∎
cos θ

2013
2)
a) We name the sector AOB and let the radius be r cm

Perimeter of the sector ¿ Length of radius OA +¿ Length of arc AB+ ¿

Length of radius BO

¿r+ r
( 6)
( π
) πr
+ r=2 r + cm
6
Hence,
πr 5
2r+ = (12+ π )
6 6

r 2+( π6 )= 56 ( 12+ π )
5
(12+ π )
6
r=
2+( )
π
6
60+5 π
¿
12+ π
5 ( 12+ π )
¿ =5 cm
12+ π

Area of the sector ¿ 2 r

1 2
( π6 )
1
¿ ( 5 )2
2
π
6 ()
=
25 π
12
c m2

b) Recall: cos 2 θ=1−sin 2 θ

Therefore, 2 ( 1−sin 2 θ ) + 3sin θ=0
2
2 sin θ−3 sin θ−2=0
( 2 sin θ+1 ) ( sin θ−2 ) =0
−1
sin θ=2∨
2

Maximum value of sin θ=1∴ sin θ=2 has no real solutions

−1
When sin θ= 2
 ° ° ° ° ° °
θ=180 +30 =210 θ=360 −30 =330

c)
tan θ−tan α
tan (θ−α )=
1+tan θ tan α

Hence,
3−tan α 1
=
1+ 3 tan α 2
6−2 tan α =1+3 tan α
tan α =1
c
−1 ° π
α =tan ( 1 )=45 ∨
4

2014
 3)
a)
1 2
i) Area of sector OAB= 2 ( 9 ) ( 0.7 )=28.35 c m2

ii) Angle AOC=90

°

AC
=tan ( 0.7 )
c
9
AC=7.580 cm
1 2
Area of the triangle OAB= 2 ( 9 ×7.580 ) =34.11c m

∴ Area of the shaded region, H=¿ Area of triangle OAC−¿ Area of sector
OAB
2
¿ ( 34.11−28.35 )=5.76 c m

b) By the compound angle formulae

( π6 )=cos x cos( π6 )−sin x sin ( π6 )

cos x +

¿ cos x ( √ )−sin x ( )
3 1
2 2
1
¿ ( √ 3 cos x−sin x ) ∎
2

c)
sinθ
( sin θ )
tan θ sin θ cos θ
=
1−cos θ 1−cos θ
2
sin θ
¿
cos θ ( 1−cos θ )
Recall: sin2 θ=1−cos 2 θ ⇒ ( 1−cos θ ) ( 1+cos θ )
Hence,
tan θ sin θ ( 1+ cos θ ) ( 1−cos θ )
=
1−cos θ cos θ (1−cos θ )
1+ cos θ 1 cos θ
¿ = +
cos θ cos θ cos θ
1
¿ +1∎
cos θ

2015
4)
a) Area of the shaded segment ¿ Area of sector AOB−¿ Area of triangle
AOB

( )
π
6
¿ ( π r 2) − 1 r 2 sinθ
2π 2

¿( 121 π ( 4 ) )− 12 ( 4 ) sin ( π6 )
2 2

¿( −4 ) c m
4π 2
3

b) 8 sin2 θ=5−10 cos θ 2 2

∴ sin θ=1−cos θ

8 ( 1−cos2 θ )=5−10 cos θ

2
−8 cos θ+10 cos θ+3=0
2
8 cos θ−10 cos θ−3=0
( 4 cos θ+1 ) ( 2 cos θ−3 )=0
−1 3
cos θ= ∨
4 2
 3
−1 ≤cos θ ≤1 ∀ θ, hence cos θ= 2 has no real solutions

°
A=75.52
θ=180−75.52 , 180+75.52
° ° ° °
θ=104.48 , 255.52 ≈ 104.5 , 255.5

c) Recall: sin 2 θ=2 sin θ cos θ

2 2 2 2
cos 2 θ=cos θ−sin θ ,2 cos θ−1 ,1−2sin θ

sin θ+2 sin θ cos θ sinθ ( 1+2 cos θ )

=
1+ cos θ+2 cos θ−1 cos θ ( 1+ 2cos θ )
2

sin θ
¿ =tan θ∎
cos θ

2016
5)
a)

( )
2
1 2 2π πx 2
The area of the sector AOB= ( x )
2 9
=
9
cm

( )
2
1 2π π ( x+ 2 )
The area of the sector DOC= ( x +2 )2
2 9
=
9
c m2

Hence, the area of the shaded region

 [( ) ]
2
π ( x +2 ) πx
2
π 2
¿ − = (4 x + 4) c m
9 9 9

b) cos 60 °=cos [ 2 ( 30° ) ]

2 °
¿ 2 cos 30 −1

[( ) ]
2
¿ 2
√ 3 −1 = 1
2 2

cos 105 °=cos ( 60° + 45° )

° ° ° °
¿ cos 60 cos 45 −sin 60 sin 45

¿
1 √2
2 2 ( ) ( )( √22 )
−
√3
2

¿√
2−√ 3 √ 2
4

¿ [ √ 2 ( 1−√3 ) = √ 2 ( 1−√3 )
4 2 √2 √ 2 ][ ]
1−√ 3
¿
2√ 2

We rationalize the surd to obtain

[ 1−√ 3 √ 2 √ 2− √ 6
×
2 √ 2 √2
=
4 ]
c)
sin ( θ+α ) sin θ cos α + cos θ sin α
=
cos θ cos α cos θ cos α
 sin θ cos α cos θ sin α
¿ +
cos θ cos α cos θ cos α
sin θ sin α
¿ +
cos θ cos α
¿ tanθ tan α ∎

2017
6)
a)
50 π
i) Area of ACB=
3
1 2
Area of a sector ¿ 2 r θ

1 50 π
Hence, ( 10 )2 θ=
2 3
50 π
50 θ=
3
c
π
θ=
3

ii) Region ACD is used for farming

ACD=
[ π π π
− =
2 3 6]
DC
10
=cos
π
6 ()
DC= [ √]
10 3
2
=5 √ 3

Area of triangle
1
ACD= (10 ) ( 5 √ 3 ) sin
2 ()
π
6

1
()
¿ ( 10 ) ( 5 √ 3 )
2
1
2

25 √ 3 2
¿ m
2

b) Using the compound angle formulae

cos( π4 − π3 )=cos( π4 ) cos ( π3 )+ sin ( π4 ) cos( π3 )
¿ √ ( )+ √ ( √ )
2 1 2 3
2 2 2 2

¿ √ +√ √
2 2 3
4 4

sin ( 23π )=sin ( π− π3 )

¿ sin π cos ( π3 )−cos π sin ( π3 )
2 ( ( ))
¿ ( 0 ) − −1 √
1
2
3

¿
√3
2

cos ( π4 − π3 ) = ( √ 2+√4 2√ 3)
sin ( )
2π √3
3 2

¿√
2+ √ 6
4 ( √23 )
¿√
2+ √ 6
∎
2 √3

c) Recall: cos 2 θ=1−sin 2 θ

This is a difference of two squares, cos 2 θ=( 1−sin θ )( 1+sin θ )
So,

1−
cos 2 θ
1+sin θ
=1− (
( 1−sin θ ) ( 1+sin θ )
1+sin θ )
¿ 1−( 1−sin θ )
¿ sin θ ∎

2018
7)
 a) Circle of radius 4 cm→ Sector of radius 10 cm

The circumference of the circle will be equal to the perimeter of the

sector
Circumference of circle ¿ 2 πr
¿ 2 π ( 4 ) =8 π cm

The perimeter of the sector r +r + rθ ¿ is in radians)

8 π=10+10+ 10θ
8 π=20+ 10θ
8 π−20
θ=
10
2 ( 4 π −10 )
θ=
10

( )
c
4 π −10
θ=
5

b) Recall: cos 2 θ=1−2 sin2 θ

sin2 θ+3 ( 1−2sin 2 θ ) =2
2
−5 sin θ=−1
2
5 sin θ=1

sin θ=±
√ 1
5
sin θ=
√ 1
5
sin θ=−
√ 1
5

θ=0.46 θ=−0.46 [No solutions for the required range]

θ=π −0.46
¿ 2.67
θ=0.46 , 2.67

θ=0.5 ,2.7
c c
for 0 ≤ θ ≤ π

c)
( 1+ sin x ) −( 1−sin x ) 2 sin x
=
(1−sin x )( 1+sin x ) (1−sin x )( 1+sin x )
2 sin x
¿ 2
1−sin x

Recall: cos 2 x=1−sin2 x , so

2sin x
¿ 2
cos x
sin x
Recall: tan x= cos x , hence

¿
2sin x 1
cos x cos x( )
2 tan x
¿ ∎
cos x

2019
8)
 a)
1 2
i) A= θ r
2
1 2
11.32= θ r
2
22.64=θ ( 36 )
c
θ=0.63

ii) Perimeter of the sector ¿ r +r + rθ

Perimeter ¿ 12+6 ( 0.63 )
¿ 15.78 cm

b)
i)

2
( √ 5 ) + x 2=25
2
x + 5=25
2
x =20
x=2 √ 5

2 √5
cos θ=
5
 ii) sin 2 θ=2 sin θ cos θ

¿2( √55 )( 2 √55 )

¿4( )
5
25
4
sin 2 θ=
5

c)
2
tan θ+1
2
sin θ
¿ 2
+1
cos θ
2 2
sin θ+ cos θ
¿ 2
cos θ
1
¿ 2
∎
cos θ

2021
9)

Using Pythagoras’ theorem:

opp=√ ( 25 ) −( 24 )
2 2
¿ √ 49=7

7
tan M =
24
2 tan M
tan2 M = 2
1−tan M

¿
(2
24 )
7

1−( )
2
7
24
336
¿
527

2022
10)
1 2
a) Area of sector ¿ 2 r θ

1 2
¿ ( 10 ) ( 0.5 )
2
2
¿ 25 c m

1 2
Area of triangle ¿ 2 r sinθ

1 2
¿ ( 10 ) sin ( 0.5 )
2
2
¿ 23.97 c m

∴ Area of shaded region ¿ Area of sector −¿Area of triangle

¿ 25−23.97
2
¿ 1.03 c m
 b)
i) cos ( A+ A )=cos A cos A−sin A sin A
2 2
cos (2 θ ) =cos θ−sin θ

Recall: sin2 θ=1−cos 2 θ

cos 2 θ=cos2 θ−( 1−cos 2 θ )
2 2
¿ cos θ−1+cos θ
2
¿ 2 cos θ−1

ii) cos 2 θ+cos θ+1=0

2
2 cos θ−1+ cosθ+1=0
2
2 cos θ+cos θ=0
cos θ ( 2 cos θ+ 1 )=0

cos θ=0 2 cos θ+1=0

π 3π −1
θ= , cos θ=
2 2 2
π
θ=
3

θ= π −( π
3)( )
, π+
π
3
 2π 4π
θ= ,
3 3