Trigonometry

Higher Unit 2
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Higher

Outcome 3

Trigonometry identities of the form sin(A+B)
Double Angle formulae
Trigonometric Equations
Radians & Trig Basics
More Trigonometric Equations
Exam Type Questions

Trig Identities
Outcome 3


Higher

Supplied on
a formula
sheet !!

The following relationships are always true for
two angles A and B.
1a.
1b.

sin(A + B) = sinAcosB + cosAsinB
sin(A - B) = sinAcosB - cosAsinB

2a.

cos(A + B) = cosAcosB – sinAsinB

2b.

cos(A - B) = cosAcosB + sinAsinB

Quite tricky to prove but some of following
examples should show that they do work!!

Trig Identities
Higher

Examples 1

Outcome 3


(1) Expand cos(U – V).

(use formula 2b )

cos(U – V) = cosUcosV + sinUsinV
(2) Simplify

sinf°cosg° - cosf°sing°

(use formula 1b )

sinf°cosg° - cosf°sing° = sin(f – g)°
(3) Simplify cos8 θ sinθ + sin8 θ cos θ (use formula 1a )
cos8 θ sin θ + sin8 θ cos θ = sin(8 θ + θ) = sin9 θ

Trig Identities
Higher

Example 2

Outcome 3


By taking A = 60° and B = 30°,
prove the identity for cos(A – B).
NB:

cos(A – B) = cosAcosB + sinAsinB

LHS = cos(60 – 30 )° = cos30° =

√3

/2

RHS = cos60°cos30° + sin60°sin30°
=(½

X

√3

=
Hence LHS = RHS !!

√3

/4 +

=

√3

/2

/2 ) + (√3/2 X ½)
√3

/4

Trig Identities
Higher

Example 3

Outcome 3


Prove that sin15° = ¼(√6 - √2)
sin15° = sin(45 – 30)°

= sin45°cos30° - cos45°sin30°
= (1/√2
=

X

√3

/2 ) - (1/√2 X ½)

(√3/2√2 -

1

=

(√3 - 1)
2√2

=

(√6 - √2)
4

= ¼(√6 - √2)

/2√2)
X

√2
√2

Trig Identities

Higher

Example 4

NAB type
Question

Outcome 3

y
β

41

x

α

40

Show that

3

4

cos(α - β) =

187

/205

Triangle2

Triangle1
If missing side = x

If missing side = y

Then x2 = 412 – 402 = 81

Then y2 = 42 + 32 = 25

So

So

x=9

y=5

sinα = 9/41 and cosα = 40/41 sin β = 3/5 and cosβ = 4/5

Higher

Trig Identities
Outcome 3


sinα = 9/41 and cosα = 40/41 sin β = 3/5 and cosβ = 4/5
cos(α - β) = cosαcosβ + sinαsinβ

= (40/41

X

/5) + (9/41

4

=

160

/205 +

=

187

X

/5 )

3

/205

/205

27

Remember this is a NAB type Question

Trig Identities
Higher

Example 5

Solve

Outcome 3

sinx°cos30° + cosx°sin30° = -0.966

ALWAYS
work out
By
Quad 1 rule 1a
first
sin(x


NAB type
Question

where 0o < x < 360o
sinx°cos30° + cosx°sin30° = sin(x + 30)°

S
180-xo

+ 30)° = -0.966

Quad 3 and Quad 4
sin-1 0.966 = 75°
Quad 3: angle = 180o + 75o
x + 30o = 255o
x = 225o

A
xo

180+x
T

360-xo
C

o

Quad 4: angle = 360o – 75o
x + 30o = 285o
x = 255o

Trig Identities
Higher

Solve


Outcome 3

Example 6

sin5 θ cos3 θ - cos5 θ sin3 θ = √3/2

By rule 1b.

where 0 < θ < π

sin5θ cos3θ - cos5θ sin3θ =sin(5θ - 3θ) = sin2θ

sin2θ = √3/2
Repeats every π

Quad 1 and Quad 2
sin-1 √3/2 = π/3

S
π-θ

A
θ

π+θ 2π-θ
T
C

Quad 1: angle = π/3 Quad 2: angle = π - π/3 In this example
repeats lie out
2 θ = π/ 3
2 θ = 2π/3
θ = π /6

θ = π/ 3

with limits

Trig Identities
Higher

Example 7

Outcome 3


Find the value of x that minimises the expression
cosx°cos32° + sinx°sin32°
Using rule 2(b) we get
cosx°cos32° + sinx°sin32° = cos(x – 32)°
cos graph is roller-coaster

min value is -1 when angle = 180°
ie

x – 32o = 180o
ie

x = 212o

Paper 1 type
questions

Trig Identities

Higher

Example 8

Outcome 3

Simplify

sin(θ - π/3) + cos(θ + π/6) + cos(π/2 - θ)

sin(θ - π/3) + cos(θ + π/6) + cos(π/2 - θ)
=

sin θ cosπ/3 – cos θ sinπ/3
+ cos θ cosπ/6 – sin θ sinπ/6
+ cosπ/2 cos θ + sinπ/2 sin θ

= 1/2 sin θ – √3/2cos θ + √3/2 cos θ – 1/2sin θ + 0 x cos θ + 1 X sin θ

=

sin θ

Paper 1 type
questions

Trig Identities
Higher

Example 9

Outcome 3


Prove that
(sinA + cosB)2 + (cosA - sinB)2 = 2(1 + sin(A - B))
LHS = (sinA + cosB)2 + (cosA - sinB)2
= sin2A + 2sinAcosB + cos2B + cos2A – 2cosAsinB + sin2B
= (sin2A + cos2A) + (sin2B + cos2B) + 2sinAcosB - 2cosAsinB
= 1 + 1 + 2(sinAcosB - cosAsinB)
= 2 + 2sin(A – B)
= 2(1 + sin(A – B)) = RHS

Double Angle Formulae
Outcome 3


Higher

sin 2 A = 2sin A cos A
cos 2 A = cos 2 A − sin 2 A
= 2cos 2 A − 1
2
= 1 − 2sin A
Two further formulae derived from the cos 2 A formulae.

cos A = 1 (1 + cos 2 A)
2
2

sin A = 1 (1 − cos 2 A)
2
2



Higher

Mixed Examples:

Outcome 3

4
Given that A is an acute angle and tan A = , calculate sin 2 A and cos 2 A.
3
sin A 4
=
cos A 3

2

sin 2 A + ( 3 sin A) 2 = 1
4
sin A = ±

Similarly:
sin 2 A = 2sin A cos A =

Substitute form the
tan (sin/cos) equation

sin A + cos A = 1
2

cos A =

16
=
25

3
5

9 − 16
=
25

+ve because A is acute
3-4-5 triangle !

24
25

cos 2 A = cos 2 A − sin 2 A =

4
5

−7
25

A is greater than 45
degrees – hence 2A is
greater than 90 degrees.

Outcome 3

Higher


Find the exact value of sin 75o.

2

sin(75o ) = sin(45 + 30)

o

sin(75 ) = sin 45cos30 + cos 45sin 30

45
1

o

1 3 1 1
1+ 3
=
+
=
2 2
22
2 2

Prove that

1

2 30

o

1

sin(α + β )
= tan α + tan β
cos α cos β

sin(α + β )
sin α cos β + cos α sin β
=
cosα cos β
cosα cos β

=

sin α sin β
+
cosα cos β

= tan α + tan β

3

Outcome 3

Higher


For the diagram opposite show that cos LMN =

5
.
5

cos LMN = cos(α + β )
Length of LM =
Length of MN =

M

18 = 3 2
10

3 2

cos(α + β ) = cosα cos β − sin α sin β
=
=
=

1 3
1 1
−
2 10
2 10
2
2
=
=
20
4 5
5
5

1
5

α

3
L

3

β

10

1 N

Outcome 3

Higher


Prove that,

cos 4 α − sin 4 α = cos 2α .

cos 4 α − sin 4 α = (cos 2 α ) 2 − (sin 2 α ) 2

Using x 2 − y 2 = ( x − y )( x + y )

= (cos 2 α − sin 2 α )(cos 2 α + sin 2 α )

cos 2 α + sin 2 α = 1

= cos 2 α − sin 2 α

cos 2α = cos 2 α − sin 2 α

= cos 2α



Higher

Outcome 3

Double angle formulae (like cos2A or sin2A) often occur in
trig equations. We can solve these equations by substituting
the expressions derived in the previous sections.

Rules for solving equations
sin2A = 2sinAcosA when replacing sin2Aequation
cos2A = 2cos2A – 1 if cosA is also in the equation
cos2A = 1 – 2sin2A if sinA is also in the equation

Outcome 3

Higher

cos 2 x o − 4sin x o + 5 = 0 for 0 ≤ x ≤ 360o.


Solve:

cos2x and sin x,

(1 − 2sin x) − 4sin x + 5 = 0
2

6 − 4sin x − 2sin x = 0
2

so substitute 1-2sin2x
compare with 6 − 4 z − 2 z 2 = 0

(6 + 2sin x)(1 − sin x) = 0

sin x = 1 or sin x = −3

x = 90

o

0 ≤ sin x ≤ 1 for all real angles

Outcome 3

Higher


Solve:

5cos 2 x o = cos x o − 2

for 0 ≤ x ≤ 360o

cos 2x and cos x,

5(2cos 2 x − 1) = cos x − 2

so substitute 2cos2 -1
π
2

10cos 2 x − cos x − 3 = 0
(5cos x − 3)(2cos x + 1) = 0
3
1
cos x =
or cos x = −
5
2

π

x = 90 + 30 = 120

x = 51.3o and
x = 360 − 51.3 = 308.7

90o

o

o

x = 270 − 30 = 240o

180

o

and

S

A

T

C

270o
3π
2

0o



Higher

Outcome 3

Outcome 3

Higher


The diagram shows the graphs of

f ( x ) = a sin bx o

and g ( x ) = c sin x o

for 0 ≤ x ≤ 360o.
4

y

y

y = f ( x)

2

360

0
-2

o

x
x

y = g ( x)

-4

Three problems concerning this graph follow.

Outcome 3

Higher


i)

State the values of a, b and c.

y

y

y = f ( x)

f ( x) = a sin bx o g ( x) = c sin x o
The max & min values of sinbx
are 1 and -1 resp.
The max & min values of
asinbx are 3 and -3 resp.

360o

y = g ( x)

a=3

f(x) goes through 2 complete cycles from 0 – 360o b =

2

g ( x) = c sin x o
The max & min values of csinx are 2 and -2 resp.

c=2

x
x

Outcome 3

Higher

f ( x ) = g ( x) algebraically.


ii) Solve the equation

From the previous problem we now have:

f ( x) = 3sin 2 x

g ( x) = 2sin x

and

Hence, the equation to solve is:

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

Expand sin 2x

6sin x cos x − 2sin x = 0

3sin x cos x − sin x = 0

sin x(3cos x − 1) = 0
sin x = 0 or

1
cos x =
3

3sin 2 x = 2sin x
Divide both sides by 2

Spot the common factor in the terms?

Is satisfied by all values of x for which:

Outcome 3

Higher


iii) find the coordinates of the points of intersection
of the graphs for 0 ≤ x ≤ 360o.

From the previous problem we have:

sin x = 0 and

sin x = 0
Hence

x

= 0o

x

= 180o

x

= 360o

1
cos x =
3

1
cos x =
3
x = 70.5o
x = (360 − 70.5)o

= 289.5o

Radian Measurements
Outcome 3


Higher

Reminders

i) Radians

π radians = 180

Converting between degrees and radians:
120o = 120.

π
18 0

5π 5π 180
=
.
6
6 π

2π
=
radians
3

= 150o

o

Degree Measurements
Outcome 3

Higher

Equilateral triangle:


ii) Exact Values
45o right-angled
triangle:
1

2

60o
1

2
45o
1

cos 45o = sin 45o =

tan 45 = 1
o

30o

2

3
1

1
2

sin 60o =
1
2

3
2

sin 30o =

cos 60o =

1
2

cos30o =

3
2

tan 30o =

1
3

tan 60o =

3

Radians / Degrees
Outcome 3


Higher

degrees

0o

30o

45o

radians

0

π

π

π

π

0

cos

1

4
1
2
1
2

3
3
2
1
2

2

sin

0

tan

0

6
1
2
3
2
1
3

1

3

∞

Example:

60o

90o
1

What is the exact value of sin 240o ?

240 = 180 + 60

sin(180 + α ) = − sin α

sin 240o = −

3
2

Sine Graph

Higher

Outcome 3

Period = 360o

Amplitude = 1

Cosine Graph
Outcome 3


Higher

Period = 360o

Amplitude = 1

Tan Graph

Higher

Outcome 3

Period = 180o

Amplitude cannot be found for tan function

Solving Trigonometric Equations
Outcome 3


Higher

Solve 2cos3 x − 1 = 0
(0 ≤ x ≤ 360o )
Example:
Step 2: consider what solutions
Step 1: Re-Arrange

are expected
π
2

2cos3 x − 1 = 0

2cos3 x = 1
1
cos3 x =
2

90o

π

180

o

S

A

T

C

270o
3π
2

0o

Outcome 3


Higher

cos 3x is positive so solutions
1
in the first and fourth quadrants
cos3 x =

2

0 ≤ x ≤ 360o

Since
x3
Then

has 2 solutions
x3

0 ≤ 3 x ≤ 1080o

has 6 solutions

Outcome 3

Higher


Step 3: Solve the equation

1
cos3 x =
2

1
3 x = cos  ÷ = 60o
2
−1

1st quad 4th quad

3x = 60o

300o

x = 20o

100o

cos wave repeats every 360o
420o 660o 780o 1020o
140o 220o 260o

340o

Higher

Outcome 3


Graphical solution for

1
cos3 x =
2

Outcome 3

Higher


Example:

Solve 1 + 2 sin 6t = 0

Step 1: Re-Arrange

(0 ≤ t ≤ 180o )

Step 2: consider what solutions
are expected

−1
sin 6t =
2

π
2

90o

sin 6t is negative so solutions in
the third and fourth quadrants
Since

0 ≤ t ≤ 180o

x6

x6
Then

has 2 solutions

0 ≤ 6t ≤ 1080o

has 12 solutions

π

180

o

S

A

T

C

270o
3π
2

0o

Outcome 3

Higher



−1
sin 6t =
2

 −1 
6t = sin −1 
÷
2


3rd quad 4th quad

6t = 225o
x = 39.1o

315o

 1 
o
st
sin −1 
÷ = 45 always 1 Quad first
 2

sin wave repeats every 360o
585o 675o 945o 1035o

52.5o 97.5o 112.5o 157.5o 172.5o



Higher

Outcome 3

−1
Graphical solution for sin 6t =
2

The solution is to be in radians – but
work in Trigonometric at the
Solvingdegrees and convertEquations
end.
Outcome 3

Higher


Example:

π
Solve 2sin(2 x − ) = 1
3

(0 ≤ x ≤ 2π )

Step 2: consider what
solutions are expected

Step 1: Re-Arrange

1
sin(2 x − 60 ) =
2

π
2

o

90o

(2x – 60o ) = sin-1(1/2)
Since

0 ≤ x ≤ 360

has 2 solutions

o

x2
Then

π
x2

0 ≤ 2 x ≤ 720

o

has 4 solutions

180

o

S

A

T

C

270o
3π
2

0o

Outcome 3

Higher


sin(2 x − 60o ) =

1
2

1
2 x − 60o = sin −1  ÷
2

1
sin −1  ÷ = 30o
2

(1st quadrant)

2 x − 60o = 30o and 150o

1st quad 2nd quad

2x = 90o

x = 45o
π
4

210o

sin wave repeats every 360o
450o 570o

105o 225o
7π
5π
12
4

285o
19π
12



Higher

Outcome 3


1
sin(2 x − 60 ) =
2
o

work in degrees and convert at the end.

Outcome 3

Higher


Harder Example: Solve

tan 2 x = 3

(0 ≤ x ≤ 2π )
Step 2: consider what
solutions are expected

Step 1: Re-Arrange

π
2

tan x = ± 3

90o
tan x = + 3

tan x = − 3

2 solutions
1 and 3
st

rd

quads

2 solutions
2

nd

and 4

th

quads

π

180

o

S

A

T

C

270o
3π
2

0o

Outcome 3

Higher



tan x = + 3 tan x = − 3

π
3

π
3

(60o in the 1st quadrant)

120o

tan wave repeats every 180o
240o 300o

2π
3

4π
3

1st quad 2nd quad

x = 60o

tan −1 3 =

5π
3



Higher

Outcome 3


tan x = 3
2

Outcome 3

Higher



3sin 2 x − 4sin x + 1 = 0 (0 ≤ x ≤ 360o )

Step 1: Re-Arrange

Step 2: Consider what solutions

(3sin x − 1)(sin x − 1) = 0
1
sin x =
3

are expected

π
2

90o

sin x = 1

π
Two solutions

One solution

180

o

S

A

T

C

270o
3π
2

0o

Outcome 3

Higher


sin x =

1
3

sin x = 1

Two solutions
1stquad

2nd quad

x = 19.5o 160.5o
Overall solution

One solution
90o

x = 19.5o , 90o and 160.5o

Higher

Outcome 3



3sin 2 x − 4sin x + 1 = 0

work in degrees and convert at the end.

Outcome 3

Higher



5sin 2 x − 2 = 2cos x

Step 1: Re-Arrange

Step 2: Consider what solutions

5(1 − cos x) − 2 = 2cos x
2

are expected

π
2

Remember o
this !
90
2
2

3 − 2cos x − 5cos 2 x = 0
(3 − 5cos x)(1 + cos x) = 0

3
cos x =
5

(0 ≤ x ≤ 2π )

cos x = −1

sin α + cos α = 1
cos 2 α = 1 − sin 2 α
A2
o 2 S
180 α = 1 − cos α
sin

π

C

T

270o
Two solutions

One solution

3π
2

0o

Outcome 3

Higher


cos x =

3
5

Two solutions
1stquad

3rd quad

x = 53.1o 306.9o
Overall solution in radians

cos x = −1
One solution
180o

x = 0.93 , π and 5.35

Higher

Outcome 3



5 sin 2 x − 2 = 2cos x

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Higher Maths
Strategies

Compound Angles

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Higher

The following questions are on

Compound Angles
Non-calculator questions will be indicated
You will need a pencil, paper, ruler and rubber.

Click to continue

Maths4Scotland

Higher

This presentation is split into two parts
Using Compound angle formula for
Exact values
Solving equations

Choose by clicking on the appropriate button

Quit

Quit

Maths4Scotland

Higher

A is the point (8, 4). The line OA is inclined at an angle p radians to the
x-axis
a) Find the exact values of:
i) sin (2p)
ii) cos (2p)
The line OB is inclined at an angle 2p radians to the x-axis.
b) Write down the exact value of the gradient of OB.
Draw triangle

80

Pythagoras

4

p
8

8
sin p =
80
4
8
⇒ 2×
×
⇒
80
80

cos p =

Write down values for cos p and sin p
Expand sin (2p)

sin 2 p = 2sin p cos p

Expand cos (2p)

cos 2 p = cos p − sin p ⇒

Use m = tan (2p)

tan 2 p =

Previous

2

2

sin 2 p
cos 2 p
Quit

⇒

( ) ( )
8
80

2

−

4
80

2

⇒

4
80
64
4
⇒
80
5
64 − 16
3
⇒
80
5

4 3
4
÷
⇒
5 5
3
Quit

Hint
Next

Maths4Scotland

Higher

In triangle ABC show that the exact value of

sin(a + b) is
Use Pythagoras

sin a =

sin a, cos a, sin b, cos b

Substitute values
Simplify

Previous

10

2

AC = 2 CB = 10

Write down values for

Expand sin (a + b)

2
5

1
2

cos a =

1
2

sin b =

1
10

cos b =

3
10

sin( a + b) = sin a cos b + cos a sin b

sin( a + b) =

sin(a + b) =

3
20

+

1
20

Quit

1
2

×

→

3
10

+

4
20

Quit

1
2
→

×

1
10
4
4
→
→
4×5
2 5

2
5

Hint
Next

Maths4Scotland

Higher

Using triangle PQR, as shown, find the
11

exact value of cos 2x
Use Pythagoras

PR = 11


cos x and sin x

2
cos x =
11

7
sin x =
11

Expand cos 2x

cos 2 x = cos 2 x − sin 2 x

Substitute values

( ) −( )

Simplify

Previous

2
11

cos 2x =

4
7
cos 2 x =
−
11
11
Quit

2

7
11

→ −

2

3
11

Quit

Hint
Next

Maths4Scotland

Higher

On the co-ordinate diagram shown, A is the point (6, 8) and

10

B is the point (12, -5). Angle AOC = p and angle COB = q
Mark up triangles
Find the exact value of sin (p + q).
OA = 10 OB = 13
Use Pythagoras

sin p =

sin p, cos p, sin q, cos q
Expand sin (p + q)
Substitute values
Simplify

Previous

8
,
10

cos p =

6
,
10

sin q =

8
6
12

5

13
5
,
13

cos q =

12
13

sin ( p + q ) = sin p cos q + cos p sin q
sin ( p + q) =

sin ( p + q) =

96
130

+

30
130

Quit

8 12
×
10 13

→

+

6
5
×
10 13

126
130

Quit

→

63
65

Hint
Next

Maths4Scotland

Higher

A and B are acute angles such that tan A =

3
4

Draw triangles

Use Pythagoras

Write down sin A, cos A, sin B, cos B

sin A =

sin 2 A = 2sin A cos A

Expand cos 2A

cos 2 A = cos A − sin A
2

Expand sin (2A + B)

13

3

A

5

B
4

12

Hypotenuses are 5 and 13 respectively

Expand sin 2A

Previous

5

Find the exact value of
cos 2A b)
sin(2 A + B)
c)

sin 2A a)

Substitute

5

and tan B = 12 .

2

3
,
5

cos A =

4
,
5

sin B =
3
5

sin 2 A = 2 × ×
cos 2A =

2

4
5

5
,
13

⇒
2

4
 3
 ÷ − ÷
5
5

⇒

cos B =

12
13

24
25
16
9
−
25 25

⇒

7
25

sin ( 2 A + B ) = sin 2 A cos B + cos 2 A sin B

sin ( 2 A + B ) =

24 12
7
5
323
×
+
×
=
25 13
25 13 325
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Hint
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Maths4Scotland

Higher

tan x°
If x° is an acute angle such that =

4
3

5

4 3 +3
sin( x + 30)° is
show that the exact value of
10

4

x
3

Draw triangle

Use Pythagoras

Write down sin x and cos x
Expand sin (x + 30)

sin x =

Hypotenuse is 5
4
,
5

cos x =

3
5

sin( x + 30) = sin x cos 30 + cos x sin 30

Substitute

sin( x + 30) =

4
3
3 1
×
+ ×
5 2
5 2

Simplify

sin( x + 30) =

4 3
3
+
10
10

⇒

4 3 +3
10
Hint

Previous
Table of exact values

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Maths4Scotland

Higher

The diagram shows two right angled triangles
24

ABD and BCD with AB = 7 cm, BC = 4 cm and CD = 3 cm.
Angle DBC = x° and angleyABD is 20 − 6 6
cos( x + )° is y°.
35
Show that the exact value of
BD = 5, AD = 24
Use Pythagoras
Write down
sin x, cos x, sin y, cos y.
Expand cos (x + y)

sin x =

3
,
5

cos x =

4
,
5

sin y =

5

24
,
7

cos y =

5
7

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

Substitute

cos( x + y ) =

4 5
3
24
× − ×
5 7
5
7

Simplify

cos( x + y ) =

20 − 3 4 × 6
20 − 6 6
20 3 24
⇒
⇒
−
35
35
35
35

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Maths4Scotland

Higher

The framework of a child’s swing has dimensions
as shown in the diagram. Find the exact value of sin x°
Draw triangle

Use Pythagoras

Draw in perpendicular

h= 5
3

Use fact that sin x = sin ( ½ x + ½ x)

sin

Write down sin ½ x and cos ½ x

sin

Expand sin ( ½ x + ½ x)
Substitute
Simplify

sin

(

x x
+
2 2

sin x =

)=

(

x
x
2 h5

x x
+
2 2
2
3

( ) = , cos ( ) =
) = sin cos + sin

2× ×

x
2

2
3

x
2

x
2

x
2

5
3
x
x
cos =
2
2

3

2
4
x
2

2sin cos

x
2

5
3

4 5
9

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Hint
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Maths4Scotland

Higher

11
π
, 0<α <
3
2
sin 2α
find the exact value of

tan
Given that α =

Draw triangle

11

a

Use Pythagoras


cos a and sin a

20

hypotenuse

3
cos a =
20

= 20

3

11
sin a =
20

Expand sin 2a

sin 2a = 2 sin a cos a

Substitute values

sin 2a = 2 ×

Simplify

Previous

11
3
×
20
20

6 11
sin 2a =
20
Quit

⇒

Quit

3 11
10

Hint
Next

Maths4Scotland

Higher

Find algebraically the exact value of

sin θ ° + sin ( θ + 120 ) ° + cos(θ + 150)°

Expand sin (θ +120)

sin ( θ + 120 ) = sin θ cos120 + cos θ sin120

Expand cos (θ +150)

cos ( θ + 150 ) = cos θ cos150 − sin θ sin150

Use table of exact values

cos 120 = − cos 60 = −
sin 120 =

Combine and substitute
Simplify

sin 60 =

sin θ + sin θ .

1
2

sin θ − sin θ +

3
2

1
2
3
2

cos 150 = − cos 30 = −
sin 150 =

sin 30 =

3
2
1
2

( ) + cosθ .( ) + cos θ .( ) − sin θ . ( )

cos θ −

−

1
2

3
2

3
2

−

3
2

1
2

1
2

cos θ − sin θ

=0
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If cos θ =

Higher

4
π
, 0 ≤θ ≤
5
2

sin 2θ a)

find the exact value of

sin 4θb)

Draw triangle

cos θ =

cos θ and sin θ

Expand cos 2θ
Find sin 4θ
Previous

Opposite side = 3

4
5

sin θ =

sin 2θ = 2 sin θ cos θ

Expand sin 4θ (4θ = 2θ + 2θ)

4

3
5

3 4
24
⇒ 2× × ⇒
5 5
25

sin 4θ = 2 sin 2θ cos 2θ

cos 2θ = cos θ − sin θ
2

sin 4θ = 2 ×

3

θ

Use Pythagoras


Expand sin 2θ

5

24 7
×
25 25
Quit

2

⇒

16 9
7
⇒
−
⇒
25 25
25

336
625
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Hint
Next

Maths4Scotland

Higher

For acute angles sin P =
P and Q

12
and
13

sin Q =

3
5

13

63
sin ( P +Q) =
65
Show that the exact value of

Draw triangles

Use Pythagoras

Write down sin P, cos P, sin Q, cos Q
Expand sin (P + Q)

5

12

P

3

Q

5

4

Adjacent sides are 5 and 4 respectively

sin P =

12
,
13

cos P =

5
,
13

sin Q =

3
,
5

cos Q =

4
5

sin ( P + Q ) = sin P cos Q + cos P sin Q

Substitute

sin ( P + Q ) =

12 4
5 3
× +
×
13 5
13 5

Simplify

sin ( P + Q ) =

48
15
+
65
65

Previous

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⇒

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63
65

Hint
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Maths4Scotland

Higher

You have completed all

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12 questions in this section

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Maths4Scotland

Higher

Using Compound angle formula for

Solving Equations

Continue

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Maths4Scotland

Higher

Solve the equation 3cos(2 x) + 10 cos( x) − 1 = 0 for 0 ≤ x ≤ π correct to 2 decimal places
Replace cos 2x with
Substitute
Simplify

cos 2 x = 2 cos 2 x − 1
3 ( 2 cos x − 1) + 10 cos x − 1 = 0
2

Determine quadrants
S

A

T

6 cos x + 10 cos x − 4 = 0
2

C

3cos 2 x + 5cos x − 2 = 0
Factorise
Hence

( 3cos x − 1) ( cos x + 2 ) = 0
cos x =

x = 1.23

Find acute x

Previous

acute

x = 1.23 rad

Quit

2π − 1.23

rads

x = 5.05

cos x = −2 Discard

or

x = 1.23

1
3

rads

rads

Hint
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Maths4Scotland

Higher

The diagram shows the graph of a cosine function from 0 to π.
a) State the equation of the graph.
b) The line with equation y = -√3 intersects this graph
at points A and B. Find the co-ordinates of B.

Equation

y = 2 cos 2 x
Determine quadrants

2 cos 2 x = − 3

Solve simultaneously

cos 2x = −

Rearrange
Check range

0≤ x ≤π

Find acute 2x
Deduce 2x

acute

2x =

2x =

S

T

x =


6π π
+
rads
6
6

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5π
7π
or
12
12

B

π
6

Previous

C

3
2

⇒ 0 ≤ 2 x ≤ 2π

6π π
−
or
6
6

A

(

is

B

7π
12

,− 3

)
Next

Hint

Maths4Scotland

Higher

Functions f and g are defined on suitable domains by f(x) = sin (x) and g(x) = 2x
a)
i)
1 expression
st

2nd expression
Form equation

Find expressions for:
f(g(x))

g(f(x)) Determine x

ii)

for 0 ≤ x sin360° 0 ⇒ x = 0°, 360°
≤ x=

f b)g ( xSolve f (2f(g(x))sing(f(x))
(
)) = 2 x) = = 2 x

cos x =

g ( f ( x )) = g (sin x) = 2sin x
2sin 2 x = 2sin x → sin 2 x = sin x

Replace sin 2x

2sin x cos x − sin x = 0

Common factor
Hence

or

2 cos x − 1 = 0 ⇒ cos x =

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acute

x = 60°

S

A

T

C

Determine

x = 0°, 60°, 300°, 360°

sin x ( 2 cos x − 1) = 0

sin x = 0

⇒

quadrants
x = 60°, 300°

2sin x cos x = sin x

Rearrange

1
2

1
2

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Hint
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Maths4Scotland

Higher

Functions f ( x) = sin x, g ( x) = cos x
a)
b)

i)

2 expression
nd

h( x ) = x +

Find expressions for
1
1

f (h( x)) =

Show that

1st expression

and

sin x +

2

2

π
4

are defined on a suitable set of real numbers

i) f(h(x))

cos x

ii) Find a similar expression for g(h(x))
f (h( x )) − g (h( x )) = 1 for 0 ≤ x ≤ 2π
iii) Hence solve the equation
2
π
π
sin x = 1
f (h( x)) = f x + = sin x +
Simplifies to
2
4
4

( ) ( )
g (h( x)) = g ( x + ) = cos ( x + )
π
4

f

Simplify 1 expr.
st

π
4

π
(h( x)) = sin x cos
4
1
2

Rearrange: sin x =

π
+ cos x sin
4
1
2

Use exact values

f ( h( x)) =

Similarly for 2nd expr.

g (h( x )) = cos x cos − sin x sin

sin x +

g (h( x)) =

Form Eqn.

ii) g(h(x))

1
2

acute x

cos x

π
4

cos x −

1
2

Determine
π
4

sin x

f ( h( x)) − g ( h( x)) = 1

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acute

2
=
2

x=

2
1
=
2 2
2

π
4

S

A

T

C

quadrants
x=

π 3π
,
4 4

Hint
Next

Maths4Scotland
a)

Higher

Solve the equation sin 2x - cos x = 0 in the interval 0 ≤ x ≤ 180°
b)

The diagram shows parts of two trigonometric graphs,
y = sin 2x and y = cos x. Use your solutions in (a) to
write down the co-ordinates of the point P.

Replace sin 2x

2sin x cos x − cos x = 0

Common factor

cos x ( 2sin x − 1) = 0

Hence

cos x = 0

Determine x

or

Solutions for where graphs cross

2sin x − 1 = 0 ⇒ sin x =

1
2

cos x = 0 ⇒ x = 90°, ( 270° out of range)
sin x =

1
2

⇒

acute

x = 30°

S

A

Determine quadrants
T
Previous

x =150°
y = cos150°

Find y value
Coords, P

C
Quit

By inspection (P)

y=−

x = 30°, 150°

for sin x

x = 30°, 90°, 150°

Quit

(

P 150°, −

3
2

)

3
2

Hint
Next

Maths4Scotland

Higher

Solve the 3cos(2 x) + cos( x) = − 1
equation

for 0 ≤ x ≤ 360°

cos 2 x = 2 cos 2 x − 1

Replace cos 2x with

Determine quadrants

3 ( 2 cos x − 1) + cos x = −1
2

Substitute

cos x = −

2
3

cos x =

1
2

6 cos 2 x + cos x − 2 = 0

Factorise

x = 48°

acute

x = 60°

S

Simplify

acute

A

S

A

( 3cos x + 2 ) ( 2 cos x − 1) = 0
cos x = −

Hence
Find acute x

acute

2
3

x = 48°

cos x =
acute

1
2

x = 60°

T
C
x = 132°
x = 228°

T
C
x = 60°
x = 300°

Solutions are: x= 60°, 132°, 228° and 300°
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Maths4Scotland

Higher

(

π
6

)

2sin x −
Solve the2equation= 1
sin

Rearrange

)

(

)

(

π
=
6

Determine quadrants

T

2x −

)

π
6

=

and for range

(

0 ≤ x ≤ 2π ⇒ 0 ≤ 2 x ≤ 4π
S

0 ≤ 2 x ≤ 2π

for range

1
=
2

π
acute 2 x −
6

Find acute x
Note range

(

π
2x −
6

for 0 ≤ x ≤ 2π

2x −

A

π
6

)

=

π
6

(

2x −

π
6

π
6

)

=

5π
6

)

=

17π
6

2π ≤ 2 x ≤ 4π
13π
6

(

2x −

Solutions are:

x=

C

π π 7π 3π
, ,
,
6 2
6
2
Hint

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Maths4Scotland

Higher

a) Write the equation cos 2θ + 8 cos θ + 9 = 0 in terms of cos θ
and show that for cos θ it has equal roots.
b) Show that there are no real roots for θ
Replace cos 2θ with
Rearrange
Divide by 2
Factorise
Deduction

cos 2θ = 2 cos 2 θ − 1

Try to solve:

( cos θ + 2 ) = 0

2 cos 2 θ + 8cos θ + 8 = 0

cos θ = −2

cos 2 θ + 4 cos θ + 4 = 0

No solution

Hence there are no real solutions for θ

( cos θ + 2 ) ( cos θ + 2 ) = 0
Equal roots for cos θ

Hint
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Maths4Scotland

Higher

Solve algebraically, the equation sin 2x + sin x = 0, 0 ≤ x ≤ 360
Replace sin 2x

2sin x cos x + sin x = 0

Determine quadrants

sin x ( 2 cos x + 1) = 0

for cos x
S
A

Common factor
Hence

sin x = 0
or

Determine x

2 cos x + 1 = 0

⇒

cos x = −

1
2

1
2

acute

x = 60°

x = 0°,
Previous

C

x = 120°, 240°

sin x = 0 ⇒ x = 0°, 360°
cos x = − ⇒

T

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120°, 240°, 360°

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Hint
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Maths4Scotland

Higher

Find the exact solutions of 4sin2 x = 1, 0 ≤ x ≤ 2π

sin 2 x =

Rearrange
Take square roots
Find acute x

1
4

sin x = ±

1
2

x =

π
6

acute

Determine quadrants for sin x
S

T

+ and – from the square root requires all 4 quadrants

A

C

π 5π 7π 11π
x = ,
,
,
6
6
6
6
Hint

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Maths4Scotland

Higher

Solve the equation x + cos x = 0
cos 2
Replace cos 2x with

for 0 ≤ x ≤ 360°

cos 2 x = 2 cos 2 x − 1

Substitute

2 cos 2 x + cos x − 1 = 0

Factorise

cos x =

2 cos 2 x − 1 + cos x = 0

Simplify

Determine quadrants
1
2

( 2 cos x − 1) ( cos x + 1) = 0
cos x =

Hence
Find acute x

acute

1
2

x = 60°

acute

S

cos x = −1

x = 60°

A

T
C
x = 60°
x = 300°

x = 180°

Solutions are: x= 60°, 180° and 300°
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Maths4Scotland

Higher

cos
Solve algebraically, the equation 2 x + 5cos x − 2 = 0
Replace cos 2x with
Substitute

cos 2 x = 2 cos 2 x − 1

for 0 ≤ x ≤ 360°
Determine quadrants

2 cos x − 1 + 5cos x − 2 = 0
2

cos x =
acute

Simplify

2 cos 2 x + 5cos x − 3 = 0

Factorise

1
2

x = 60°

( 2 cos x − 1) ( cos x + 3) = 0

Hence
Find acute x

cos x =
acute

1
2

x = 60°

S

cos x = −3
Discard above

A

T
C
x = 60°
x = 300°

Solutions are: x= 60° and 300°
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Maths4Scotland

Higher

You have completed all

Previous

12 questions in this presentation

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Back to start

Maths4Scotland

Higher


30°

sin
cos
tan

Return

45°

60°

π
6
1
2

π
4

π
3

1
2
1
2

3
2

3
2
1
3

1

1
2

3

Trigonometry

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