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Higher Unit 2
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
Hence LHS = RHS !!
X
=
3/
4
=
3/
2
3/
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
=
(3/22 -
=
(3 - 1)
22
=
(6 - 2)
4
= ¼(6 - 2)
) - (1/2 X ½)
1/
22)
X
2
2
Trig Identities
Higher
Example 4
NAB type
Question
Outcome 3
y

41
x

40
Show that
3
4
cos( - ) =
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
sin = 9/41 and cos =
40/
41
y=5
sin  = 3/5 and cos = 4/5
Higher
Trig Identities
Outcome 3
sin = 9/41 and cos =
40/
41
sin  = 3/5 and cos = 4/5
cos( - ) = coscos + sinsin
= (40/41 X 4/5) + (9/41
=
160/
205
=
187/
205
+
X
3/
5
27/
205
Remember this is a NAB type Question
)
Trig Identities
Higher
Example 5
NAB type
Question
Outcome 3
Solve
sinxcos30 + cosxsin30 = -0.966
ALWAYS
where 0o < x < 360o
work out
By1 rule 1a
sinxcos30 + cosxsin30 = sin(x + 30)
Quad
first
S
A
sin(x + 30) = -0.966
180-xo
Quad 3 and Quad 4
sin-1 0.966 = 75
Quad 3: angle = 180o + 75o
x + 30o = 255o
x = 225o
xo
o
180+xo 360-x
T
C
Quad 4: angle = 360o – 75o
x + 30o = 285o
x = 255o
Trig Identities
Higher
Outcome 3
Example 6
Solve
sin5 θ cos3 θ - cos5 θ sin3 θ =
By rule 1b.
3/
2
where 0 < θ < 
sin5θ cos3θ - cos5θ sin3θ =sin(5θ - 3θ) = sin2θ
sin2θ = 3/2
Repeats every 
Quad 1 and Quad 2
sin-1
3/
2
= /3
S
-θ
A
θ
+θ 2-θ
T
C
Quad 1: angle = /3 Quad 2: angle =  - /3 In this example
repeats lie out
2 θ = /3
2 θ = 2/3
θ = /6
θ = /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 - /3) + cos(θ + /6) + cos(/2 - θ)
sin(θ - /3) + cos(θ + /6) + cos(/2 - θ)
=
sin θ cos/3 – cos θ sin/3
+ cos θ cos/6 – sin θ sin/6
+ cos/2 cos θ + sin/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
Find the exact value of sin 75o.
sin(75o )  sin(45  30)
2
sin(75 )  sin 45cos30  cos 45sin 30
o
o
45
1
1 3 1 1
1 3



2 2
22
2 2
Prove that
1
2 30
o
1
sin(   )
 tan   tan 
cos cos 
sin(   )
sin  cos   cos sin 

cos cos 
cos cos 

sin  sin 

cos cos 
 tan   tan 
3
Double Angle formulae
Outcome 3
Higher
For the diagram opposite show that cos LMN 
5
.
5
cos LMN  cos(   )
Length of LM 
18  3 2
Length of MN 
10
M
3 2
cos(   )  cos cos   sin  sin 



1 3
1 1

2 10
2 10
2
2


20
4 5
5
5
1
5

3
L
3

10
1 N
Double Angle formulae
Outcome 3
Higher
Prove that,
cos 4   sin 4   cos 2 .
cos 4   sin 4   (cos 2  ) 2  (sin 2  ) 2
Using x 2  y 2  ( x  y )( x  y )
 (cos 2   sin 2  )(cos 2   sin 2  )
cos 2   sin 2   1
 cos 2   sin 2 
cos 2  cos 2   sin 2 
 cos 2
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  12 (1  cos 2 A)
2
sin A  12 (1  cos 2 A)
2
Double Angle formulae
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  ( 43 sin A)2  1
sin A  
Similarly:
sin 2 A  2sin A cos A 
Substitute from 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
Trigonometric Equations
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
Trigonometric Equations
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
Trigonometric Equations
Outcome 3
Higher
Solve:
5cos 2 x o  cos x o  2
for 0  x  360o
cos 2x and cos x,
so substitute 2cos2 -1
5(2cos 2 x  1)  cos x  2

2
10cos 2 x  cos x  3  0
(5cos x  3)(2cos x  1)  0
3
1
cos x 
or cos x  
5
2

x  90  30  120
x  51.3o and
x  360  51.3  308.7
90o
o
o
x  270  30  240o
o
180
and
S
A
T
C
270o
3
2
0o
Trigonometric Equations
Higher
Outcome 3
Trigonometric Equations
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
o
0
-2
xx
y  g ( x)
-4
Three problems concerning this graph follow.
Trigonometric Equations
Outcome 3
Higher
y
i)
State the values of a, b and c.
y
y  f ( x)
f ( x)  a sin bx o g ( x)  c sin x o
360o
x
x
The max & min values of sinbx
are 1 and -1 resp.
The max & min values of
asinbx are 3 and -3 resp.
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
Trigonometric Equations
Outcome 3
Higher
ii) Solve the equation
f ( x)  g ( x) algebraically.
From the previous problem we now have:
f ( x)  3sin 2 x
and
g ( x)  2sin x
Hence, the equation to solve is:
3(2sin x cos x)  2sin x
6sin x cos x  2sin x  0
3sin 2 x  2sin x
Expand sin 2x
Divide both sides by 2
3sin x cos x  sin x  0 Spot the common factor in the terms?
sin x(3cos x  1)  0 Is satisfied by all values of x for which:
sin x  0 or
1
cos x 
3
Trigonometric Equations
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
x
 180o
 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
 radians  180
Converting between degrees and radians:
120o  120.

180
5 5 180

.
6
6 
2

radians
3
 150o
o
Degree Measurements
Outcome 3
Higher
Equilateral triangle:
ii) Exact Values
45o right-angled
triangle:
1
2
o
60
1
2
o
45
1
cos 45o  sin 45o 
tan 45  1
o
30o
1
2
2
3
1
1
2
sin 60o 
3
2
sin 30o 
cos60o 
1
2
cos30o 
3
2
tan 30o 
1
3
tan 60o 
3
Radians / Degrees
Outcome 3
Higher
0o
30o
45o
60o
90o
radians
0




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

degrees
Example:
1
What is the exact value of sin 240o ?
240  180  60
sin(180   )   sin 
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
Example: Solve 2cos3x  1  0
(0  x  360o )
Step 2: consider what solutions
Step 1: Re-Arrange
are expected

2
2cos3x  1  0
2cos3x  1
1
cos3x 
2
90o

180
o
S
A
T
C
270o
3
2
0o
Solving Trigonometric Equations
Outcome 3
Higher
cos 3x is positive so solutions
1 in the first and fourth quadrants
cos3x 
2
0  x  360o
Since
x3
Then
has 2 solutions
x3
0  3x  1080o
has 6 solutions
Solving Trigonometric Equations
Outcome 3
Higher
Step 3: Solve the equation
1
3 x  cos    60o
2
1
cos3x 
2
1
o
1st quad 4th quad cos wave repeats every 360
3x = 60o
x = 20o
300o 420o 660o 780o 1020o
100o
140o 220o 260o
340o
Solving Trigonometric Equations
Higher
Outcome 3
Graphical solution for
1
cos3x 
2
Solving Trigonometric Equations
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

2
90o
sin 6t is negative so solutions in
the third and fourth quadrants
Since
0  t  180o
has 2 solutions
x6
x6
Then

0  6t  1080o
has 12 solutions
180
o
S
A
T
C
270o
3
2
0o
Solving Trigonometric Equations
Outcome 3
Higher
Step 3: Solve the equation
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
Solving Trigonometric Equations
Higher
Outcome 3
1
Graphical solution for sin 6t 
2
The solution is to be in radians – but
work in degrees
and convertEquations
at the
Solving
Trigonometric
end.
Outcome 3
Higher
Example:

(0  x  2 )
Solve 2sin(2 x  )  1
3
Step 2: consider what
solutions are expected
Step 1: Re-Arrange

2
1
sin(2 x  60 ) 
2
o
90o
(2x – 60o ) = sin-1(1/2)
Since
0  x  360
has 2 solutions
o
x2
Then

x2
0  2 x  720
o
has 4 solutions
180
o
S
A
T
C
270o
3
2
0o
Solving Trigonometric Equations
Outcome 3
Higher
Step 3: Solve the equation
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

4
210o
sin wave repeats every 360o
450o 570o
105o 225o 285o
7
5
19
12
4
12
Solving Trigonometric Equations
Higher
Outcome 3
Graphical solution for sin(2 x  60o ) 
1
2
The solution is to be in radians – but
work in degrees and convert at the end.
Solving Trigonometric Equations
Outcome 3
Higher
Harder Example: Solve
tan 2 x  3
(0  x  2 )
Step 2: consider what
solutions are expected
Step 1: Re-Arrange

2
tan x   3
90o
tan x   3
tan x   3
2 solutions
1st
and
3rd
quads
2 solutions
2nd
and
4th
quads

180
o
S
A
T
C
270o
3
2
0o
Solving Trigonometric Equations
Outcome 3
Higher
Step 3: Solve the equation
tan x   3 tan x   3
tan 1 3 

3
(60o in the 1st quadrant)
o
1st quad 2nd quad tan wave repeats every 180
x = 60o

3
120o
240o 300o
2
3
4
3
5
3
Solving Trigonometric Equations
Higher
Outcome 3
Graphical solution for
tan x  3
2
Solving Trigonometric Equations
Outcome 3
Higher
Harder Example: Solve
Step 1: Re-Arrange
3sin 2 x  4sin x  1  0 (0  x  360o )
Step 2: Consider what solutions
(3sin x  1)(sin x  1)  0
1
sin x 
3
are expected

2
90o
sin x  1

Two solutions
One solution
180
o
S
A
T
C
270o
3
2
0o
Solving Trigonometric Equations
Outcome 3
Higher
Step 3: Solve the equation
sin x 
sin x  1
1
3
Two solutions
1stquad
2nd quad
x = 19.5o 160.5o
Overall solution
One solution
90o
x = 19.5o , 90o and 160.5o
Solving Trigonometric Equations
Higher
Outcome 3
Graphical solution for
3sin 2 x  4sin x  1  0
The solution is to be in radians – but
work in degrees and convert at the end.
Solving Trigonometric Equations
Outcome 3
Higher
Harder Example:
Solve 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

2
Remember othis !
90
2
2
3  2cos x  5cos 2 x  0
(3  5cos x)(1  cos x)  0
3
cos x 
5
(0  x  2 )
cos x  1
sin   cos   1

2
cos 2   1  sin

A
S
o 2
180sin
  1  cos 2 
C
T
270o
Two solutions
One solution
3
2
0o
Solving Trigonometric Equations
Outcome 3
Higher
Step 3: Solve the equation
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
Solving Trigonometric Equations
Higher
Outcome 3
Graphical solution for
5 sin 2 x  2  2cos x
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Higher
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Higher
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Using Compound angle formula for
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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
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
1
3

2
10


4
20
1
1

2
10

4
4
2


45
2 5
5
Hint
Previous
Quit
Quit
Next
Maths4Scotland
Higher
Using triangle PQR, as shown, find the
exact value of cos 2x
Use Pythagoras
11
PR  11
Write down values for
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
cos 2x 
Simplify
Previous
cos 2 x 
  
2
11
4
7

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
Find the exact value of sin (p + q). Mark up triangles
Use Pythagoras
OA  10
Write down values for
sin p, cos p, sin q, cos q
Expand sin (p + q)
Substitute values
Simplify
Previous
6
12
OB  13
sin p 
8
,
10
8
5
13
cos p 
6
,
10
sin q 
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
and tan B 
Find the exact value of
cos 2A b)
sin(2 A c)B)
sin 2A a)
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
Substitute
Previous
.
5
13
3
A
5
B
4
12
Hypotenuses are 5 and 13 respectively
Expand sin 2A
2
Expand sin (2A + B)
5
12
2
3
,
5
cos A 
4
,
5
sin B 
3
5
sin 2 A  2  
cos 2A 
2
4
5
 4  3
   
 5 5
5
,
13

2

cos B 
12
13
24
25
16 9

25 25

7
25
sin  2 A  B   sin 2 A cos B  cos 2 Asin B
sin  2 A  B  
24 12
7
5
323




25 13
25 13 325
Quit
Quit
Hint
Next
Maths4Scotland
Higher
tan x 
If x° is an acute angle such that
4
3
4 3 3
show that the exact value of sin( x  30) is
10
Draw triangle
Use Pythagoras
Write down sin x and cos x
Expand sin (x + 30)
sin x 
5
4
x
3
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
Quit
Quit
Next
Maths4Scotland
Higher
The diagram shows two right angled triangles
24
ABD and BCD with AB = 7 cm, BC = 4 cm and CD = 3 cm.
20  6 6
Angle DBC = x° and
cos(angle
x  yABD
) isis 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)
Substitute
Simplify
Previous
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
cos( x  y ) 
4 5
3
24
  
5 7
5
7
20  3 4  6
20  6 6
20 3 24


cos( x  y ) 

35
35
35
35
Quit
Quit
Hint
Next
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 
  sin
Write down sin ½ x and cos ½ x
Expand sin ( ½ x + ½ x)
Substitute
Simplify
sin
x x

2 2
sin x 
sin
x
2
x x

2 2
 
x
x
2 h5
2
3
2 
2
3
 , cos

x
2
x
x
cos
2
2

2
5
3
x
2
 sin cos
3
4
x

2
x
2
2sin cos
x
2
5
3
4 5
9
Previous
Table of exact values
Hint
Quit
Quit
Next
Maths4Scotland
Higher
tan  
Given that
find the exact value of
Draw triangle
cos a and sin a
sin 2
Use Pythagoras
Write down values for
11

, 0  
3
2
11
a
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
6 11
sin 2a 
20
Previous
20
11
3

20
20
Quit

Quit
3 11
10
Hint
Next
Maths4Scotland
Higher
Find algebraically the exact value of
sin q   sin q  120   cos(q  150)
Expand sin (q +120)
sin q  120  sin q cos120  cosq sin120
Expand cos (q +150)
cos q  150  cosq cos150  sin q sin150
Use table of exact values
Combine and substitute
Simplify
1
2
3
2
cos 120   cos 60  
cos 150   cos 30  
sin 120 
sin 150 
sin 60 
sin q  sin q .
sin q  sin q 
1
2
3
cos q
2
sin 30 
3
2
1
2
   cosq .   cosq .   sin q . 


1
2
3
2
3
cos q
2

3
2
1
2
 sin q
1
2
0
Previous
Table of exact values
Quit
Quit
Hint
Next
Maths4Scotland
If cos q 
Higher
4

, 0 q 
5
2
sin 2q a)
Previous
4
3
sin q 
5
3 4
24
 2  
5 5
25
sin 2q  2 sin q cos q
Expand sin 4q (4q = 2q + 2q)
Find sin 4q
Opposite side = 3
4
cos q 
5
cos q and sin q
3
q
Use Pythagoras
Write down values for
Expand cos 2q
5
sin 4qb)
Draw triangle
Expand sin 2q
find the exact value of
sin 4q  2 sin 2q cos 2q
cos 2q  cos q  sin q
2
24 7
sin 4q  2  
25 25
Quit
2

16 9
7


25 25
25
336

625
Quit
Hint
Next
Maths4Scotland
Higher
For acute angles P and Q
sin P 
Show that the exact value of
Draw triangles
Write down sin P, cos P, sin Q, cos Q
13
12
P
63
65
5
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
3
5
sin Q 
sin ( P Q) 
Use Pythagoras
Expand sin (P + Q)
12
and
13
Quit

Quit
63
65
Hint
Next
Maths4Scotland
Higher
You have completed all 12 questions in this section
Previous
Quit
Quit
Back to start
Maths4Scotland
Higher
Using Compound angle formula for
Solving Equations
Continue
Quit
Quit
Maths4Scotland
Higher
Solve the equation 3cos(2 x)  10cos( x)  1  0 for 0 ≤ x ≤  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
6 cos x  10 cos x  4  0
2
Determine quadrants
S
A
T
C
3cos 2 x  5cos x  2  0
Factorise
Hence
3cos x 1 cos x  2  0
cos x 
x  1.23
1
3
cos x  2 Discard
Find acute x
Previous
acute
x  1.23
x  5.05
x  1.23 rad
Quit
or
2  1.23
rads
rads
rads
Hint
Quit
Next
Maths4Scotland
Higher
The diagram shows the graph of a cosine function from 0 to .
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
2cos 2 x   3
Solve simultaneously
cos 2x  
Rearrange
Check range
0 x 
Find acute 2x
Deduce 2x
acute
2x 
2x 

Table of exact values
T
C
x 
5
7
or
12
12
B

6
Previous
A
3
2
 0  2 x  2
6 

or
6
6
S
6 

rads
6
6
Quit
Quit
is
B
7
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)
1st
expression
2nd expression
Form equation
b)
Find expressions for:
f(g(x))
f ( g ( x))  f (2 x)  sin 2 x
cos x 
g ( f ( x))  g (sin x)  2sin x
2sin 2x  2sin x  sin 2 x  sin x
2sin x cos x  sin x
Rearrange
2sin x cos x  sin x  0
Hence
x  0  x  0, 360
for 0  x sin
 360°
Solve 2 f(g(x)) = g(f(x))
Replace sin 2x
Common factor
g(f(x)) Determine x
ii)
1
2

acute
x  60
S
A
T
C
Determine
quadrants
x  60, 300
x  0, 60, 300, 360
sin x  2cos x 1  0
sin x  0
or
2 cos x  1  0  cos x 
Previous
Table of exact values
Quit
1
2
Quit
Hint
Next
Maths4Scotland
Higher
Functions f ( x)  sin x, g ( x)  cos x
a)
b)
i)
expression
Simplify
1st
h( x )  x 

4
Find expressions for
1
1
f (h( x)) 
sin x 
cos x
2
2
Show that
1st expression
2nd
and
are defined on a suitable set of real numbers
i) f(h(x))
ii) Find a similar expression for g(h(x))
f (h( x))  g (h( x))  1 for 0  x  2
iii) Hence solve the equation
2


sin x  1
f (h( x))  f x   sin x 
Simplifies
to
2
4
4
   
g (h( x))  g  x    cos  x  


4
4
Rearrange: sin x 


4
4
f (h( x))  sin x cos  cos x sin
expr.
1
1
sin x 
2
2
Use exact values
f (h( x)) 
Similarly for 2nd expr.
g (h( x))  cos x cos  sin x sin
g (h( x)) 
Form Eqn.
ii) g(h(x))
1
2
acute x
cos x
Determine


4
4
cos x 
1
sin x
2
f (h( x))  g (h( x))  1
Previous
Table of exact values
Quit
acute
Quit
2

2
x
2
1

2 2
2

4
S
A
T
C
quadrants
x
 3
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
x  30, 90, 150
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
Table of exact values
x  150
y  cos150
Find y value
y
Coords, P
x  30, 150
for sin x
By inspection (P)

P 150, 
3
2

Hint
C
Quit
Quit
3
2
Next
Maths4Scotland
Higher
3cos(2 x)  cos( x)  1
Solve the equation
cos 2 x  2 cos 2 x  1
Replace cos 2x with
Determine quadrants
3  2 cos x  1  cos x  1
2
Substitute
Simplify
6 cos 2 x  cos x  2  0
Factorise
3cos x  2 2cos x 1  0
cos x  
Hence
Find acute x
acute
2
3
x  48
cos x 
acute
for 0 ≤ x ≤ 360°
1
2
x  60
cos x  
2
3
cos x 
acute
x  48
acute
x  60
S
A
S
A
T
C
x  132
x  228
T
C
x  60
x  300
Solutions are: x= 60°, 132°, 228° and 300°
Previous
Table of exact values
Quit
Quit
1
2
Hint
Next
Maths4Scotland
Higher



Solve the equation 2sin 2 x  6  1
Rearrange
sin
Find acute x
Note range


2x 
6
acute

2x 


6
for 0 ≤ x ≤ 2




Determine quadrants
2x 

6


and for range
6

0  x  2  0  2 x  4
S
0  2 x  2
for range
1

2
2x 

6



6

2x 

6


5
6


17
6
2  2 x  4
13
6

2x 

6
A
Solutions are:
T
x
C

6
,

2
,
7 3
,
6
2
Hint
Previous
Table of exact values
Quit
Quit
Next
Maths4Scotland
Higher
a) Write the equation cos 2q + 8 cos q + 9 = 0 in terms of cos q
and show that for cos q it has equal roots.
b) Show that there are no real roots for q
Replace cos 2q with
Rearrange
Divide by 2
Factorise
Deduction
cos 2q  2 cos 2 q  1
Try to solve:
 cosq  2  0
2 cos 2 q  8cos q  8  0
cosq  2
cos 2 q  4 cos q  4  0
No solution
Hence there are no real solutions for q
 cosq  2 cosq  2  0
Equal roots for cos q
Hint
Previous
Quit
Quit
Next
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  2cos 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
Table of exact values
C
x  120, 240
sin x  0  x  0, 360
cos x   
T
Quit
120°, 240°, 360°
Quit
Hint
Next
Maths4Scotland
Higher
Find the exact solutions of 4sin2 x = 1, 0  x  2
Rearrange
Take square roots
Find acute x
sin 2 x 
1
4
sin x  
1
2
x 

acute
Determine quadrants for sin x
S
6
+ and – from the square root requires all 4 quadrants
A

T
C
5
7 11
x  ,
,
,
6
6
6
6
Hint
Previous
Table of exact values
Quit
Quit
Next
Maths4Scotland
Higher
cos2x  cos x  0
Solve the equation
Replace cos 2x with
cos 2 x  2 cos 2 x  1
Determine quadrants
cos x 
2 cos 2 x  1  cos x  0
Substitute
Simplify
2 cos 2 x  cos x  1  0
Factorise
 2cos x 1 cos x  1  0
cos x 
Hence
Find acute x
acute
1
2
x  60
for 0 ≤ x ≤ 360°
cos x  1
1
2
acute
x  60
S
A
T
C
x  60
x  300
x  180
Solutions are: x= 60°, 180° and 300°
Previous
Table of exact values
Quit
Quit
Hint
Next
Maths4Scotland
Higher
Solve algebraically, the equation
Replace cos 2x with
Substitute
cos2x  5cos x  2  0
cos 2 x  2 cos 2 x  1
Determine quadrants
2 cos x  1  5cos x  2  0
2
cos x 
acute
Simplify
2 cos 2 x  5cos x  3  0
Factorise
 2cos x 1 cos x  3  0
Hence
Find acute x
cos x 
acute
1
2
x  60
for 0 ≤ x ≤ 360°
S
cos x  3
Discard above
1
2
x  60
A
T
C
x  60
x  300
Solutions are: x= 60° and 300°
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Maths4Scotland
Higher
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Maths4Scotland
Higher
Table of exact values
sin
cos
tan
Return
30°
45°
60°

6
1
2

4

3
1
2
1
2
3
2
3
2
1
3
1
1
2
3
Maths4Scotland
Higher
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