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Higher Maths
Strategies
Compound Angles
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Maths4Scotland
Higher
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Compound Angles
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Maths4Scotland
Higher
This presentation is split into two parts
Using Compound angle formula for
Exact values
Solving equations
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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
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Maths4Scotland
Higher
In triangle ABC show that the exact value of
sin(a  b) is
Use Pythagoras
Substitute values
Simplify
10
2
AC  2 CB  10
Write down values for
sin a, cos a, sin b, cos b
Expand sin (a + b)
2
5
sin a 
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
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Maths4Scotland
Higher
Using triangle PQR, as shown, find the
exact value of cos 2x
11
PR  11
Use Pythagoras
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
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Maths4Scotland
Higher
On the co-ordinate diagram shown, A is the point (6, 8) and
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
10
6
12
OB  13
sin p 
8
,
10
cos p 
6
,
10
sin q 
5
,
13
sin ( p  q)  sin p cos q  cos p sin q
Substitute values
sin ( p  q) 
Previous
sin ( p  q) 
96
130

30
130
Quit
5
13
Expand sin (p + q)
Simplify
8
8 12

10 13


12
13
6
5

10 13
126
130
Quit
cos q 

63
65
Hint
Next
Maths4Scotland
Higher
A and B are acute angles such that tan A 
Find the exact value of
a) sin 2A
b) cos 2A
Draw triangles
c)
and tan 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
sin(2 A  B )
Use Pythagoras
Write down sin A, cos A, sin B, cos B
3
4
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
cos B 

24
25
2
16 9

25 25

12
13

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
If x° is an acute angle such that tan x 
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)
Hypotenuse is 5
sin x 
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
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Table of exact values
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Maths4Scotland
Higher
The diagram shows two right angled triangles
ABD and BCD with AB = 7 cm, BC = 4 cm and CD = 3 cm.
Angle DBC = x° and angle ABD is y°.
20  6 6
Show that the exact value of cos( x  y ) is
35
24
5
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 
24
,
7
cos y 
5
7
cos( x  y )  cos x cos y  sin x sin y
4 5
3
24
  
5 7
5
7
Substitute
cos( x  y ) 
Simplify
20  3 4  6
20  6 6
20 3 24


cos( x  y ) 

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
h 5
Use Pythagoras
Draw in perpendicular
3
Use fact that sin x = sin ( ½ x + ½ x)

sin 
  sin
Write down sin ½ x and cos ½ x
Substitute
Simplify
sin

x x

2 2
sin x 
sin
x
2
x x

2 2
Expand sin ( ½ x + ½ x)

x
x
2 h5
2
3
2 

2
3
, cos

x
x
cos
2
2
x
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
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Maths4Scotland
Given that
tan  
Higher
11

, 0  
3
2
find the exact value of
sin 2
Draw triangle
Use Pythagoras
Write down values for
cos a and sin a
20
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
11
11
3

20
20
Quit

Quit
3 11
10
Hint
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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
Simplify
cos 150   cos 30  
sin 120 
sin 150 
sin 60 
sin q  sin q .
Combine and substitute
1
2
3
2
cos 120   cos 60  
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
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Hint
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Maths4Scotland
If cos q 
a)
sin 2q
Higher
4

, 0 q 
5
2
b)
Draw triangle
Find sin 4q
Previous
3
q
Use Pythagoras
Opposite side = 3
4
cos q 
5
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)
Expand cos 2q
5
sin 4q
Write down values for
cos q and sin q
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
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Maths4Scotland
Higher
For acute angles P and Q
sin P 
12
and
13
sin Q 
Show that the exact value of sin ( P Q ) 
Draw triangles
Use Pythagoras
Write down sin P, cos P, sin Q, cos Q
Expand sin (P + Q)
3
5
63
65
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
13
Quit

Quit
63
65
Hint
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Maths4Scotland
Higher
You have completed all 12 questions in this section
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Maths4Scotland
Higher
Using Compound angle formula for
Solving Equations
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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
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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) Find expressions for:
i) f(g(x))
ii) g(f(x))
Determine x
b) Solve 2 f(g(x)) = g(f(x)) for 0  x  360°
1st
expression
2nd expression
Form equation
Replace sin 2x
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
Common factor
1
2

acute
x  60
S
A
T
C
Determine
quadrants
x  60, 300
2sin x cos x  sin x  0
Rearrange
Hence
sin x  0  x  0, 360
x  0, 60, 300, 360
sin x  2cos x 1  0
sin x  0
or
2 cos x  1  0  cos x 
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Table of exact values
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1
2
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Maths4Scotland
Higher
Functions f ( x)  sin x, g ( x)  cos x
a)
Find expressions for
b)
i)
2nd
expression
expression
Simplify
h( x )  x 

4
are defined on a suitable set of real numbers
i) f(h(x)) ii) g(h(x))
1
1
sin x 
cos x ii) Find a similar expression for g(h(x))
2
2
Hence solve the equation f (h( x))  g (h( x))  1 for 0  x  2
Show that
iii)
1st
and
1st
f (h( x)) 
   
g (h( x))  g  x    cos  x  

f (h( x))  f x 
 sin x 
4

4


4
4
Rearrange:


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.
1
2
acute x
cos x


4
4
cos x 
2
2
Simplifies to
1
sin x
2
acute
Determine
quadrants
x
 3
4
,
sin x 
sin x  1
2

2
x

4
S
A
T
C
4
Hint
f (h( x))  g (h( x))  1
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Table of exact values
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2
1

2 2
2
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Maths4Scotland
a)
b)
Higher
Solve the equation sin 2x - cos x = 0 in the interval 0  x  180°
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
for sin x
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Table of exact values
x  150
y  cos150
Find y value
y
Coords, P
x  30, 150
T
By inspection (P)

P 150, 
3
2

Hint
C
Quit
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3
2
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Maths4Scotland
Solve the equation
Higher
3cos(2 x)  cos( x)  1
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
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
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
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1
2
Hint
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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
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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
cos 2q  2 cos 2 q  1
Rearrange
2 cos 2 q  8cos q  8  0
Divide by 2
cos 2 q  4 cos q  4  0
Factorise
Deduction
Try to solve:
 cosq  2  0
cosq  2
No solution
Hence there are no real solutions for q
 cosq  2 cosq  2  0
Equal roots for cos q
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
Common factor
sin x  2cos x  1  0
Hence
S
A
T
C
sin x  0
or
Determine x
Determine quadrants
for cos x
2 cos x  1  0

cos x  
1
2
x  120, 240
sin x  0  x  0, 360
1
2
cos x   
acute
x  60
x = 0°,
Previous
Table of exact values
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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  2
Rearrange
sin 2 x 
1
4
Take square roots
sin x  
1
2
Find acute x
acute
x 

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
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Table of exact values
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Maths4Scotland
Higher
Solve the equation
cos 2x  cos x  0
Replace cos 2x with
cos 2 x  2 cos 2 x  1
for 0 ≤ x ≤ 360°
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
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°
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Higher
cos2x  5cos x  2  0
Solve algebraically, the equation
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
 2cos x 1 cos x  3  0
Hence
Find acute x
cos x 
acute
1
2
x  60
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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Table of exact values
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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