The document provides information about trigonometric identities of the form sin(A+B) and double angle formulae. It includes:
1) Trigonometric identity formulas for sin(A+B), sin(A-B), cos(A+B), and cos(A-B).
2) Examples of using the identity formulas to simplify trigonometric expressions and prove identities.
3) Double angle formulas for sin(2A), cos(2A) and their uses in finding exact trig values and solving trig equations.
1. 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
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2. Trig Identities
Outcome 3
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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!!
6. Trig Identities
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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
7. Higher
Trig Identities
Outcome 3
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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
8. 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
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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
9. Trig Identities
Higher
Solve
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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
10. Trig Identities
Higher
Example 7
Outcome 3
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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
11. Paper 1 type
questions
Trig Identities
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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
13. Double Angle Formulae
Outcome 3
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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
14. Double Angle formulae
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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.
15. Double Angle formulae
Outcome 3
Higher
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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
16. Double Angle formulae
Outcome 3
Higher
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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
17. Double Angle formulae
Outcome 3
Higher
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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
18. Trigonometric Equations
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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
19. Trigonometric Equations
Outcome 3
Higher
cos 2 x o − 4sin x o + 5 = 0 for 0 ≤ x ≤ 360o.
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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
20. Trigonometric Equations
Outcome 3
Higher
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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
23. Trigonometric Equations
Outcome 3
Higher
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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
24. Trigonometric Equations
Outcome 3
Higher
f ( x ) = g ( x) algebraically.
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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:
25. Trigonometric Equations
Outcome 3
Higher
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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
32. Solving Trigonometric Equations
Outcome 3
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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
33. Solving Trigonometric Equations
Outcome 3
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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
34. Solving Trigonometric Equations
Outcome 3
Higher
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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
36. Solving Trigonometric Equations
Outcome 3
Higher
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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
37. Solving Trigonometric Equations
Outcome 3
Higher
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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
39. The solution is to be in radians – but
work in Trigonometric at the
Solvingdegrees and convertEquations
end.
Outcome 3
Higher
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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
40. Solving Trigonometric Equations
Outcome 3
Higher
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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
π
4
210o
sin wave repeats every 360o
450o 570o
105o 225o
7π
5π
12
4
285o
19π
12
42. The solution is to be in radians – but
work in degrees and convert at the end.
Solving Trigonometric Equations
Outcome 3
Higher
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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
43. Solving Trigonometric Equations
Outcome 3
Higher
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Step 3: Solve the equation
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
45. Solving Trigonometric Equations
Outcome 3
Higher
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Harder Example: Solve
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
46. Solving Trigonometric Equations
Outcome 3
Higher
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Step 3: Solve the equation
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
48. The solution is to be in radians – but
work in degrees and convert at the end.
Solving Trigonometric Equations
Outcome 3
Higher
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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
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
49. Solving Trigonometric Equations
Outcome 3
Higher
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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
53. 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
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54. 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
55. 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
56. Maths4Scotland
Higher
Using triangle PQR, as shown, find the
11
exact value of cos 2x
Use Pythagoras
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
( ) −( )
Simplify
Previous
2
11
cos 2x =
4
7
cos 2 x =
−
11
11
Quit
2
7
11
→ −
2
3
11
Quit
Hint
Next
57. 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
Write down values for
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
58. 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
Quit
Quit
Hint
Next
59. 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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Next
60. 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
Previous
Quit
Quit
Hint
Next
61. 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
Previous
Table of exact values
Hint
Quit
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62. 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
Write down values for
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
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63. 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
Previous
Table of exact values
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Hint
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64. Maths4Scotland
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
Write down values for
Expand sin 2θ
5
24 7
×
25 25
Quit
2
⇒
16 9
7
⇒
−
⇒
25 25
25
336
625
Quit
Hint
Next
65. 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
Quit
⇒
Quit
63
65
Hint
Next
68. 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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69. 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 =
Table of exact values
6π π
+
rads
6
6
Quit
Quit
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
70. 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 =
Previous
Table of exact values
Quit
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
Quit
Hint
Next
71. 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
Previous
Table of exact values
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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
72. 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
Table of exact values
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
73. 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°
Previous
Table of exact values
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Hint
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74. 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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Table of exact values
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75. 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
Previous
Quit
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76. 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
Table of exact values
C
x = 120°, 240°
sin x = 0 ⇒ x = 0°, 360°
cos x = − ⇒
T
Quit
120°, 240°, 360°
Quit
Hint
Next
77. 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
Previous
Table of exact values
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78. 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°
Previous
Table of exact values
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Hint
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79. 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°
Previous
Table of exact values
Quit
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Hint
Next