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Higher
Mathematics
UNIT 2 OUTCOME 3
Trigonometry
Contents
Trigonometry
1
2
3
4
5
Solving Trigonometric Equations
Trigonometry in Three Dimensions
Compound Angles
Double-Angle Formulae
Further Trigonometric Equations
HSN22300
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Higher Mathematics
Unit 2 – Trigonometry
OUTCOME 3
Trigonometry
1
Solving Trigonometric Equations
You should already be familiar with solving some trigonometric equations.
EXAMPLES
1. Solve sin x ° = 12 for 0 < x < 360 .
sin x ° = 12
180° − x °
x°
S A
Since sin x ° is positive
T
C
180° + x °
360° − x °
First quadrant solution:
x = sin −1 ( 12 )
= 30.
x = 30
or 180 − 30
x = 30
or 150.
Remember
The exact value triangle:
30°
2
60°
1
2. Solve cos x ° = − 1 for 0 < x < 360 .
5
cos x ° = − 1
180° − x °
x°
S A
Since cos x ° is negative
T C 360° − x °
180° + x °
5
( 5)
x = cos −1 1
= 63·435 (to 3 d.p.).
x = 180 − 63·435
or 180 + 63·435
x = 116·565
243·435.
or
3. Solve sin x ° = 3 for 0 < x < 360 .
There are no solutions since −1 ≤ sin x ° ≤ 1 .
Note that −1 ≤ cos x ° ≤ 1 , so cos x ° = 3 also has no solutions.
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3
Higher Mathematics
Unit 2 – Trigonometry
4. Solve tan x ° = −5 for 0 < x < 360 .
180° − x °
x°
tan x ° = −5
S A
Since tan x ° is negative
T
C
180° + x °
360° − x °
x = tan −1 ( 5 )
= 78·690 (to 3 d.p.).
x = 180 − 78·690
or
x = 101·310
281·310.
or
360 − 78·690
Note
All trigonometric equations we will meet can be reduced to problems like
those above. The only differences are:
the solutions could be required in radians – in this case, the question will
not have a degree symbol, e.g. “Solve 3 tan x = 1 ” rather than “ 3tan x ° = 1 ”;
exact value solutions could be required in the non-calculator paper – you
will be expected to know the exact values for 0, 30, 45, 60 and 90 degrees.
Questions can be worked through in degrees or radians, but make sure the
final answer is given in the units asked for in the question.
EXAMPLES
5. Solve 2sin 2 x ° − 1 = 0 where 0 ≤ x ≤ 360 .
180° − 2 x °
2x°
2sin2 x ° = 1
0 ≤ x ≤ 360 Remember
S A
1
sin2 x ° = 2
T C 360° − 2 x ° 0 ≤ 2 x ≤ 720 The exact value triangle:
180° + 2 x °
2 x = sin −1 ( 12 )
= 30.
2 x = 30 or 180 − 30
30°
2
3
60°
1
or 360 + 30 or 360 + 180 − 30
or 360 + 360 + 30
2 x = 30 or 150 or 390 or 510
x = 15 or 75 or 195 or 255.
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Note
There are more solutions
every 360°, since
sin(30°) = sin(30°+360°) = …
So keep adding 360 until
2x > 720.
HSN22300
Higher Mathematics
6. Solve
Unit 2 – Trigonometry
2 cos 2 x = 1 where 0 ≤ x ≤ π .
cos 2x = 1
2
2x
π − 2x
0≤ x ≤π
S A
π + 2 x T C 2 π − 2 x 0 ≤ 2 x ≤ 2π
2 x = cos
−1
( )
1
2
Remember
The exact value triangle:
= π4 .
2 x = π4 or 2π − π4
π
4 1
2
π
4
1
or 2π + π4
2 x = π4 or 74π
x = π8 or 78π .
7. Solve 4cos 2 x = 3 where 0 < x < 2 π .
( cos x )2 = 34
cos x = ± 34
cos x = ± 23
S A Since cos x can be positive or negative
T C
Remember
The exact value triangle:
( )
x = cos −1 23
= π6 .
π
6
2
x = π6 or π − π6 or π + π6 or 2π − π6
3
π
3
or 2π + π6
1
x = π6 or 56π or 76π or 116π .
8. Solve 3tan ( 3 x° − 20° ) = 5 where 0 ≤ x ≤ 360 .
3tan ( 3 x ° − 20° ) = 5
tan ( 3 x ° − 20° ) = 53
S A
T C
0 ≤ x ≤ 360
0 ≤ 3 x ≤ 1080
−20 ≤ 3 x − 20 ≤ 1060
()
3 x − 20 = tan −1 53
= 59·036 (to 3 d.p.).
3 x − 20 = 59·036 or 180 + 59·036
or 360 + 59·036 or 360 + 180 + 59·036
or 360 + 360 + 59·036 or 360 + 360 + 180 + 59·036
or 360 + 360 + 360 + 59·036
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Unit 2 – Trigonometry
3 x − 20 = 59·036 or 239·036 or 419·036
or 599·036 or 779·036 or 959·036
3 x = 79·036 or 259·036 or 439·036
or 619·036 or 799·036 or 979·036
x = 26·35 or 86·35 or 146·35 or 206·35 or 266·35 or 326·35.
(
)
9. Solve cos 2 x + π3 = 0·812 for 0 < x < 2 π .
(
)
S A
T C
cos 2 x + π3 = 0·812
0 < x < 2π
0 < 2 x < 4π
π < 2 x + π < 4π + π
3
3
3
1·047 < 2 x + π3 < 13·614 (to 3 d.p.)
Remember
2 x + π3 = cos −1 ( 0·812 )
Make sure your
= 0·623 (to 3 d.p.).
calculator uses radians.
2 x + π3 = 0·623 or 2π − 0·623
or 2π + 0·623 or 2π + 2π − 0·623
or 2π + 2π + 0·623 or 2π + 2π + 2π − 0·623
2 x + π3 = 5.660 or 6.906 or 11.943 or 13.189
2 x = 4.613 or 5.859 or 10.896 or 12.142
x = 2.307 or 2.930 or 5.448 or 6.071.
2
Trigonometry in Three Dimensions
It is possible to solve trigonometric problems in three dimensions using
techniques we already know from two dimensions. The use of sketches is
often helpful.
The angle between a line and a plane
The angle a between the plane P and the line ST is calculated by adding a
line perpendicular to the plane and then using basic trigonometry.
T
a
P
S
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Higher Mathematics
Unit 2 – Trigonometry
EXAMPLE
1. The triangular prism ABCDEF is shown below.
E
3 cm
F
B
C
6 cm
A
10 cm
Calculate the acute angle between:
(a) The line AF and the plane ABCD.
(b) AE and ABCD.
(a) Start with a sketch:
F
tan a =
a
10 cm
Opposite 3
=
Adjacent 10
( )
3
a = tan −1 10
3 cm
A
D
= 16.699° (or 0.291 radians) (to 3 d.p.).
D
Note
Since the angle is in a right-angled triangle, it must be acute so there is
no need for a CAST diagram.
(b) Again, make a sketch:
E
3 cm
C
b
angle to be
A
calculated
We need to calculate the length of AC first using Pythagoras’s Theorem:
C
AC = 102 + 6 2
6 cm
= 136.
A
D
10 cm
E
Therefore:
3 cm
tan b =
C
b
A
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Opposite
= 3
Adjacent 136
b = tan −1
( 1363 )
= 14.426° (or 0.252 radians) (to 3 d.p.).
136 cm
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Unit 2 – Trigonometry
The angle between two planes
The angle a between planes P and Q is calculated by adding a line
perpendicular to Q and then using basic trigonometry.
P
a
Q
EXAMPLE
2. ABCDEFGH is a cuboid with dimensions 12 × 8 × 8 cm as shown below.
H
G
F
E
D
8 cm
C
8 cm
A
12 cm
B
(a) Calculate the size of the angle between the planes AFGD and ABCD.
(b) Calculate the size of the acute angle between the diagonal planes
AFGD and BCHE.
(a) Start with a sketch:
F
tan a =
8 cm
A
Opposite 8
=
Adjacent 12
( )
a = tan −1 23
Note
Angle GDC is the same
size as angle FAB.
= 33.690° (or 0.588 radians) (to 3 d.p.).
a
12 cm
B
(b) Again, make a sketch:
Let AF and BE intersect at T.
= TBA
ɵ = 33.690° .
E
F ATB is isosceles, so TAB
= 180° − ( 33.690° + 33.690° )
ATB
Note
= 112.620°.
T
A
•
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•
The angle could also
B
So the acute angle is:
have been calculated
.
BTF = ATE = 180° − 112 620° using rectangle DCGH.
= 67.380° (or 1.176 radians) (to 3 d.p.).
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3
Unit 2 – Trigonometry
Compound Angles
When we add or subtract angles, the result is called a compound angle.
For example, 45° + 30° is a compound angle. Using a calculator, we find:
• sin ( 45° + 30° ) = sin ( 75° ) = 0.966 ;
•
sin ( 45° ) + sin ( 30° ) = 1.207 (both to 3 d.p.).
This shows that sin ( A + B ) is not equal to sin A + sin B . Instead, we can
use the following identities:
sin ( A + B ) = sin A cos B + cos A sin B
sin ( A − B ) = sin A cos B − cos A sin B
cos ( A + B ) = cos A cos B − sin A sin B
cos ( A − B ) = cos A cos B + sin A sin B .
These are given in the exam in a condensed form:
sin ( A ± B ) = sin A cos B ± cos A sin B
cos ( A ± B ) = cos A cos B ∓ sin A sin B .
EXAMPLES
1. Expand and simplify cos ( x ° + 60° ) .
cos ( x ° + 60° ) = cos x ° cos 60° − sin x ° sin 60°
Remember
The exact value triangle:
= 12 cos x ° − 23 sin x °.
30°
2
3
60°
1
2. Show that sin ( a + b ) = sin a cos b + cos a sin b for a = π6 and b = π3 .
LHS = sin ( a + b )
(
= sin π6 + π3
)
RHS = sin a cos b + cos a sin b
= sin π6 cos π3 + cos π6 sin π3
(
= sin ( π2 )
= ( 12 × 12 ) + 23 × 23
= 1.
= 14 + 34 = 1.
)
Since LHS = RHS , the claim is true for a = π6 and b = π3 .
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Remember
The exact value triangle:
π
6
2
3
π
3
1
HSN22300
Higher Mathematics
Unit 2 – Trigonometry
3. Find the exact value of sin75° .
sin 75° = sin ( 45° + 30° )
= sin 45° cos30° + cos 45° sin30°
(
) (
= 1 × 23 + 1 × 12
2
2
)
= 3 + 1
2 2 2 2
= 23 +21
= 6 +4 2 .
Finding Trigonometric Ratios
You should already be familiar with the following formulae (SOH CAH TOA).
Hypotenuse
Opposite
a
Adjacent
sin a =
Opposite
Hypotenuse
cos a =
Adjacent
Hypotenuse
tan a =
Opposite
.
Adjacent
p
If we have sin a = q where 0 < a < π2 , then we can form a right-angled
triangle to represent this ratio.
Since sin a =
q
a
p
Opposite
p
= q then:
Hypotenuse
the side opposite a has length p;
the hypotenuse has length q.
The length of the unknown side can be found using Pythagoras’s Theorem.
Once the length of each side is known, we can find cos a and tan a using
SOH CAH TOA.
The method is similar if we know cos a and want to find sin a or tan a .
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Unit 2 – Trigonometry
EXAMPLES
5 . Show that
4. Acute angles p and q are such that sin p = 54 and sin q = 13
sin ( p + q ) = 63
65 .
5
sin p = 54
4
13
q
12
cos p = 35
p
3
5
5
sin q = 13
cos q = 12
13
sin ( p + q ) = sin p cos q + cos p sin q
(
) (
3 5
= 54 × 12
13 + 5 × 13
)
48 + 15
= 65
65
= 63
65 .
Note
Since “Show that” is
used in the question, all
of this working is
required.
Confirming Identities
EXAMPLES
5. Show that sin ( x − π2 ) = − cos x .
sin ( x − π2 )
= sin x cos π2 − cos x sin π2
= sin x × 0 − cos x × 1
= − cos x .
sin ( s + t )
= tan s + tan t for cos s ≠ 0 and cos t ≠ 0 .
cos s cos t
sin ( s + t ) sin s cos t + cos s sin t
=
cos s cos t
cos s cos t
sin s cos t cos s sin t
=
+
cos s cos t cos s cos t
Remember
sin s sin t
sin x
= tan x .
=
+
cos x
cos s cos t
= tan s + tan t .
6. Show that
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4
Unit 2 – Trigonometry
Double-Angle Formulae
Using the compound angle identities with A = B , we obtain expressions for
sin 2 A and cos2 A . These are called double-angle formulae.
sin 2 A = 2 sin A cos A
cos 2 A = cos 2 A − sin 2 A
= 2 cos 2 A − 1
= 1 − 2 sin 2 A.
Note that these are given in the exam.
EXAMPLES
1. Given that tanθ = 34 , where 0 < θ < π2 , find the exact value of sin2θ and
cos2θ .
5
4
θ
sin 2θ = 2sin θ cos θ
sin θ = 34
cos θ = 35
3
cos 2θ = cos 2 θ − sin 2 θ
2
2
= 2 × 54 × 35
= 35
= 24
25 .
9 − 16 Note
= 25
25
( ) − ( 54 )
7 .
= − 25
Any of the cos2A
formulae could have
been used here.
5 , where
2. Given that cos2 x = 13
0 < x < π , find the exact values of sin x
and cos x .
Since cos2 x = 1 − 2sin 2 x ,
5
1 − 2 sin 2 x = 13
8
2 sin 2 x = 13
8
sin 2 x = 26
4
= 13
sin x = ±
2 .
13
We are told that 0 < x < π , so only sin x =
13
2
x
a=
2
13
is possible.
2
13 − 2 2 = 13 − 4 = 9 = 3.
a
So cos x =
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3 .
13
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Unit 2 – Trigonometry
Further Trigonometric Equations
We will now consider trigonometric equations where double-angle formulae
can be used to find solutions. These equations will involve:
sin 2x and either sin x or cos x ;
Remember
The double-angle
cos 2x and cos x ;
formulae are given in
cos 2x and sin x .
the exam.
Solving equations involving sin2x and either sinx or cosx
EXAMPLE
1. Solve sin 2 x ° = − sin x ° for 0 ≤ x < 360 .
2 sin x ° cos x ° = − sin x °
2 sin x ° cos x ° + sin x ° = 0
sin x ° ( 2 cos x ° + 1) = 0
sin x ° = 0
Replace sin 2x using the double angle
formula
Take all terms to one side, making the
equation equal to zero
Factorise the expression and solve
2cos x ° + 1 = 0
x = 0 or 180 or 360
S A
T C
cos x ° = − 12
x = 180 − 60 or 180 + 60
x = cos −1 ( 12 )
= 60
= 120 or 240.
So x = 0 or 120 or 180 or 240 .
Solving equations involving cos2x and cosx
EXAMPLE
2. Solve cos 2 x = cos x for 0 ≤ x ≤ 2 π .
cos 2 x = cos x
2 cos2 x − 1 = cos x
2 cos2 x − cos x − 1 = 0
( 2 cos x + 1)( cos x − 1) = 0
2 cos x + 1 = 0
cos x = − 12
x = π − π3 or π + π3
= 23π or 43π
Replace cos 2x by 2cos2 x − 1
Take all terms to one side, making a
quadratic equation in cos x
Solve the quadratic equation (using
factorisation or the quadratic formula)
S A
T C
x = cos −1 ( 12 )
cos x − 1 = 0
cos x = 1
x = 0 or 2π .
= π3
So x = 0 or 23π or 43π or 2π .
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Unit 2 – Trigonometry
Solving equations involving cos2x and sinx
EXAMPLE
3. Solve cos 2 x = sin x for 0 < x < 2 π .
cos2 x = sin x
1 − 2 sin 2 x = sin x
2 sin 2 x + sin x − 1 = 0
( 2 sin x − 1)( sin x + 1) = 0
2 sin x − 1 = 0
sin x = 12
x = π6 or π − π6
= π6
or
5π
6
Replace cos 2x by 1 − 2 sin 2 x
Take all terms to one side, making
a quadratic equation in sin x
Solve the quadratic equation (using
factorisation or the quadratic
formula)
sin x + 1 = 0
sin x = −1
S A
T C
x = sin
−1
( )
1
2
x = 32π .
= π6
So x = π6 or 56π or 32π .
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