Download Trigonometry

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Trigonometry
2.1 Kick off with CAS
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2.2 Reciprocal trigonometric functions
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2.3 Trigonometric identities using reciprocal trigonometric
functions
2.4 Compound-angle formulas
2.5 Double-angle formulas
2.6 Inverse trigonometric functions
2.7 General solutions of trigonometric equations
2.8 Graphs of reciprocal trigonometric functions
2.9 Graphs of inverse trigonometric functions
2.10 Review
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2.1 Kick off with CAS
Exploring inverse trigonometric functions
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In this topic, we will investigate the inverse trigonometric functions.
1 Using CAS, determine each of the following. Remember to have the calculator in
radians mode.
2
a cos−1 acosa b b
5
−1
b cos (cos(3))
c cos−1 (cos(6))
π
7
e cos−1 acosa
PR
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d cos−1 acosa b b
8π
bb
7
TE
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PA
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4π
bb
3
g cos−1 ( cos ( −π ) )
For what values of x is cos−1 ( cos ( x ) ) = x? Confirm your result using CAS.
2 Using CAS, determine each of the following.
1
4
a tan−1 atana b b
b tan−1 atana− b b
3
5
f cos−1 acosa
c tan−1 ( tan ( 6 ) )
7π
bb
5
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e tan−1 atana
π
5
d tan−1 atana b b
π
3
f tan−1 atana− b b
4π
bb
3
For what values of x is tan−1 (tan(x)) = x? Confirm your result using CAS.
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g tan−1 atana
Please refer to the Resources tab in the Prelims section of your eBookPlUs for a comprehensive
step-by-step guide on how to use your CAS technology.
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AOS 1
Topic 2
Concept 1
Reciprocal circular
functions
Concept summary
Practice questions
History of trigonometry
The word trigonometry is derived from the Greek words trigonon and metron,
meaning ‘triangle’ and ‘measure’. Trigonometry is the branch of mathematics that
deals with triangles and the relationships between the angles and sides of a triangle.
Trigonometry was originally devised in the third century BC to meet the needs of the
astronomers of those times. Hipparchus was a Greek astronomer and mathematician
and is considered to be the founder of trigonometry, as he compiled the first
trigonometric tables in about 150 BC.
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2.2
Reciprocal trigonometric functions
Definitions of trigonometric ratios
tan(θ) =
hypotenuse
adjacent
hypotenuse
opposite
=
=
=
b
a
b
c
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cos(θ) =
opposite
a
c
c
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sin(θ) =
E
PR
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The following is a review of trigonometry, which is needed for the rest of this topic
and subsequent work in this book.
The trigonometric functions sin(x), cos(x) and tan(x) are defined in terms of
the ratio of the lengths of the sides of a right-angled triangle. Let the lengths of
the three sides of the triangle be a, b and c, and let the angle between sides a
and c be θ.
adjacent
θ
b
a
R
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Pythagoras’ theorem states that in any right-angled triangle, the square of the
hypotenuse is equal to the sum of the squares of the other two sides. That is:
a2 + b2 = c2 .
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The unit circle
y
(0, 1) θ = —2π
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An alternative definition of the trigonometric
functions is based on the unit circle, which is a
circle with radius one unit and centre at the origin.
The unit circle has the equation x2 + y2 = 1.
The coordinate of any point P (x, y) on the unit
circle is defined in terms of the trigonometric
functions OR = x = cos(θ) and RP = y = sin(θ),
where θ is the angle measured as a positive
angle, anticlockwise from positive direction of
the x-axis. The trigonometric functions are also
called circular functions as they are based on the
unit circle.
58 1
(−1, 0)
θ=π
O
θ
x
T
P
tan (θ)
y
R
(1, 0)
S x
θ=0
(0, −1) θ = 3π
—
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By substituting x = cos(θ) and y = sin(θ) into the equation x2 + y2 = 1,
we can derive the relationship sin2 (θ) + cos2 (θ) = 1.
Note that sin2 (θ) = (sin(θ)) 2 and cos2 (θ) = (cos(θ)) 2.
The vertical distance from S to T is defined as tan(θ). As the triangles ΔORP and
ΔOST are similar,
RP ST tan(θ)
=
=
OR OS
1
FS
y
sin(θ)
= tan(θ)
=
x cos(θ)
Angles of any magnitude
E
PR
O
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In the diagram of the unit circle, consider the point (0, 1) on the y-axis. This
π
point corresponds to the angle θ = 90° or radians rotated from the positive
2
end of the x-axis. Since the sine of the angle is the y-coordinate, it follows that
π
sin(90°) = sin a b = 1. Since the cosine of the angle is the x-coordinate, it follows
2
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π
that cos (90°) = cos a b = 0. The tangent is the value of sine divided by the cosine;
2
π
because we cannot divide by zero, the tan of θ = 90° or radians is undefined.
2
Similarly for the point (–1, 0), where θ = 180° or π radians, it follows that
cos(180°) = cos(π) = −1 and sin(180°) = sin(π) = 0.
The diagram can be used to obtain the trigonometric value of any multiple of 90°, and
these results are summarised in the following table.
Angle (degrees)
0°
Angle (radians)
0
sin (θ)
180°
270°
π
0
1
0
3π
2
−1
cos (θ)
1
0
−1
0
1
tan (θ)
0
Undefined
0
Undefined
0
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90°
π
2
360°
2π
0
Note: Whenever an angle measurement is shown without a degree symbol in this
topic, assume that it is measured in radians.
The first quadrant
π
The angle in the first quadrant is 0° < θ < 90° in degrees or 0 < θ < in radians.
2
In the first quadrant, x > 0 and y > 0, so cos(θ) > 0 and sin(θ) > 0; therefore,
tan(θ) > 0. The following table shows values derived from triangles in the first
quadrant using the trigonometric ratios. You should memorise these values, as they are
used extensively in this topic.
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0°
Angle (radians)
0
sin (θ)
0
cos (θ)
1
tan (θ)
0
30°
π
6
1
2
45°
π
4
60°
π
3
!3
2
!2
2
!2
2
!3
2
1
2
!3
3
1
!3
90°
π
2
1
0
Undefined
FS
Angle (degrees)
PR
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Note that sin(30° + 60°) ≠ sin(90°) and in general
sin(A + B) ≠ sin(A) + sin(B),
cos(A + B) ≠ cos(A) + cos(B) and
tan(A + B) ≠ tan(A) + tan(B).
The formulas for sin(A + B) are called compound angle formulas. They are studied in
greater depth in Section 2.4.
y
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The second quadrant
P (a, b)
π− θ
−a O
θ
a
T
b
x
tan (π − θ)
Tʹ
= sin(θ)
= −cos(θ)
= −tan(θ)
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The angle in the second quadrant is 90° < θ < 180°
π
in degrees or < θ < π in radians. In the second
Pʹ (−a, b)
2
quadrant, x < 0 and y > 0, so cos(θ) < 0 and
b
sin(θ) > 0; therefore, tan(θ) < 0.
Consider the point P (a, b) in the first quadrant.
When this point is reflected in the y-axis, it
becomes the point P′(–a, b). If P makes an angle
of θ with the x-axis, then P′ makes an angle of
180 − θ degrees or π − θ radians with the x-axis.
From the definitions of sine and cosine, we obtain
the following relationships.
sin(180° − θ) = sin(θ) sin(π − θ)
cos(180° − θ) = −cos(θ) cos(π − θ)
tan(180° − θ) = −tan(θ) tan(π − θ)
For example:
5π
π
π
1
sin a b = sin aπ − b = sin a b =
6
6
6
2
cos a
tan a
The third quadrant
!2
3π
π
π
b = cos aπ − b = −cos a b = −
4
4
4
2
2π
π
π
b = tan aπ − b = −tan a b = −!3
3
3
3
3π
in
2
radians. In the third quadrant, x < 0 and y < 0, so cos(θ) < 0 and sin(θ) < 0.
However, tan(θ) > 0.
The angle in the third quadrant is 180° < θ < 270° in degrees or π < θ <
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T
tan (π + θ)
x
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y
Consider the point P (a, b) in the first quadrant.
P (a, b)
When this point is reflected in both the x- and
y-axes, it becomes the point P′ (–a, –b). If P makes
an angle of θ with the x-axis, then P′ makes an
π+ θ
b
angle of 180 + θ degrees or π + θ radians with
θ
−a O a
the positive end of the x-axis. From the definitions
−b
of sine and cosine, we obtain the following
relationships.
Pʹ (−a, −b)
sin(180° + θ) = −sin(θ) sin(π + θ) = −sin(θ)
cos(180° + θ) = −cos(θ) cos(π + θ) = −cos(θ)
tan(180° + θ) = tan(θ) tan(π + θ) = tan(θ)
For example:
7π
π
π
1
b = sin aπ + b = −sin a b = −
6
6
6
2
tan a
4π
π
π
b = tan aπ + b = tan a b = !3
3
3
3
E
!2
π
5π
π
b = cos aπ + b = −cos a b = −
4
4
4
2
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cos a
PR
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sin a
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The fourth quadrant
y
The angle in the fourth quadrant is
T
P (a, b)
3π
270° < θ < 360° in degrees or
< θ < 2π in
2
radians. In the fourth quadrant, x > 0 and y < 0, so
b
cos(θ) > 0 and sin(θ) < 0; therefore, tan(θ) < 0.
θ
a
x
O
Consider the point P (a, b) in the first quadrant.
−b
2π − θ
−tan (θ)
When this point is reflected in the x-axis, it
becomes the point P′ (a, –b). If P makes an angle
of θ with the x-axis, then P′ makes an angle of
Pʹ (a, −b) Tʹ
360 − θ degrees or 2π − θ radians with the x-axis.
From the definitions of sine and cosine, we obtain
the following relationships.
sin(360° − θ) = −sin(θ) sin(2π − θ) = −sin(θ)
cos(360° − θ) = cos(θ) cos(2π − θ) = cos(θ)
tan(360° − θ) = −tan(θ) tan(2π − θ) = −tan(θ)
For example:
sin a
cos a
tan a
!2
7π
π
π
b = sin a2π − b = −sin a b = −
4
4
4
2
5π
π
π
1
b = cos a2π − b = cos a b =
3
3
3
2
!3
11π
π
π
b = tan a2π − b = −tan a b = −
6
6
6
3
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Summary
The trigonometric ratios sin(θ), cos(θ) and tan(θ)
are all positive in the first quadrant. Only sin(θ)
is positive in the second quadrant; only tan(θ) is
positive in the third quadrant; and finally, only cos(θ)
is positive in the fourth quadrant. This is summarised
in the diagram at right. The mnemonic CAST is often
used as a memory aid.
sin(θ) = sin(π − θ) = −sin(π + θ) = −sin(2π − θ)
y
π−θ
θ=π
π+ θ
S
A
T
C
cos(θ) = −cos(π − θ) = −cos(π + θ) = cos(2π − θ)
2π − θ
FS
y
O
P (a, b)
θ
−θ a
T
b
x
−b
−tan (θ)
Pʹ (a, −b)
Tʹ
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cos(−θ) = cos(θ)
θ=0
x
O
PR
O
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sin(−θ) = −sin(θ)
θ
––
θ = 3π
2
tan(θ) = −tan(π − θ) = tan(π + θ) = −tan(2π − θ)
Negative angles
A negative angle is one that is measured clockwise
from the positive direction of the x-axis.
Consider the point P (a, b) in the first quadrant.
When this point is reflected in the x-axis, it
becomes the point P′ (a, –b). If P makes an angle
of θ with the x-axis, then P′ makes an angle of –θ
with the x-axis. From the definitions of sine and
cosine, we obtain the following relationships.
θ = –π2
tan(−θ) = −tan(θ)
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π
A negative angle − < θ < 0 is just the equivalent angle in the fourth quadrant.
2
For positive angles greater than 360° or 2π, we can just subtract multiples of
360° or 2π.
sin(360° + θ) = sin(θ) sin(2π + θ) = sin(θ)
cos(360° + θ) = cos(θ) cos(2π + θ) = cos(θ)
tan(360° + θ) = tan(θ) tan(2π + θ) = tan(θ)
For example:
sin a−
!3
4π
4π
π
π
b = −sin a b = −sin aπ + b = sin a b =
3
3
3
3
2
tan a−
2π
2π
π
π
b = −tan a b = −tan aπ − b = tan a b = !3
3
3
3
3
cos a−
!2
7π
7π
π
π
b = cos a b = cos a2π − b = cos a b =
4
4
4
4
2
Reciprocal trigonometric functions
The reciprocal of the sine function is called the cosecant function, often abbreviated
1
to cosec. It is defined as cosec(x) =
, provided that sin(x) ≠ 0.
sin(x)
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cot (θ) =
opposite
hypotenuse
adjacent
adjacent
opposite
=
=
c
b
=
c
a
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sec (θ) =
hypotenuse
c
PR
O
cosec (θ) =
FS
The reciprocal of the cosine function is called the secant function, often abbreviated
1
to sec. It is defined as sec(x) =
, provided that cos(x) ≠ 0.
cos(x)
The reciprocal of the tangent function is called the cotangent function, often
cos(x)
1
abbreviated to cot. It is defined as cot(x) =
=
, provided that sin(x) ≠ 0.
tan(x) sin(x)
Note that these are not the inverse trigonometric functions. (The inverse trigonometric
functions are covered in Section 2.6.)
The reciprocal trigonometric functions can also be defined in terms of the sides of
a right-angled triangle.
a
b
θ
b
a
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1
Find the exact value of cosec a
5π
b.
4
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exact values
The exact values for the reciprocal trigonometric functions for angles that are
multiples of 30° and 45° can be found from the corresponding trigonometric values
by finding the reciprocals. Often it is necessary to simplify the resulting expression or
rationalise the denominator.
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1 State the required identity.
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2 Use the known results.
3 Simplify the ratio and state the
final answer.
cosec (θ) =
cosec a
5π
b=
4
cosec a
5π
b=
4
1
sin (θ)
1
sin a
5π
b
4
π
Use sin (π + θ) = −sin (θ) with θ = .
4
5π
1
cosec a b =
4
π
sin aπ + b
4
1
=
π
−sin a b
4
1
1
−
!2
= −!2
Topic 2 TrIgonoMeTry
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using triangles to find values
Triangles can be used to find the values of the required trigonometric ratios. Particular
attention should be paid to the sign of the ratio.
2
If cosec (θ) =
π
7
and < θ < π, find the exact value of cot(θ).
4
2
tHinK
WritE/draW
1 State the values of the sides of a
PR
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corresponding right-angled triangle.
7
4
7
1
=
sin(θ)
4
4
sin(θ) =
7
The hypotenuse has a length of 7 and the opposite
side length is 4.
cosec (θ) =
FS
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eXaMPLe
2 Draw the triangle and label the
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side lengths using the definition of
the trigonometric ratio. Label the
unknown side length as x.
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3 Calculate the value of the third side
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using Pythagoras’ theorem.
4 State the value of a related
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trigonometric ratio.
5 Calculate the value of the required
trigonometric value.
7
θ
x
x2 + 42
x2 + 16
x2
x2
x
= 72
= 49
= 49 − 16
= 33
4
= !33
7
θ
33
π
< θ < π, θ is in the second quadrant.
2
Although sin(θ) is positive in this quadrant, tan(θ) is
negative.
Given that
4
tan(θ) = − !33
cot (θ) =
=
1
tan (θ)
1
4
− !33
= − !33
4
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Exercise 2.2 Reciprocal trigonometric functions
1
WE1
Find the exact value of cosec a
2 Find the exact value of sec a−
5
π
and < θ < π, find the exact value of cot (θ).
2
2
3π
4 If cot (θ) = 4 and π < θ < , find the exact value of sec (θ).
2
5 Find the exact values of each of the following.
π
3π
4π
7π
a sec a b
b sec a b
c sec a b
d sec a− b
6
4
3
4
WE2
If cosec (θ) =
FS
3
Consolidate
7π
b.
6
2π
b.
3
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PRactise
π
3
a cosec a b
b cosec a
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6 Find the exact values of each of the following.
5π
b
6
c cosec a
b cot a
2π
b
3
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π
6
a cot a b
E
7 Find the exact values of each of the following.
c cot a
7π
b
4
7π
b
4
d cosec a
d cot a
π
1
and < x < π, find the exact value of sec(x).
3
2
π
b If cosec(x) = 4 and < x < π, find the exact value of cot(x).
2
3
3π
9 aIf cos(x) = − and π < x < , find the exact value of cot(x).
7
2
5
3π
b If sec(x) = − and π < x < , find the exact value of cosec(x).
2
2
3
3π
10 aIf cos(x) = and
< x < 2π, find the exact value of cosec(x).
7
2
8
3π
b If sec(x) = and
< x < 2π, find the exact value of cot(x).
5
2
π
11 aIf cosec(x) = 4 and < x < π, find the exact value of tan(x).
2
5
π
b If cot(x) = − and < x < π, find the exact value of cosec(x).
6
2
3π
12 aIf sec(x) = −7 and π < x < , find the exact value of cot(x).
2
3π
b If cot(x) = 4 and π < x < , find the exact value of cosec(x).
2
3π
13 aIf sec(x) = 6 and
< x < 2π, find the exact value of cosec(x).
2
5
3π
b If cot(x) = − and
< x < 2π, find the exact value of sec (x).
2
2
11π
b
6
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8 aIf sin(x) =
5π
b
3
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!6
π
and < x < π, find the exact value of cosec(x).
3
2
2 !6
3π
b If sec(x) =
and
< x < 2π, find the exact value of cot(x).
3
2
p
π
15 If cosec(x) = where p, q ∈ R + and < x < π, evaluate sec(x) − cot(x).
q
MastEr
2
3π
a
16 If sec(x) = where a, b ∈ R + and
< x < 2π, evaluate cot(x) − cosec(x).
b
2
14 a If cot(x) = −
Trigonometric identities using reciprocal
trigonometric functions
FS
2.3
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Identities
By mathematical convention, (sin(θ)) 2 is written as sin2 (θ), and similarly (cos(θ)) 2 is
written as cos2 (θ).
Note that sin2 (θ) + cos2 (θ) = 1 is an identity, not an equation, since it holds true for
all values of θ.
sin(θ)
holds for all values of θ for which tan(θ) is defined, that is
Similarly, tan(θ) =
cos(θ)
π
π
for all values where cos(θ) ≠ 0, or θ ≠ (2n + 1) where n ∈ Z or odd multiples of .
2
2
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AOS 1
Topic 2
Concept 3
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Trigonometric
identities
Concept summary
Practice questions
Proving trigonometric identities
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A trigonometric identity is verified by transforming one side into the other. Success
in verifying trigonometric identities relies upon familiarity with known trigonometric
identities and using algebraic processes such as simplifying, factorising, cancelling
common factors, adding fractions and forming common denominators. The following
identities must be known.
WorKeD
eXaMPLe
3
Prove the identity tan (θ) + cot (θ) = sec (θ)cosec (θ).
tHinK
1 Start with the left-hand side.
2 Substitute for the appropriate trigonometric
identities.
66
sin(θ)
cos(θ)
1
sec(θ) =
cos(θ)
1
cosec(θ) =
sin(θ)
1
cot(θ) =
tan(θ)
tan(θ) =
WritE
LHS = tan(θ) + cot(θ)
sin(θ)
cos(θ)
and cot(θ) =
cos(θ)
sin(θ)
sin(θ) cos(θ)
LHS =
+
cos(θ) sin(θ)
tan(θ) =
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3 Add the fractions, forming the lowest common
denominator.
LHS =
sin2 (θ) + cos2 (θ)
cos(θ)sin(θ)
Since sin2 (θ) + cos2 (θ) = 1,
1
LHS =
cos(θ)sin(θ)
4 Simplify the numerator.
LHS =
5 Write the expression as factors.
1
1
×
cos(θ) sin(θ)
1
1
and cosec(θ) =
cos(θ)
sin(θ)
LHS = sec(θ)cosec(θ)
= RHS
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identities. The proof is complete.
FS
sec(θ) =
6 Substitute for the appropriate trigonometric
Fundamental relations
If all terms of sin2 (θ) + cos2 (θ) = 1 are divided by sin2 (θ), we obtain
sin2 (θ)
+
cos2 (θ)
sin2 (θ)
1
and hence obtain the trigonometric identity
sin2 (θ)
=
E
sin2 (θ)
PA
G
1 + cot2 (θ) = cosec2 (θ).
If all terms of sin2 (θ) + cos2 (θ) = 1 are divided by cos2 (θ), we obtain
4
cos2 (θ)
=
1
and hence obtain the trigonometric identity
cos2 (θ)
Prove the identity
1+
cot2 (θ)
R
tHinK
tan2 (θ) + 1 = sec2 (θ).
1 + tan2 (θ)
R
WorKeD
eXaMPLe
cos2 (θ)
EC
cos2 (θ)
+
TE
D
sin2 (θ)
= tan2 (θ).
WritE
1 + tan2 (θ)
LHS =
2 Substitute the appropriate trigonometric
Replace 1 + tan2 (θ) = sec2 (θ) in the numerator
and 1 + cot2 (θ) = cosec2 (θ) in the denominator.
sec2 (θ)
LHS =
cosec2 (θ)
1
1
and cosec2 (θ) = 2
sec2 (θ) =
2
cos (θ)
sin (θ)
1
cos2 (θ)
LHS =
1
sin2 (θ)
C
O
1 Start with the left-hand side.
U
N
identities.
3 Use appropriate trigonometric identities
to express the quotient in terms of sines
and cosines.
1 + cot2 (θ)
Topic 2 TrIgonoMeTry
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20/08/15 10:41 AM
1
a b
Use = .
1 a
b
sin2 (θ)
LHS =
cos2 (θ)
cos2 (θ)
LHS = tan2 (θ)
= RHS
,
Exercise 2.3 Trigonometric identities using reciprocal
WE3
Prove the identity sec2 (θ) + cosec2 (θ) = sec2 (θ)cosec2 (θ).
2 Prove the identity
3
WE4
sin(θ)
1 + cos(θ)
+
= 2 cosec(θ).
1 + cos(θ)
sin(θ)
Prove the identity
1 + cot2 (θ)
1 + tan2 (θ)
E
1
= cot2 (θ).
PA
G
PRactise
PR
O
trigonometric functions
FS
sin2 (θ)
Since tan2 (θ) =
5 Simplify and state the final result.
O
4 Simplify the quotient.
4 Prove the identity (1 − sin2 (θ))(1 + tan2 (θ)) = 1.
For questions 5–14, prove each of the given identities.
5 a cos(θ)cosec(θ) = cot(θ)
TE
D
Consolidate
6 a sin(θ)sec(θ)cot(θ) = 1
b cos(θ)tan(θ) = sin(θ)
b cos(θ)cosec(θ)tan(θ) = 1
7 a (cos(θ) + sin(θ)) 2 + (cos(θ) − sin(θ)) 2 = 2
EC
b 2 − 3 cos2 (θ) = 3 sin2 (θ) − 1
8 a tan2 (θ)cos2 (θ) + cot2 (θ)sin2 (θ) = 1
sin(θ)
cos(θ)
+
=1
cosec(θ) sec(θ)
1
1
9 a
+
= 2 sec2 (θ)
1 − sin(θ) 1 + sin(θ)
1
1
b
+
= 2 cosec2 (θ)
1 − cos(θ) 1 + cos(θ)
U
N
C
O
R
R
b
10 a
1
1
+
=1
2
1 + sec (θ) 1 + cos2 (θ)
b (1 − tan(θ)) 2 + (1 + tan(θ)) 2 = 2 sec2 (θ)
11 a (tan(θ) + sec(θ)) 2 =
1 + sin(θ)
1 − sin(θ)
b sec4 (θ) − sec2 (θ) = tan4 (θ) + tan2 (θ)
tan(θ)
tan(θ)
+
= 2 cosec(θ)
sec(θ) − 1 sec(θ) + 1
1 + cot(θ)
sec(θ)
b
−
= cos(θ)
cosec(θ)
tan(θ) + cot(θ)
12 a
68 Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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cos(θ)
= sec(θ) + tan(θ)
1 − sin(θ)
1
1
14 a
+
=1
2
1 + sin (θ) 1 + cosec2 (θ)
cos(θ)
= sec(θ) − tan(θ)
1 + sin(θ)
1
1
b
+
=1
2
1 + cot (θ) 1 + tan2 (θ)
13 a
For questions 15 and 16, prove each of the given identities.
2.4
sin2 (θ)
a − b tan2 (θ)
1+
tan2 (θ)
= b + (a − b)cosec2 (θ) b = (a + b)cos2 (θ) − b
b
a − b sin2 (θ)
cos2 (θ)
a − b cot2 (θ)
1 + cot (θ)
Compound-angle formulas
2
= b + (a − b)sec2 (θ)
= (a + b)sin2 (θ) − b
FS
16 a
a − b cos2 (θ)
O
15 a
The compound-angle formulas are also known as trigonometric addition and
subtraction formulas.
PR
O
Master
b
Proof of the compound-angle formulas
AOS 1
The compound addition formulas state that:
E
Topic 2
Concept 4
PA
G
sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
Compound and
double angle
formulas
Concept summary
Practice questions
cos(A + B) = cos(A)cos(B) − sin(A)sin(B)
tan(A) + tan(B)
1 − tan(A)tan(B)
TE
D
tan(A + B) =
T
S
R
B
A
P
Q
U
N
C
O
R
R
EC
It is interesting to consider one method of proving these identities.
Consider the triangle OQR with a right angle at Q,
as shown in the diagram. The line segment TR is
constructed so that TR is perpendicular to OR, and the
line segment TP is constructed so that it is perpendicular
to OP and SR. Let ∠ROQ = A and ∠TOR = B so that
∠TOP = A + B.
O
Using the properties of similar triangles in ΔTSR and
ΔOQR, or the property that supplementary angles sum
to 90°, it follows that ∠STR = A.
In triangle OQR, sin(A) =
QR
OQ
and cos(A) =
.
OR
OR
In triangle RST, sin(A) =
SR
ST
and cos(A) =
.
RT
RT
In triangle ORT, sin(B) =
OR
RT
and cos(B) =
.
OT
OT
Now consider the triangle OPT.
sin(A + B) =
ST
PT PS + ST PS
=
=
+
OT
OT
OT OT
Topic 2 Trigonometry c02Trigonometry.indd 69
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PS = QR, so
QR ST
+
OT OT
QR OR ST RT
=
×
+
×
OT OR OT RT
QR OR ST RT
×
+
×
=
OR OT RT OT
sin(A + B) =
That is,
Also in the triangle OPT:
OP OQ − PQ OQ PQ
=
=
−
OT
OT
OT OT
O
cos(A + B) =
FS
sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
PR
O
PQ = SR, so
OQ SR
−
OT OT
OQ OR SR RT
×
−
×
=
OT OR OT RT
OQ OR SR RT
=
×
−
×
OR OT RT OT
cos(A + B) = cos(A)cos(B) − sin(A)sin(B)
PA
G
E
cos(A + B) =
TE
D
Proof of the compound-angle subtraction formulas
The compound subtraction formulas state that:
sin(A − B) = sin(A)cos(B) − cos(A)sin(B)
EC
cos(A − B) = cos(A)cos(B) + sin(A)sin(B)
U
N
C
O
R
R
These formulas can obtained by replacing B with –B and using cos(−B) = cos(B)
and sin(−B) = −sin(B).
Substituting into the formula sin(A + B) = sin(A)cos(B) + cos(A)sin(B), we
derive sin(A + (−B)) = sin(A)cos(−B) + cos(A)sin(−B), so that
sin(A − B) = sin(A)cos(B) − cos(A)sin(B).
Similarly, in the formula cos(A + B) = cos(A)cos(B) − sin(A)sin(B), we derive
cos(A + (−B)) = cos(A)cos(−B) − sin(A)sin(−B), so that
cos(A − B) = cos(A)cos(B) + sin(A)sin(B).
Proof of the compound-angle formulas involving tangents
Let us substitute the formulas for sin(A + B) and cos(A + B) into the identity for the
tangent ratio.
tan(A + B) =
=
70 sin(A + B)
cos(A + B)
sin(A)cos(B) + cos(A)sin(B)
cos(A)cos(B) − sin(A)sin(B)
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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In order to simplify this fraction, divide each term by cos(A)cos(B):
sin(A)sin(B)
cos(A)cos(B)
−
cos(A)cos(B) cos(A)cos(B)
sin(A)
sin(B)
+
cos(A) cos(B)
1−
=
sin(B)
sin(A)
×
cos(A) cos(B)
tan(A) + tan(B)
1 − tan(A)tan(B)
FS
=
O
tan(A + B) =
sin(A)cos(B)
cos(A)sin(B)
+
cos(A)cos(B) cos(A)cos(B)
=
tan(A) − tan(B)
1 + tan(A)tan(B)
PA
G
tan(A − B)
tan(A) + tan(−B)
1 − tan(A)tan(−B)
E
tan(A + (−B)) =
PR
O
The corresponding formula for the tangent for the difference of two angles is obtained
by replacing B with –B and using tan(−B) = −tan(B).
Summary of the compound-angle formulas
TE
D
These results are called the compound-angle formulas or addition theorems. They can
be summarised as:
sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
U
N
C
O
R
R
EC
sin(A − B) = sin(A)cos(B) − cos(A)sin(B)
WORKED
EXAMPLE
5
cos(A + B) = cos(A)cos(B) − sin(A)sin(B)
cos(A − B) = cos(A)cos(B) + sin(A)sin(B)
tan(A + B) =
tan(A) + tan(B)
1 − tan(A)tan(B)
tan(A − B) =
tan(A) − tan(B)
1 + tan(A)tan(B)
Using compound-angle formulas in problems
The compound-angle formulas can be used to simplify many trigonometric
expressions. They can be used in both directions, for example
sin(A)cos(B) + cos(A)sin(B) = sin(A + B).
Evaluate sin (22°)cos (38°) + cos (22°)sin (38°).
THINK
1 State an appropriate identity.
WRITE
sin(A)cos(B) + cos(A)sin(B) = sin(A + B)
Let A = 22° and B = 38°.
sin(22°)cos(38°) + cos(22°)sin(38°) = sin(22° + 38°)
Topic 2 TRIGONOMETRY
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sin (22°)cos (38°) + cos (22°)sin (38°) = sin (60°)
2 Simplify and use the exact values.
=
!3
2
WorKeD
eXaMPLe
6
tHinK
π
Expand 2 cos aθ + b.
3
FS
expanding trigonometric expressions with phase shifts
The compound-angle formulas can be used to expand trigonometric expressions.
WritE
cos(A + B) = cos(A)cos(B) − sin(A)sin(B)
π
Let A = θ and B = .
3
π
π
π
2 cos aθ + b = 2acos(θ)cos a b − sin(θ)sin a b b
3
3
3
PR
O
O
1 State an appropriate identity.
!3
π
π
1
Since cos a b = and sin a b =
,
3
2
3
2
PA
G
E
2 Substitute for exact values.
!3
π
1
2 cos aθ + b = 2acos(θ) × − sin(θ) ×
b
3
2
2
3 Simplify.
= cos(θ) − !3 sin(θ)
TE
D
π
2 cos aθ + b = cos(θ) − !3 sin(θ)
3
4 State the answer.
EC
simplification of sin a
nπ
nπ
± θb and cos a
± θb for n ∈ Z
2
2
U
N
C
O
R
R
π
π
Recall that cos a − θb = sin(θ) and sin a − θb = cos(θ) as complementary angles.
2
2
Compound-angle formulas can be used to simplify and verify many of these results
and similar formulas from earlier results, that is trigonometric expansions of the forms
nπ
nπ
sin a ± θb and cos a ± θb where n ∈ Z.
2
2
WorKeD
eXaMPLe
tHinK
7
Use compound-angle formulas to simplify cos a
1 State an appropriate identity.
72
3π
− θb.
2
WritE
cos(A − B) = cos(A)cos(B) + sin(A)sin(B)
3π
Let A =
and B = θ.
2
3π
3π
3π
cos a − θb = cos a bcos (θ) + sin a bsin (θ)
2
2
2
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2 Simplify and use exact values.
Since cos a
cos a
cos a
3 State the final answer.
3π
3π
b = 0 and sin a b = −1,
2
2
3π
− θ b = 0 × cos (θ) + −1 × sin (θ)
2
3π
− θ b = −sin (θ)
2
π
12
π
Exact values are known for the trigonometric ratios for all multiples of radians or
6
π
30°, and for all multiples radians or 45°. Using the compound-angle formulas the
4
exact value can be found for a trigonometric ratio of an angle that is an odd multiple
π
π
radians or 15°. This can be obtained by rewriting the multiple of
radians or
of
12
12
π
15° as a sum or difference of known fractions in terms of multiples of radians or
6
π
30° and radians or 45°.
4
WorKeD
eXaMPLe
8
Find the exact value of sin a
13π
b.
12
WritE
TE
D
tHinK
PA
G
E
PR
O
O
FS
exact values for multiples of
1 Rewrite the argument as a sum or
EC
difference of fractions.
C
O
R
R
2 State an appropriate identity.
U
N
3 Simplify and use exact values.
5π π 13π
+ =
, or in degrees, 150° + 45° = 195°.
6
4
12
13π
5π π
sin a
b = sin a + b
12
6
4
sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
5π
π
Let A =
and B = .
4
6
sin a
Substitute sin a
cos a
sin a
4 Simplify and state the final answer.
13π
5π
π
5π
π
b = sin a bcos a b + cos a bsin a b
12
6
4
6
4
sin a
!2
5π
π
1
,
b = , cos a b =
6
2
4
2
!3
!2
5π
π
and sin a b =
.
b=−
6
2
4
2
13π
1 !2 −!3 !2
+
×
b= ×
12
2
2
2
2
!2 !6
−
=
4
4
13π
1
b = (!2 − !6 )
12
4
Topic 2 TrIgonoMeTry
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using triangles to find values
By drawing triangles to find the values of trigonometric ratios of a single angle and
then using the compound-angle formulas, the trigonometric values of the addition or
subtraction of two angles may be found.
9
π
π
7
12
and sin(B) = , where 0 < A < and < B < π, find the
13
25
2
2
( − B).
(A
exact value of sin(A
If cos(A
( )=
(A
WritE/draW
FS
tHinK
adjacent
12
=
13 hypotenuse
The adjacent side length is 12 and the hypotenuse is 13.
cos(A) =
1 State the values of the sides of the
O
required right-angled triangle.
2 Use Pythagoras’ theorem to calculate
the third side length.
"132 − 122 = !169 − 144
= !25
=5
The other side length is 5.
We know that 5, 12, 13 is
a Pythagorean triad.
E
3 State the third side length of the
PR
O
WorKeD
eXaMPLe
PA
G
triangle. Draw the triangle.
13
5
A
12
R
EC
TE
D
π
Given that 0 < A < , so A is in the first quadrant,
2
trigonometric ratio.
5
sin (A) = .
13
opposite
7
sin (B) =
=
5 State the values of the sides of
25 hypotenuse
another required right-angled triangle.
The opposite side length is 7 and the hypotenuse is 25.
4 State the value of the unknown
R
6 Use Pythagoras’ theorem to calculate
O
the third side length.
C
7 State the third side length of the
U
N
triangle. Draw the triangle.
8 Calculate the value of the unknown
trigonometric ratio.
74
"252 − 72 = !625 − 49
= !576
= 24
The other side length
is 24. We know that
7
7, 24, 25 is a
Pythagorean triad.
25
24
B
π
< B < π, B is in the second quadrant, B is
2
an obtuse angle and cosine is negative in the second
quadrant.
Therefore, cos(B) = −24
25
Since
MaTHs QuesT 12 sPeCIaLIsT MaTHeMaTICs VCe units 3 and 4
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9 State and use an appropriate identity.
sin(A − B) = sin(A)cos(B) − cos(A)sin(B)
10 Substitute for the values and simplify. sin(A − B) =
11 Simplify and state the final answer.
× −24
− 12
×
25
13
5
13
7
25
sin(A − B) = −204
325
Exercise 2.4 Compound-angle formulas
1
WE5
Evaluate sin(51°)cos(9°) + cos(51°)sin(9°).
3
WE6
π
Expand 4 cos a θ + b.
6
FS
2 Find the value of cos(37°)cos(23°) − sin(37°)sin(23°).
O
PRactise
PR
O
π
4
4 Express !2 sin a θ + b as a combination of sines and cosines.
5
WE7
Use compound-angle formulas to simplify cos(π − θ).
WE8
Find the exact value of sin a
8 Find the exact value of tan a
11π
b.
12
3
8
π
π
and sin(B) =
where 0 < A < and < B < π, find the
5
17
2
2
exact value of sin(A − B).
WE9
If cos(A) =
TE
D
9
7π
b.
12
PA
G
7
E
6 Simplify sin(2π − θ).
9
3π
π
7
and cos(B) =
where π < A <
and 0 < B < , find
40
25
2
2
the exact value of cos(A + B).
EC
10 Given that tan(A) =
R
11 Evaluate each of the following.
a sin(27°)cos(33°) + cos(27°)sin(33°)
b cos(47°)cos(43°) − sin(47°)sin(43°)
c cos(76°)cos(16°) + sin(76°)sin(16°)
d cos(63°)sin(18°) − sin(63°)cos(18°)
U
N
C
O
R
Consolidate
12 Evaluate each of the following.
a
tan(52°) − tan(22°)
1 + tan(52°)tan(22°)
b
tan(32°) + tan(28°)
1 − tan(32°)tan(28°)
13 Expand each of the following.
π
4
a !2 sin a θ − b
π
6
c 2 cos a θ − b
π
3
b 2 sin a θ + b
π
4
d !2 cos a θ + b
Topic 2 Trigonometry c02Trigonometry.indd 75
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14 Use compound-angle formulas to simplify each of the following.
π
2
π
2
a sin a − θ b
b cos a − θ b
c sin(π + θ)
d cos(π − θ)
15 Use compound-angle formulas to simplify each of the following.
a sin a
3π
− θb
2
b cos a
3π
+ θb
2
c tan(π − θ)
d tan(π + θ)
16 Simplify each of the following.
π
3
π
4
π
3
π
6
c cos a + xb − cos a − xb
4
b tan a
π
b
12
18 Given that cos(A) = 5, sin(B) =
c sin a
d tan a
5π
b
12
and A and B are both acute angles, find the
PA
G
b tan(A + B).
19 Given that sin(A) =
5
13
and tan(B) =
20 Given that sec(A) =
7
,
2
3
,
2
exact value of:
a sin(A + B)
24
7
where A is obtuse and B is acute, find the
b cos (A + B).
cosec(B) =
TE
D
exact value of:
a cos(A + B)
11π
b
12
E
exact value of:
a cos(A − B)
12
,
13
PR
O
7π
b
12
π
6
d cos a − xb − cos a + xb
17 Find each of the following in exact simplest surd form.
a cos a
FS
π
3
π
4
b tan ax + b tan ax − b
O
π
3
a sin ax + b − sin ax − b
and A is acute but B is obtuse, find the
b sin(A − B).
1
1
a
Master
b
0 < b < 1, evaluate tan(A + B).
a
a
22 Given that sin(A) =
and cos(B) =
where A and B are both acute,
a+1
a+2
evaluate tan(A + B).
Double-angle formulas
U
N
C
2.5
O
R
R
EC
21 Given that cosec(A) = , sec(B) = , A and B are both acute, and 0 < a < 1 and
In this section we consider the special cases of the addition formulas when B = A.
Double-angle formulas
In the formula sin(A + B) = sin(A)cos(B) + cos(A)sin(B), let B = A.
sin(2A) = sin(A)cos(A) + sin(A)cos(A)
sin(2A) = 2 sin(A)cos(A)
In the formula cos(A + B) = cos(A)cos(B) − sin(A)sin(B), let B = A.
cos(2A) = cos(A)cos(A) − sin(A)sin(A)
cos(2A) = cos2(A) − sin2(A)
76 Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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PR
O
O
FS
Since sin2(A) + cos2(A) = 1, it follows that cos2(A) = 1 − sin2(A). This formula can
be rewritten in terms of sin(A) only.
cos(2A) = (1 − sin2(A)) − sin2(A)
cos(2A) = 1 − 2 sin2(A)
Alternatively, if we substitute sin2(A) = 1 − cos2(A), then this formula can also be
rewritten in terms of cos(A) only.
cos(2A) = cos2(A) − (1 − cos2(A))
cos(2A) = 2 cos2(A) − 1
There are thus three equivalent forms of the double-angle formulas for cos(2A).
tan(A) + tan(B)
If we let B = A in the formula tan(A + B) =
, we obtain
1 − tan(A)tan(B)
tan(A) + tan(A)
tan(2A) =
1 − tan(A)tan(A)
2 tan(A)
tan(2A) =
1 − tan2(A)
All of these formulas can be summarised as follows:
PA
G
E
sin(2A) = 2 sin(A)cos(A)
cos(2A) = cos2(A) − sin2(A)
cos(2A) = 1 − 2 sin2(A)
cos(2A) = 2 cos2(A) − 1
TE
D
tan(2A) =
2 tan(A)
1 − tan2(A)
EC
using double-angle formulas in simplifying expressions
The double-angle formulas can be used to simplify many trigonometric expressions
C
tHinK
R
10
Find the exact value of sin a
O
WorKeD
eXaMPLe
R
and can be used both ways; for example, sin(A)cos(A) = 12 sin(2A).
U
N
1 State an appropriate identity.
2 Simplify.
7π
7π
bcos a b.
12
12
WritE
sin(A)cos(A) =
Let A =
7π
.
12
1
sin(2A)
2
7π
7π
bcos a b =
12
12
7π 7π
Since 2 ×
= ,
12
6
7π
7π
sin a bcos a b =
12
12
sin a
7π
1
sin a2 × b
2
12
7π
1
sin a b
2
6
Topic 2 TrIgonoMeTry
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20/08/15 10:42 AM
Substitute sin a
3 Use the exact values to substitute into the
expression.
sin a
4 State the answer.
7π
1
b=− :
6
2
7π
1
1
1
sin a b = × −
2
6
2
2
7π
7π
1
bcos a b = −
12
12
4
11
If cos(A
( ) = 14, determine the exact values of:
(A
2 )
2A
a sin (2A
2 )
2A
b cos (2A
WRITE/DRAW
1 State the values of the sides of the
required right-angled triangle.
cos(A) =
E
THINK
2 ).
2A
c tan (2A
adjacent
1
=
4 hypotenuse
PA
G
WORKED
EXAMPLE
PR
O
O
FS
Finding trigonometric expressions involving double-angle formulas
We can use the double-angle formulas to obtain exact values for trigonometric
expressions.
The adjacent side length is 1 and the hypotenuse is 4.
2 Draw the triangle and label the
R
EC
TE
D
side lengths using the definition of
the trigonometric ratio. Label the
unknown side length as x.
R
3 Use Pythagoras’ theorem to
C
O
calculate the third unknown
side length.
U
N
4 Redraw the triangle.
4
x
A
1
12 + x2 = 42
x2 = 16 − 1
x = !15
4
15
A
1
5 Apply the definitions of the sine and
tangent functions.
78
cos(A) = 14 , sin(A) =
!15
4
and tan(A) = !15
MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c02Trigonometry.indd 78
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a sin(2A) = 2 sin(A)cos(A)
a 1 Use the required identity.
=2×
2 Substitute the known values
and simplify.
=
!15
8
!15
4
×
1
4
b cos(2A) = cos2(A) − sin2(A)
b 1 Using the required identity,
choose any one of the three
choices for cos(2A).
2
cos(2A) = a14b − a !15
b
4
2 Substitute the known values
and simplify.
2
=
2 Substitute the known values and
"15
8
−78
E
simplify the ratio.
sin(2A)
cos(2A)
PR
O
c tan(2A) =
c 1 State the required identity.
O
= −78
FS
1
= 16
− 15
16
3 As an alternative method, use the
double-angle formulas for tan.
4 Substitute for the known value
tan(2A) =
2 tan(A)
1 − tan2(A)
tan(A) = !15
2 tan(A)
tan(2A) =
1 − tan2(A)
2 !15
− ( !15)2
2 !15
−14
− !15
7
=1
=
=
R
R
EC
TE
D
and simplify.
PA
G
= −"15
7
U
N
C
O
solving trigonometric equations involving double-angle formulas
Trigonometric equations are often solved over a given domain, usually x ∈ [0, 2π].
In this section we consider solving trigonometric equations that involve using the
double-angle formulas.
WorKeD
eXaMPLe
tHinK
12
Solve for x if sin(2x) + !3
! cos(x) = 0 for x ∈ [0, 2π] .
1 Expand and write the equation in terms of one
argument only.
WritE
Use sin(2x) = 2 sin(x)cos(x)
sin(2x) + !3cos(x) = 0
2 sin(x)cos(x) + !3cos(x) = 0
Topic 2 TrIgonoMeTry
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2 Factorise by taking out the common factor.
cos(x)(2 sin(x) + !3) = 0
cos(x) = 0 or 2 sin(x) + !3 = 0
3 Use the Null Factor Law.
sin(x) = − !3
2
4 Solve the first equation.
cos(x) = 0
π 3π
x= ,
2 2
5 Solve the second equation.
sin(x) = − !3
2
4π 5π
,
3 3
π 4π 3π 5π
x= , , ,
2 3 2 3
O
PR
O
6 State all solutions of the original equation.
FS
x=
13
Prove the identity
tHinK
cos(2A
2 )cos(A
2A
( ) + sin(2A
(A
2 )sin(A
2A
( ) 1
(A
= cosec(A
( ).
(A
sin(3A)cos(A
( ) − cos(3A)sin(A
(A
( )
(A
2
WritE
1 Start with the left-hand side.
TE
D
WorKeD
eXaMPLe
PA
G
E
Trigonometric identities using double-angle formulas
Previously we used the fundamental trigonometric relationships to prove
trigonometric identities using the reciprocal trigonometric functions. In this section
we use the compound-angle formulas and the double-angle formulas to prove more
trigonometric identities.
EC
2 Simplify the numerator and denominator
R
by recognising these as expansions of
appropriate compound-angle identities.
O
R
3 Simplify.
C
4 Expand the denominator using the
U
N
double-angle formula.
5 Simplify by cancelling the common factor.
The proof is complete.
LHS =
=
cos(2A)cos(A) + sin(2A)sin(A)
sin(3A)cos(A) − cos(3A)sin(A)
cos(2A − A)
sin(3A − A)
cos(A)
sin(2A)
cos(A)
=
2 sin(A)cos(A)
=
1
2 sin(A)
1
Since
= cosec(A),
sin(A)
=
1
2 sin(A)
= 1 cosec(A)
LHS =
2
= RHS as required.
80
MaTHs QuesT 12 sPeCIaLIsT MaTHeMaTICs VCe units 3 and 4
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Half-angle formulas
A
If we replace A with , the double-angle formulas can be written as the
2
half-angle formulas.
A
A
sin(A) = 2 sina bcosa b
2
2
A
A
cos(A) = cos2 a b − sin2 a b
2
2
PR
O
These can also be rearranged and are often used as:
O
A
= 1 − 2 sin2 a b
2
FS
A
= 2 cos2 a b − 1
2
A
1 − cos(A) = 2 sin2 a b
2
14
tHinK
A
( ) − cot(A
(A
( ) = tana b.
(A
Prove the identity cosec(A
2
1 Start with the left-hand side.
cos(A)
1
and cot(A) =
.
sin(A)
sin(A)
EC
2 Use cosec(A) =
TE
D
WorKeD
eXaMPLe
PA
G
E
A
1 + cos(A) = 2 cos2 a b
2
R
R
3 Form the common denominator.
LHS = cosec(A) − cot(A)
=
cos(A)
1
−
sin(A) sin(A)
=
1 − cos(A)
sin(A)
=
U
N
C
O
4 Use appropriate half-angle formulas.
WritE
5 Simplify by cancelling the common factors.
The proof is complete.
=
A
2 sin2 a b
2
A
A
2 sina bcosa b
2
2
A
sina b
2
A
cosa b
2
A
= tana b
2
= RHS
Topic 2 TrIgonoMeTry
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Multiple-angle formulas
There are many other trigonometric formulas for multiple angles. For example:
sin(3A) = 3 sin(A) − 4 sin3(A)
cos(3A) = 4 cos3(A) − 3 cos(A)
3 tan(A) − tan3(A)
1 − 3 tan2(A)
sin(4A) = cos(A)(4 sin(A) − 8 sin3(A))
cos(4A) = 8 cos4(A) − 8 cos2(A) + 1
4 tan(A)(1 − tan2(A))
1 − 6 tan2(A) + tan4(A)
O
tan(4A) =
FS
tan(3A) =
Prove the identity cos(3A) = 4 cos3 (A
( ) − 3 cos(A
( ).
(A
tHinK
WritE
E
15
PA
G
WorKeD
eXaMPLe
PR
O
Some of these proofs are provided in the next worked example; the remaining ones
are left for the exercises.
LHS = cos(3A)
2 Expand the multiple argument.
cos(2A + A) = cos(A)cos(2A) − sin(A)sin(2A)
3 Since we want the right-hand
cos(2A) = 2 cos2 (A) − 1
sin(2A) = 2 sin(A)cos(A)
cos(3A) = cos(A)(2 cos2 (A) − 1) − sin(A)(2 sin(A)cos(A))
side in terms of cosines, substitute
using appropriate trigonometric
identities.
EC
4 Expand the brackets.
TE
D
1 Start with the left-hand side.
5 Rearrange the expression in terms
R
of cosines.
R
6 Expand the brackets and simplify.
Substitute sin2(A) = 1 − cos2(A):
cos(3A) = 2 cos3(A) − cos(A) − 2(1 − cos2(A))cos(A)
cos(3A) = 2 cos3(A) − cos(A) − 2(cos(A) − cos3(A))
= 4 cos3(A) − 3 cos(A)
= RHS
C
O
The proof is complete.
cos(3A) = 2 cos3(A) − cos(A) − 2 sin2(A)cos(A)
U
N
ExErcisE 2.5 Double-angle formulas and trigonometric identities
PractisE
5π
5π
bcos a b.
8
8
2
2 Find the exact value of 1 − 2 cos (157°30′).
1
Find the exact value of sin a
If cos(A) = 13, determine the exact values of:
a sin(2A)
b cos(2A)
c tan(2A).
4
4 Given that tan(A) = 7, determine the exact values of:
a sin(2A)
b cos(2A)
c tan(2A).
5 WE12 Solve for x if sin(2x) − !3 cos(x) = 0 for x ∈ [0, 2π].
3
82
WE10
WE11
MaTHs QuesT 12 sPeCIaLIsT MaTHeMaTICs VCe units 3 and 4
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6 Find the values of x ∈ [0, 2π] if sin(2x) + cos(x) = 0.
7
WE13
Prove the identity
sin(3A)cos(A) − cos(3A)sin(A)
= 2 sin(A).
cos(2A)cos(A) + sin(2A)sin(A)
π
4
π
4
8 Prove the identity tan a + Ab + tan a − Ab = 2 sec(2A).
A
Prove the identity cosec(A) + cot(A) = cot a b.
2
1 − cos(A)
A
10 Prove the identity
= tan a b.
Å 1 + cos(A)
2
WE14
FS
9
11 WE15 Prove the identity sin(3A) = 3 sin(A) − 4 sin3(A).
Consolidate
O
3 tan(A) − tan3(A)
.
1 − 3 tan2(A)
PR
O
12 Prove the identity tan(3A) =
13 Evaluate each of the following expressions, giving your answers in exact form.
π
8
π
8
b cos2 (112°30′) − sin2 (112°30′)
8
PA
G
c 2 sin2 (22°30′) − 1
E
a sin a bcos a b
d
2 tan a
1 − tan2 a
14 Given that sec(A) = 3, find the exact values of:
b cos(2A)
TE
D
a sin(2A)
π
b
12
π
b
12
c tan(2A).
15 Solve each of the following equations for x ∈ [0, 2π].
a sin(2x) = sin(x)
c sin(x) = cos(2x)
b cos(x) = cos(2x)
d sin(2x) = cos(x)
EC
16 Solve each of the following equations for x ∈ [0, 2π].
R
a tan(x) = sin(2x)
b sin(2x) = !3 cos(x)
c sin(4x) = sin(2x)
d cos(2x) = sin(4x)
For questions 17–21, prove each of the given identities.
sin(2A)cos(A) − cos(2A)sin(A)
= tan(A)
cos(2A)cos(A) + sin(2A)sin(A)
b
cos(2A)cos(A) + sin(2A)sin(A)
= cot(A)
sin(2A)cos (A) − cos(2A)sin(A)
U
N
C
O
R
17 a
sin(3A) cos(3A)
−
=2
sin(A)
cos(A)
cos(3A) sin(3A)
d
+
= 2 cot(2A)
sin(A)
cos(A)
c
18 a
tan(3A) + tan(A)
= 2 cos(2A)
tan(3A) − tan(A)
b
tan(A) − tan(B) sin(A − B)
=
tan(A) + tan(B) sin(A + B)
Topic 2 Trigonometry c02Trigonometry.indd 83
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20/08/15 10:42 AM
c
sin(A) − cos(A) sin(A) + cos(A)
−
= 2 tan(2A)
sin(A) + cos(A) sin(A) − cos(A)
d
cos(A) + sin(A) cos(A) − sin(A)
+
= 2 sec(2A)
cos(A) − sin(A) cos(A) + sin(A)
19 a
c
sin(A)
A
= cot a b
1 − cos(A)
2
b
1 − cos(2A) + sin(2A)
= tan(A)
1 + cos(2A) + sin(2A)
d
sin(A)
A
= tan a b
1 + cos(A)
2
sin(A) + sin(2A)
= tan(A)
1 + cos(2A) + cos(A)
cot(A)cot(B) − 1
cot(A) + cot(B)
d cot(A − B) =
cot(A)cot(B) + 1
cot(B) − cot(A)
2 tan(A)
1 + tan2(A)
b cos(2A) =
1 − tan2(A)
1 + tan2(A)
PA
G
21 a sin(2A) =
E
c cot(A + B) =
O
tan2(A) − tan2(B)
1 − tan2(A)tan2(B)
PR
O
b tan(A + B)tan(A − B) =
FS
20 a sin(A + B)sin(A − B) = sin2(A) − sin2(B)
cos3(A) − sin3(A)
1
= 1 + sin(2A)
cos(A) − sin(A)
2
d
cos3(A) + sin3(A)
1
= 1 − sin(2A)
cos(A) + sin(A)
2
EC
TE
D
c
22 In a triangle with side lengths a, b and c, where C is a right angle and c the
U
N
C
O
R
R
hypotenuse, show that:
2ab
a sin(2A) =
c2
Master
84 A
2
d sin a b =
c−b
Å 2c
b cos(2A) =
A
2
e cos a b =
b 2 − a2
c2
c tan(2A) =
c+b
Å 2c
f tan a b =
23 Chebyshev (1821–1894) was a famous Russian
mathematician. Although he is known more for his
work in the fields of probability, statistics, number
theory and differential equations, Chebyshev also
devised recurrence relations for trigonometric multiple
angles. One such result is
cos(nx) = 2 cos(x)cos((n − 1)x) − cos((n − 2)x).
Using this result, show that:
a cos(4A) = 8 cos4(A) − 8 cos2(A) + 1
b cos(5A) = 16 cos5(A) − 20 cos3(A) + 5 cos(A)
c cos(6A) = 32 cos6(A) − 48 cos4(A) + 18 cos2(A) − 1.
A
2
2ab
− a2
b2
c−b
.
Åc + b
Pafnuty Chebyshev
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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24 Chebyshev’s recurrence formula for multiple angles of the sine function is
sin(nx) = 2 cos(x)sin((n − 1)x) − sin((n − 2)x). Using this result, show that:
a sin(4A) = cos(A)(4 sin(A) − 8 sin3(A))
b sin(5A) = 16 sin5(A) − 20 sin3(A) + 5 sin(A)
c sin(6A) = cos(A)(32 sin5(A) − 32 sin3(A) + 6 sin(A)).
2.6
Inverse trigonometric functions
Inverse functions
Topic 2
Concept 5
Restricted circular
functions
Concept summary
Practice questions
The inverse sine function
The sine function, y = sin(x), is a many-to-one function.
y
2
PR
O
AOS 1
O
FS
All circular functions are periodic and are many-to-one functions. Therefore, the
inverses of these functions cannot be functions. However, if the domain is restricted
so that the circular functions are one-to-one functions, then their inverses are
functions.
3π
– ––
2
−π
PA
G
E
1
– –π2
0
π
π
–
2
3π
––
2
x
TE
D
–1
–2
R
R
EC
Therefore, its inverse does not exist as a function. However there are many restrictions
3π π
π 3π
π π
of the domain, such as c − , − d , c − , d or c , d , that will ensure it is a one-to2 2
2 2
2 2
π π
one function. For convenience, let c − , d be the domain and [–1, 1] the range of the
2 2
restricted sine function.
C
O
π π
f : c − , d → [−1, 1]
2 2
U
N
y
(1, )
π
–
2
1
– –π2
(
–1, – –π2
)
0
π
–
2
x
–1
Therefore, it is a one-to-one function and its inverse exists.
The inverse of this function is denoted by sin−1. (An alternative notation is arcsin.)
Topic 2 Trigonometry c02Trigonometry.indd 85
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The graph of y = sin−1 (x) is obtained from the graph of y = sin(x) by reflection in
the line y = x.
π π
f : [ −1, 1 ] → c − , d , f (x) = sin−1(x)
2 2
y
2
(1, )
π
–
2
π
–
2
1
)
– –π2
–2
E
(
–1, – –π2
x
PR
O
–1
2
O
0
–1
–2
FS
1
16
Find each of the following.
a sin−1 (2)
R
tHinK
EC
WorKeD
eXaMPLe
TE
D
PA
G
π
π
There are an infinite number of solutions to sin(x) = 12, for example , 2π + and
6
6
π
4π + , since we can always add any multiple of 2𝜋 to any angle and get the same
6
π π
result. However, sin−1 a12 b means sin(x) = 12 and x ∈ c − , d , so there is only one
2 2
π
solution in this case: .
6
C
O
2 State the result.
R
a 1 Write an equivalent statement.
U
N
b 1 Use the known results.
2 Write an equivalent statement and
state the result.
c State the result.
86
b sin−1 asin a
WritE
a
5π
bb
6
c sin(sin−1 (0.5))
x = sin−1(2)
sin(x) = 2
This does not exist. There is no solution to sin ( x ) = 2.
5π
1
b= ,
6
2
5π
1
x = sin−1 asin a b b = sin−1 a b
6
2
π π
sin(x) = 12 and x ∈ c − , d
2 2
π
The only solution is x = .
6
5π
π
sin−1 asin a b b =
6
6
b Since sin a
c sin(sin−1(0.5)) = 0.5
MaTHs QuesT 12 sPeCIaLIsT MaTHeMaTICs VCe units 3 and 4
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General results for the inverse sine function
In general, we have the following results for the inverse sine function:
π π
f : [ −1, 1 ] → c − , d , f (x) = sin−1(x)
2 2
sin(sin−1(x)) = x if x ∈ [−1, 1]
FS
π π
sin−1(sin(x)) = x if x ∈ c − , d
2 2
The inverse cosine function
The cosine function, y = cos(x), is a many-to-one function.
O
y
PR
O
2
1
–π
– –π2
0
E
3π
– ––
2
π
–
2
π
3π
––
2
x
PA
G
–1
TE
D
–2
EC
Therefore, its inverse does not exist as a function. However, there are many
restrictions of the domain, such as [−π , 0] or [0 , π] or [π , 2π], that will ensure it is a
one-to-one function. Let [0, 𝜋] be the domain and [–1, 1] the range of the restricted
cosine function.
f : [0, π] → [−1, 1] where f(x) = cos(x).
R
R
y
U
N
C
O
2
1
−π
– –π2
0
–1
(0, 1)
π
–
2
π
x
(π, –1)
–2
Therefore, it is a one-to-one function and its inverse exists.
The inverse of this function is denoted by cos−1. (An alternative notation is arccos.)
Topic 2 Trigonometry c02Trigonometry.indd 87
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The graph of y = cos−1 (x) is obtained from the graph of y = cos(x) by reflection in
the line y = x.
f : [−1, 1] → [0, π], f(x) = cos−1(x)
y
4
3 π
2
π
–
2
1
–4 –3 –2 –1 0
–1
(1, 0)
1
2
3
4
x
FS
(–1, π)
O
–2
–3
PR
O
–4
PA
G
E
π
π
There are an infinite number of solutions to cos(x) = !2
, for example , 2π + ,
2
4
4
π
π
π
4π + , 2π − and 4π − , since we can always add any multiple of 2𝜋 to any
4
4
4
!2
angle. However, cos−1 a !2
and x ∈ [0 , π], so there is only one
b means cos(x) =
2
2
π
solution, namely .
4
tHinK
Find each of the following.
5π
3
a cos−1 a b
b cos−1 acos a b b
2
4
TE
D
17
EC
WorKeD
eXaMPLe
a 1 Write an equivalent statement.
R
R
2 State the result.
U
N
C
O
b 1 Use the known results.
2 Write an equivalent statement and
state the result.
c State the result.
88
WritE
c cos acos−1 a
π
bb
12
3
a x = cos−1 a2b
This does not exist. There is no solution to cos(x) = 32 .
!2
5π
,
b=−
4
2
!2
5π
x = cos−1 acos a b b = cos−1 a−
b
4
2
b Since cos a
cos(x) = − !2
and x ∈ [0, π]
2
3π
The only solution is x = .
4
5π
3π
cos−1 acos a b b =
4
4
c cos acos−1 a
π
π
bb =
12
12
MaTHs QuesT 12 sPeCIaLIsT MaTHeMaTICs VCe units 3 and 4
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WORKED
EXAMPLE
18
Find the exact value of cos a sin−1 a 13 bb
b .
WRITE/DRAW
are angles.
Let θ = sin−1 a13b so that sin(θ) = 13 .
O
1 The inverse trigonometric functions
2 Draw a right-angled triangle and label
the side lengths using the definition of
the trigonometric ratios.
3
= 2 !2
3
R
3
1
θ
2 2
U
N
C
O
R
= 32
=9
=9−1
=8
= 2 !2
cos asin−1 a13bb = cos(θ)
EC
4 State the required value.
TE
D
using Pythagoras.
x2 + 12
x2 + 1
x2
x2
x
x
PA
G
3 Calculate the value of the third side
1
E
θ
PR
O
THINK
FS
General results for the inverse cosine function
In general, we find that:
f : [−1, 1] → [0, π], f (x) = cos−1 (x)
cos(cos−1(x)) = x if x ∈ [−1, 1]
cos−1(cos(x)) = x if [0, π]
Double-angle formulas
Sometimes we may need to use the double-angle formulas.
sin(2A) = 2 sin(A)cos(A)
cos(2A) = cos2 (A) − sin2 (A)
= 2 cos2 (A) − 1
= 1 − 2 sin2 (A)
Topic 2 TRIGONOMETRY
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19
Find the exact value of sin a 2 cos−1 a 25 b b .
WRITE/DRAW
Let θ = cos−1 a25b so that cos(θ ) = 25 .
x2 + 22
lengths using the definition of the trigonometric x2 + 4
x2
ratios. Calculate the value of the third side using
x2
Pythagoras’ theorem.
x
2 Draw a right-angled triangle and label the side
= 52
= 25
= 25 − 4
= 21
= !21
5
E
21
FS
1 The inverse trigonometric functions are angles.
O
THINK
PR
O
WORKED
EXAMPLE
PA
G
θ
2
sin(2θ ) = 2 sin(θ )cos(θ )
3 Use an appropriate double-angle
formula.
=2×
!21
5
× 25
TE
D
sin a2 cos−1 a25bb = 4 !21
25
4 State what is required.
EC
The inverse tangent function
The tangent function, y = tan(x), is a many-to-one function.
y
R
R
4
U
N
C
O
2
−2π
3π
– ––
2
−π
0
– –π2
π
–
2
π
3π
––
2
2π
x
–2
–4
Therefore, its inverse does not exist as a function. However, there are many
3π π
π π
π 3π
restrictions of the domain, such as a− , − b or a− , b or a , b, that will
2 2
2 2
2 2
ensure it is a one-to-one function.
90
MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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π π
Let a− , b be the domain and R the range of the restricted tangent function. Note
2 2
π
that we must have an open interval, because the function is not defined at x = ± ; at
2
these points we have vertical asymptotes.
π π
f : a− , b → R, f (x) = tan(x)
2 2
4
2
x
O
0
–2
–4
PR
O
π
–
2
E
– –π2
FS
y
U
N
C
O
R
R
EC
TE
D
PA
G
Therefore, it is a one-to-one function and its inverse exists.
The inverse of this function is denoted by tan−1. (An alternative notation is arctan.)
The graph of y = tan−1 (x) is obtained from the graph of y = tan(x) by reflection in
the line y = x.
π π
f : R → a− , b where f (x) = tan−1(x)
2 2
π
Note that there horizontal asymptotes at y = ± .
2
y
3
2
1
π
–
2
–4 –3 –2 –1 0
1
–1 – –π
2
–2
2
3
4
x
–3
π
π
There are an infinite number of solutions to tan(x) = !3, for example, , 2π +
3
3
π
and 4π + , since we can always add any multiple of 2𝜋 to any angle. However,
3
π π
π
tan−1( !3) means tan(x) = !3 and x ∈ a− , b, so there is only one solution: .
2 2
3
Topic 2 Trigonometry c02Trigonometry.indd 91
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20/08/15 10:42 AM
20
tHinK
a
Find:
a tan−1 atan a
11π
bb
6
b tan(tan−1 (2)).
WritE
!3
11π
b=−
6
3
!3
11π
Let tan−1 atan a
b b = tan−1 a−
b=x
6
3
a tan a
1 Use the known results.
2 Write an equivalent statement and
b
PR
O
O
state the result.
!3
π π
and x ∈ a− , b
2 2
3
π
The only solution is x = − .
6
11π
π
−1
tan atan a
bb = −
6
6
−1
b tan(tan (2)) = 2
tan(x) = −
FS
WorKeD
eXaMPLe
3 State the result.
PA
G
E
general results for the inverse tan function
In general, we find that:
π π
f : R → a− , b, f (x) = tan−1(x)
2 2
tan(tan−1(x)) = x if x ∈ R
TE
D
π π
tan−1(tan(x)) = x if x ∈ a− , b
2 2
2 tan(A)
.
1 − tan2 (A)
Find the exact value of tan a 2 tan−1 a 12 b b .
R
21
R
WorKeD
eXaMPLe
EC
Double-angle formulas
It may be necessary to use the double-angle formulas, such as tan(2A) =
WritE
O
tHinK
1
1
C
1 The inverse trigonometric functions are angles. Let θ = tan−1 a b so that tan(θ ) = .
2
2
U
N
2 Use the double-angle formulas.
tan(2θ ) =
=
=
=
3 State the result.
92
2 tan(θ )
1 − tan2(θ )
2×
1
2
1 − a12b
1
2
1 − 14
1
3
4
−1
tan a 2 tan a 21 b b
= 43
MaTHs QuesT 12 sPeCIaLIsT MaTHeMaTICs VCe units 3 and 4
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Compound-angle formulas
We may also need to use the compound-angle formulas:
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)
tan(A) + tan(B)
1 − tan(A)tan(B)
tan(A − B) =
tan(A) − tan(B)
1 + tan(A)tan(B)
WORKED
EXAMPLE
22
−1 3
Evaluate cos a sin−1 a 12
b − tan a b b .
13
4
THINK
WRITE/DRAW
2 Draw the right-angled triangle and state
and tan(B) = 34 .
Thus, sin(A) = 12
13
TE
D
the unknown side length using well-known
Pythagorean triads.
PA
G
angles. Use the definitions of the inverse
trigonometric functions.
−1 3
Let A = sin−1 a12
b and B = tan a b.
13
4
E
1 The inverse trigonometric functions are
PR
O
O
tan(A + B) =
FS
cos(A − B) = cos(A)cos(B) + sin(A)sin(B)
EC
13
12
A
U
N
C
O
R
R
5
3 State the ratios from the triangles.
3
5
B
4
5
sin(A) = 12
, cos(A) = 13
13
sin(B) = 35 , cos(B) = 45
4 Substitute the ratios into the
compound-angle formulas.
5 State the required result.
cos(A − B) = cos(A)cos(B) + sin(A)sin(B)
5
= 13
× 45 + 12
× 35
13
12
56
−1 3
cos a sin−1 a 13
b − tan a b b =
4
65
Topic 2 TRIGONOMETRY
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Determining maximal domains and ranges
π π
For y = sin−1(x), the domain is [–1, 1] and the range is c − , d .
2 2
For y = cos−1(x), the domain is [–1, 1] and the range is [0, π].
π π
For y = tan−1(x), the domain is R and the range is a− , b.
2 2
State the domain and range of:
tHinK
3x − 2
b−3
5
b y = 4 tan−1 a
WritE
a 1 y = cos−1(x) has a domain of [–1, 1].
3x − 2
` ≤1
5
3x − 2
−1 ≤
≤1
5
a `
PA
G
2 Use the definition of the
2x − 7
b + 1.
6
O
a y = 2 cos−1 a
PR
O
23
E
WorKeD
eXaMPLe
FS
For inverse trigonometric functions that have been dilated or translated, we can
apply these dilations and translations to determine the domain and range of the
transformed function.
modulus function.
3 Solve the inequality.
−5 ≤ 3x − 2 ≤ 5
−3 ≤ 3x ≤ 7
y = 2 cos−1 a
TE
D
4 State the domain.
R
O
6 State the range.
R
EC
5 y = cos−1(x) has a range of [0, π].
U
N
C
b 1 y = tan−1(x) has a domain of R.
π π
b.
2 2
2 y = tan−1(x) has a range of a− ,
3 State the range.
94
3x − 2
b − 3 has a maximal domain of
5
−1 ≤ x ≤ 73 or c −1, 73 d .
There is a dilation by a factor of 3 parallel to the
y-axis and a translation of 2 units down parallel to
the y-axis. The range is from 2 × 0 − 3 to 2 × π − 3.
y = 2 cos−1 a
b y = 4 tan−1 a
4x − 3
b − 3 has a range of [−3, 2π − 3].
5
2x − 7
b + 1 has a domain of R.
6
There is a dilation by a factor of 4 parallel to the
y-axis and a translation of 1 unit up parallel to the
−π
π
+ 1 to 4 × + 1,
y-axis. The range is from 4 ×
2
2
not including the end points.
2x − 7
b + 1 has a range of
6
(−2π + 1, 2π + 1).
y = 4 tan−1 a
MaTHs QuesT 12 sPeCIaLIsT MaTHeMaTICs VCe units 3 and 4
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Exercise 2.6 Inverse trigonometric functions
WE16
Find each of the following.
a sin−1(1.1)
2 Find each of the following.
b sin−1 asin a
5π
bb
3
7π
6
b sin−1 asin a b b
6
5
WE17
3
Find each of the following.
7π
a cos−1(1.2)
b cos−1 acos a b b
6
4 Find each of the following.
11π
4
a cos−1 a b
b cos−1 acos a
bb
3
3
a sin−1 a− b
WE18
Find the exact value of cos a sin−1 a 51 b b .
3
6 Find the exact value of sin a cos−1 a 7 b b .
7
WE19
1
3
c sin asin−1 a b b
π
6
c cos acos−1 a b b
1
4
c cos acos−1 a b b
PR
O
5
c sin (sin−1 (0.9))
FS
1
O
PRactise
Find the exact value of sin a 2 cos−1 a 74 b b .
3
WE20
PA
G
9
Find:
a tan−1 atan a
10 Find:
7π
bb
6
5π
bb
3
b tan (tan−1 (1.1)).
5
4
TE
D
a tan−1 atan a
E
8 Find the exact value of cos a 2 sin−1 a 8 b b .
1
b tan atan−1 a b b.
11 WE21 Find the exact value of tan a 2 tan−1 a 3 b b
1
EC
12 Find the exact value of cot a 2 tan−1 a 4 b b
3
5
R
13 WE22 Evaluate sin a cos−1 a 5 b − tan−1 a 12 b b .
3
5
R
14 Evaluate tan a sin−1 a 5 b − cot−1 a 12 b b .
U
N
C
O
15 WE23 State the domain and range of:
Consolidate
2x − 5
b − 2π
4
16 State the domain and range of:
4
a y = cos−1(3x + 5) − 3
π
17 Evaluate each of the following.
a sin−1(1)
b sin−1(1.3)
a y = 3 sin−1 a
d cos−1(−1)
g tan−1( !3)
1
e cos−1 a −2 b
b y =
b y =
6
3x − 5
tan−1 a
b + 2.
π
4
8
tan−1(10x) + 3.
π
"3
b
2
c sin−1 a −
f cos−1(−1.2)
"3
b
3
h tan−1 a −
Topic 2 Trigonometry c02Trigonometry.indd 95
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20/08/15 10:43 AM
18 Evaluate each of the following.
π
5
a sin−1(sin(1.2))
5π
bb
6
5π
f cos−1 acos a b b
3
b sin−1 asin a b b
c sin−1 asin a
π
bb
10
4π
h tan−1 atan a b b
3
d cos−1(cos(0.5))
e cos−1 acos a
π
8
g tan−1 atan a b b
19 Evaluate each of the following.
b cos a sin−1 a −
d sin(tan−1(−1))
!3
bb
2
c tan a sin−1 a −2 b b
2
c tan a sin−1 a −6 b b
1
FS
1
a sin a cos−1 a 2 b b
1
f tan a cos−1 a −
e cos a tan−1 a − !3 b b
b tan a cos−1 a −3 b b
5
2
d sin a tan−1 a 8 b b
e cos a sin−1 a 5 b b
21 Evaluate each of the following.
1
3
b tan a 2 sin−1 a 4 b b
2
2
f cos a 2 sin−1 a 5 b b
PA
G
e tan a 2 cos−1 a 5 b b
22 Evaluate each of the following.
12
3
TE
D
a sin a cos−1 a 13 b + sin−1 a 5 b b
15
9
c cos a tan−1 a 8 b + cos−1 a 41 b b
23 Show that:
15
8
π
b + tan−1 a b =
17
15
2
R
R
c sin−1 a
π
7
7
b + tan−1 a b =
24
25
2
EC
a cos−1 a
3
5
O
e tan−1 (4) − tan−1 a b =
U
N
C
24 Show that:
3
24
a 2 sin−1a5b = sin−1a25b
1
7
c 2 cos−1a4b = cos−1a−8b
1
3
e 2 tan−1a3b = tan−1a4b
π
4
1
c cos a 2 tan−1 a 3 b b
1
d sin a 2 tan−1 a 3 b b
7
f cos a tan−1 a −4 b b
E
a sin a 2 cos−1 a 4 b b
5
PR
O
2
a sin a cos−1 a 9 b b
O
20 Evaluate each of the following.
!2
bb
2
4
5
8
60
b cos a cos−1 a 5 b − sin−1 a 13 b b
d sin a tan−1 a 15 b − sin−1 a 61 b b
b sin−1 a
5
π
12
b + tan−1 a b =
13
12
2
1
2
π
4
2
3
π
4
d tan−1 (3) − tan−1 a b =
f tan−1 (5) − tan−1 a b = .
7
336
2
1
b 2 sin−1a25b = sin−1a625b
d 2 cos−1a3b = cos−1a−9b
1
8
f 2 tan−1a4b = tan−1a15b.
25 State the implied domain and range of each of the following.
a y = 2 sin−1(x − 1)
b y = 3 cos−1(x − 2)
c y = 4 tan−1(x − 3)
x
3
d y = 5 sin−1 a b
96 x
4
e y = 6 cos−1 a b
x
5
f y = 7 tan−1 a b
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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26 State the implied domain and range of each of the following.
a y = 2 sin−1(3x − 1) + π
c y = 5 tan−1(4x + 3) −
e y =
π
2
d y =
5
4 − 3x
cos−1 a
b−4
π
7
f y =
4 −1 3 − 4x
sin a
b+2
π
5
8
5x − 3
tan−1 a
b+3
π
4
27aState a sequence of transformations that, when applied to y = sin−1(x),
x
produce the graph of y = a + b sin−1 a b. Hence, state the domain and
c
x
range of y = a + b sin−1 a b.
c
b State a sequence of transformations that, when applied to y = cos−1(x),
produce the graph of y = a + b cos−1(cx). Hence, state the domain and
range of y = a + b cos−1(cx).
c State a sequence of transformations that, when applied to y = tan−1(x),
x
produce the graph of y = a + b tan−1 a b. Hence, state the domain and
c
x
range of y = a + b tan−1 a b.
c
28 Show that:
PA
G
E
PR
O
O
FS
Master
b y = 3 cos−1(2x − 5) − π
a sin−1(x) = cos−1("1 − x2) for x ∈ [0, 1]
1
x
π
for x > 0
2
TE
D
b tan−1(x) + tan−1 a b =
2
c cos−1(x) = tan−1 q "1 − x r for x ∈ (0, 1)
x
a
π
b
r + tan−1 a b = for a > 0 and b > 0
a
2
"a + b
EC
d sin−1 q
U
N
C
O
R
R
e cos−1 q
2
2
a
π
a
r + tan−1 a b = for a > 0 and b > 0
b
2
"a + b
2
2
π
x−1
b = for x > −1
x+1
4
2
2
x −1
2x
−1
−1 x − 1 for x > 0.
g sin−1 a
=
cos
=
tan
b
b
a
a
b
2x
x2 + 1
x2 + 1
f tan−1(x) − tan−1 a
2.7 General solutions of trigonometric equations
In this section consideration is given to the general solutions of trigonometric
equations, rather than finding the solutions over a specified domain.
General solutions of trigonometric equations
Trigonometric equations can have an infinite number of solutions. To express the
possible solutions mathematically, we derive formulas that will give the general
solution in terms of any natural number n, where n ∈ Z.
Topic 2 Trigonometry c02Trigonometry.indd 97
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20/08/15 10:43 AM
PR
O
O
FS
general solutions involving cosines
Consider the equation cos(x) = a. One answer is x = cos−1 (a).
π
If 0 < a < 1, then 0 < x < , so x is in the first quadrant.
2
θ
Because cosine is positive in the first and fourth quadrant, there
2π
‒
θ
is also another answer, x = 2π − cos−1 (a).
We can add or subtract any multiple of 2π to either answer and
obtain an equivalent angle.
cos(x) = a
x = cos−1 (a), 2π + cos−1 (a), 4π + cos−1 (a), …
x = 2π − cos−1 (a), 4π − cos−1 (a), 6π − cos−1 (a), …
The totality of solutions can be represented as x = 2nπ ± cos−1 (a), where n ∈ Z.
Although we have demonstrated this result for 0 < a < 1, it is in fact true
for −1 ≤ a ≤ 1.
WorKeD
eXaMPLe
24
PA
G
E
The general solution of cos(x ) = a where −1 ≤ a ≤ 1
The general solution of cos(x) = a where −1 ≤ a ≤ 1 is given by
x = 2nπ ± cos−1 (a), where n ∈ Z.
1
Find the general solution to the equation cos (x) = .
2
WritE
TE
D
tHinK
1 State one solution.
EC
2 State the general solution.
3 Take out a common factor so that the general
R
R
solution can be written in simplest form.
π
1
x = cos−1 a b =
2
3
π
x = 2nπ ±
3
π
x = (6n ± 1) where n ∈ Z
3
U
N
C
O
general solutions involving sines
Consider the equation sin(x) = a. One answer is x = sin−1 (a),
π‒θ
π
and if 0 < a < 1, then 0 < x < , so x is in the first quadrant.
2
θ
Since sine is positive in the first and second quadrants, there is
also another answer, x = π − sin−1 (a).
We can add or subtract any multiple of 2π to either answer and
obtain an equivalent angle.
sin(x) = a
x = sin−1 (a), 2π + sin−1 (a), 4π + sin−1 (a), …
x = π − sin−1 (a), 3π − sin−1 (a), 5π − sin−1 (a), …
If n is any integer, then 2n is an even integer and 2n + 1 is an odd integer.
98
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The totality of solutions can be represented as x = 2nπ + sin−1 (a) or
x = (2n + 1)π − sin−1 (a), where n ∈ Z.
Although we have demonstrated this result for 0 < a < 1, it is true for −1 ≤ a ≤ 1.
The general solution of sin(x ) = a where −1 ≤ a ≤ 1
The general solution of sin(x) = a where −1 ≤ a ≤ 1 is given by
x = 2nπ + sin−1 (a), (2n + 1)π − sin−1 (a), where n ∈ Z.
25
Find the general solution to the equation sin(x) =
3 Take out common factors in the first solution
4 Take out common factors in the
EC
TE
D
second solution
PA
G
so that the general solution can be written in
simplest form.
π
2π
x = (6n + 1), (3n + 1) where n ∈ Z
3
3
R
R
E
2 State the general solution.
5 State the general solution.
!3
π
b=
2
3
π
π
x = 2nπ + or x = (2n + 1)π −
3
3
π
x = 2nπ +
3
π
= (6n + 1)
3
π
x = (2n + 1)π −
3
π
= 2nπ + π −
3
2π
= 2nπ +
3
2π
= (3n + 1)
3
x = sin−1 a
1 State one solution.
O
WritE
PR
O
tHinK
!
!3
.
2
FS
WorKeD
eXaMPLe
U
N
C
O
general solutions involving tangents
Consider the equation tan(x) = a. One answer is x = tan−1 (a),
π
and if a > 0, then 0 < x < , so x is in the first quadrant. Since
2
θ
tangent is positive in the first and third quadrants, there is also
π+θ
another answer, x = π + tan−1 (a).
We can add or subtract any multiple of 2π to either answer and
obtain an equivalent angle.
tan(x) = a
x = tan−1 (a), 2π + tan−1 (a), 4π + tan−1 (a), …
x = π + tan−1 (a), 3π + tan−1 (a), 5π + tan−1 (a), …
The totality of solutions can be represented as one solution: x = nπ + tan−1 (a),
where n ∈ Z.
Topic 2 TrIgonoMeTry
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The general solution of tan(x ) = a
The general solution of tan(x) = a where a ∈ R is given by
x = nπ + tan−1 (a), where n ∈ Z.
Although we have demonstrated this result only for a > 0, it is true for a ∈ R.
26
tHinK
Find the general solution to the equation tan(x) = !3
! .
WritE
π
3
x = tan−1 (!3) =
2 State the general solution.
x = nπ +
3 Take out a common factor so that
π
x = (3n + 1) where n ∈ Z
3
PR
O
the general solution can be written in
simplest form.
π
3
FS
1 State one solution.
O
WorKeD
eXaMPLe
PA
G
E
general solution of trigonometric equations
When solving more complicated trigonometric equations, often multiple solutions
exist. We may be required to find all solutions to each part of the equation being
considered.
27
Find the general solution to the equation 4 cos2 (2x) − 3 = 0.
tHinK
TE
D
WorKeD
eXaMPLe
WritE
EC
1 Make the trigonometric function
U
N
C
O
R
R
the subject.
2 Use the formula to find the general
solution of the first equation.
100
4 cos2 (2x) − 3 = 0
cos2 (2x) = 34
cos(2x) = ± !3
2
So that:
(1) cos(2x) =
!3
2
or
(2) cos(2x) = − !3
2
cos(2x) =
!3
2
2x = 2nπ ± cos−1 a !3
b
2
π
2x = 2nπ ±
6
π
2x = (12n ± 1)
6
π
x = (12n ± 1)
12
MaTHs QuesT 12 sPeCIaLIsT MaTHeMaTICs VCe units 3 and 4
c02Trigonometry.indd 100
20/08/15 10:43 AM
3 Use the formula to find the general
solution of the second equation.
cos(2x) = − !3
2
2x = 2nπ ± cos−1 a− !3
b
2
2x = 2nπ ±
FS
π
2x = (12n ± 5)
6
π
x = (12n ± 5)
12
π
π
x = (12n ± 1) or x = (12n ± 5) where n ∈ Z
12
12
O
4 State the final general solutions.
5π
6
PR
O
general solutions involving phase shifts
When solving trigonometric equations involving phase shifts, we must solve the
resulting equations for the unknown values of x.
π
! sina3x + b + 1 = 0.
Find the general solution of !2
4
E
28
PA
G
WorKeD
eXaMPLe
tHinK
WritE
R
EC
TE
D
1 Make the trigonometric function the subject.
C
O
R
2 Use the formula to state the general solution.
U
N
3 Solve the first equation.
π
!2 sina3x + b + 1 = 0
4
π
!2 sina3x + b = −1
4
π
1
sina3x + b = −
4
!2
(1) 3x +
(2) 3x +
π
1
= 2nπ + sin−1 a−
b or
4
!2
π
1
= (2n + 1)π − sin−1 a−
b
4
!2
π
1
= 2nπ + sin−1 a−
b
4
!2
π
π
3x + = 2nπ −
4
4
π
3x = 2nπ −
2
π
3x = (4n − 1)
2
π
x = (4n − 1)
6
3x +
Topic 2 TrIgonoMeTry
c02Trigonometry.indd 101
101
20/08/15 10:43 AM
π
1
= (2n + 1)π − sin−1 a−
b
4
!2
π
π
3x + = 2nπ + π +
4
4
3x = π(2n + 1)
π
x = (2n + 1)
3
π
π
x = (4n − 1) or (2n + 1) n ∈ Z
6
3
3x +
4 Solve the second equation.
FS
5 State the final solutions.
WorKeD
eXaMPLe
29
Find the general solution of the equation 2 sin2 (2x) + sin(2x) − 1 = 0.
tHinK
WritE
2 sin2 (2x) + sin(2x) − 1 = 0
Let u = sin(2x).
PA
G
E
1 Use a substitution.
2u2 + u − 1 = 0
(2u − 1)(u + 1) = 0
2 Factorise.
4 Solve the trigonometric equation
using the Null Factor Law.
C
O
R
R
first equation.
U
N
(1) sin(2x) = 12
or
(2) sin(2x) = −1
sin(2x) = 12
EC
5 Find the general solution of the
TE
D
(2 sin(2x) − 1)(sin(2x) + 1) = 0
3 Substitute back for u.
102
PR
O
O
equations reducible to quadratics
Equations can often be reduced to quadratics under a suitable substitution.
2x = 2nπ + sin−1 a 12 b
π
2x = 2nπ +
6
π
2x = (12n + 1)
6
π
x = (12n + 1)
12
1
sin(2x) = 2
2x = (2n + 1)π − sin−1 a 12 b
π
2x = 2nπ + π −
6
5π
2x = 2nπ +
6
π
2x = (12n + 5)
6
π
x = (12n + 5)
12
MaTHs QuesT 12 sPeCIaLIsT MaTHeMaTICs VCe units 3 and 4
c02Trigonometry.indd 102
20/08/15 10:43 AM
FS
sin(2x) = −1
2x = 2nπ + sin−1 (−1)
π
2x = 2nπ −
2
π
2x = (4n − 1)
2
π
x = (4n − 1)
4
sin(2x) = −1
2x = (2n + 1)π − sin−1 (−1)
π
2x = 2nπ + π +
2
3π
2x = 2nπ +
2
π
2x = (4n + 3)
2
π
x = (4n + 3)
4
PR
O
second equation.
O
6 Find the general solution of the
Let n = 0, 1, 2, 3, 4.
π
x = (4n − 1)
4
⇒ x = −π , 3π , 7π , 11π , 15π
4 4 4
4
4
π
x = (4n + 3)
4
3π
7π 11π 15π
⇒ x=
,
,
,
4
4
4
4
E
7 Sometimes some parts of the
EC
TE
D
PA
G
solution are already included
in some other parts. Give n
some values.
U
N
C
R
O
the equation.
R
8 State all the general solutions of
π
We can see that the solution x = (4n + 3) incorporates all
4
π
the solutions from x = (4n − 1).
4
π
π
π
x = (4n − 1) or (12n + 1) or (12n + 5) where n ∈ Z.
4
12
12
WorKeD
eXaMPLe
30
Trigonometric equations involving multiple angles
We can find the general solutions to trigonometric equations involving multiple angles
by applying the general solution formulas rather than expanding the multiple angles.
Find the general solution to cos(4x) = sin(2x).
tHinK
1 Rewrite using one trigonometric function.
Convert sines into cosines, since the solution
for cosine is easier to work with.
WritE
π
Use sina − Ab = cos(A).
2
cos(4x) = sin(2x)
π
cos(4x) = cosa − 2xb
2
Topic 2 TrIgonoMeTry
c02Trigonometry.indd 103
103
20/08/15 10:43 AM
π
4x = 2nπ ± a − 2xb
2
2 Solve using an appropriate general solution.
π
(1) 4x = 2nπ + a − 2xb or
2
PR
O
O
FS
3 Solve the first equation.
π
(2) 4x = 2nπ − a − 2xb
2
π
4x = 2nπ + − 2x
2
π
6x = 2nπ +
2
π
6x = (4n + 1)
2
π
x = (4n + 1)
12
π
4x = 2nπ − + 2x
2
π
2x = 2nπ −
2
π
2x = (4n − 1)
2
π
x = (4n − 1)
4
π
π
x = (4n + 1) or (4n − 1) where n ∈ Z.
12
4
TE
D
5 State the general solutions of the equation.
PA
G
E
4 Solve the second equation.
U
N
C
O
R
R
EC
Comparison of examples
Note that the last two worked examples, 29 and 30, are in fact the same, as
cos(4x) = sin(2x)
cos(2(2x)) = sin(2x) by double-angle formulas
1 − 2 (sin(2x)) 2 = sin(2x)
2 sin2 (2x) + sin(2x) − 1 = 0
and therefore they should have the same general solution. The two given answers do
π
not appear to be the same, although one answer, x = (4n − 1), is common to both.
4
This situation is very common in these types of problems. However, if we substitute
values of n, the two results generate the same particular solutions.
When n = 0, 1, 2, 3, 4, 5, 6 from Worked example 29:
π
π
x = (4n − 1) ⇒ x = − ,
4
4
π
π
x = (12n + 1) ⇒ x = ,
12
12
π
5π
x = (12n + 5) ⇒ x = ,
12
12
104 3π
,
4
13π
,
12
17π
12
7π 11π 15π 19π 23π
,
,
,
,
4
4
4
4
4
25π
12
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c02Trigonometry.indd 104
20/08/15 10:43 AM
When n = 0, 1, 2, 3, 4, 5, 6 from Worked example 30:
FS
π
π 3π 7π 11π 15π 19π 23π
x = (4n − 1) ⇒ x = − ,
,
,
,
,
,
4
4 4 4
4
4
4
4
π
π 5π 9π 13π 17π 21π 25π
x = (4n + 1) ⇒ x = ,
, ,
,
,
,
12
12 12 12 12 12 12 12
It is interesting to compare these results to those obtained by CAS calculators. In
some cases a calculator will not solve the equation for the general solution, and in
other cases it will. The solution obtained by CAS may be in a different form to our
answers above. The results may be given differently depending on the MODE, which
could be set to either Exact or Auto.
WE24
Find the general solution to the equation cos(x) =
2 Determine the general solution of 2 cos(2x) + !3 = 0.
3
WE25
!2
.
2
PR
O
PRactise
1
O
Exercise 2.7 General solutions of trigonometric equations
Find the general solution to the equation sin(x) = 12.
4 Determine the general solution of 2 sin(2x) + !3 = 0.
WE26
Find the general solution to the equation tan(x) = 1.
E
5
7
WE27
PA
G
6 Find the general solution to tan(2x) + !3 = 0.
Find the general solution to the equation 4 cos2 (2x) − 1 = 0.
8 Find the general solution to the equation 3 tan2 (2x) − 1 = 0.
WE28
π
Find the general solution of 2 sin a3x + b − 1 = 0.
6
TE
D
9
π
6
EC
10 Find the general solution of 2 cos a2x − b + !3 = 0.
11 WE29 Find the general solution to 2 sin2 (2x) − 3 sin(2x) + 1 = 0.
R
12 Find the general solution to the equation 2 cos2 (2x) + cos(2x) − 1 = 0.
R
13 WE30 Find the general solution to cos(3x) = sin(2x).
O
14 Find the general solution to cos(4x) = sin(3x).
U
N
C
Consolidate
15 Find the general solution to each of the following equations.
a 2 cos(3x) − !3 = 0
c !2 sin(2x) + 1 = 0
b 2 cos(2x) + 1 = 0
d 2 sin(3x) + 1 = 0
16 Find the general solution to each of the following equations.
a 4 sin2 (2x) − 3 = 0
c 4 cos2 (3x) − 1 = 0
b 2 sin2 (2x) − 1 = 0
d 2 cos2 (3x) − 1 = 0
17 Find the general solution to each of the following equations.
a tan(x) + !3 = 0
c tan2 (2x) − 3 = 0
b !3 tan(3x) − 1 = 0
d 3 tan2 (2x) − 1 = 0
a 2 sin2 (2x) + sin(2x) = 0
c 2 cos2 (2x) + !3 cos(2x) = 0
b cos2 (2x) − cos(2x) = 0
d 2 sin2 (2x) − !3 sin(2x) = 0
18 Find the general solution to each of the following equations.
Topic 2 Trigonometry c02Trigonometry.indd 105
105
20/08/15 10:43 AM
19 Find the general solution to each of the following equations.
a 2 sin2 (2x) + 3 sin(2x) + 1 = 0
b 2 cos2 (2x) − 3 cos(2x) + 1 = 0
20 Find the general solution to each of the following equations.
π
4
π
6
a !2 sina3x − b − 1 = 0
b 2 sina2ax + b b + 1 = 0
21 Find the general solution to each of the following equations.
b 2 cosa3ax −
π
bb − 1 = 0
12
22 Find the general solution to each of the following equations.
π
4
b !3 tana2ax −
a tan2 (x) + (!3 + 1)tan(x) + !3 = 0
b tan2 (x) + (!3 − 1)tan(x) − !3 = 0
PR
O
23 Find the general solution to each of the following equations.
π
bb + 1 = 0
12
O
a tana3x + b − 1 = 0
FS
π
6
a 2 cosa2x + b + !3 = 0
24 Find the general solution to each of the following equations.
25 Find the general solution to each of the following equations.
a 2 sin3 (x) + sin2 (x) − 2 sin(x) − 1 = 0
b 2 cos3 (x) − cos2 (x) − 2 cos(x) + 1 = 0
PA
G
Master
b cos(x) = cos(2x)
E
a sin(2x) = sin(x)
26 Find the general solution to each of the following equations.
1
. We can
f(x)
1
1
1
use this method to graph sec(x) =
, cosec(x) =
and cot(x) =
.
cos(x)
sin(x)
tan(x)
Topic 1 described how the graph of f(x) can be used to find the graph of R
AOS 1
Graphs of reciprocal trigonometric functions
EC
2.8
TE
D
a tan3 (x) − tan2 (x) − tan(x) + 1 = 0
b tan4 (x) − 4 tan2 (x) + 3 = 0
Topic 2
R
The graph of y = sec(x )
Consider the graph of y = cos(x).
y
C
Sketch graphs of
reciprocal circular
functions
Concept summary
Practice questions
O
Concept 2
U
N
4
3
2
1
3π
– ––
2
−π
– –π2
0
–1
π
–
2
π
3π
––
2
x
–2
–3
–4
106 Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c02Trigonometry.indd 106
20/08/15 10:43 AM
3π
3π π π
, − , and . This
2
2 2
2
3π
π
means that the reciprocal function will have vertical asymptotes at x = − , x = − ,
2
2
3π
π
x = and x = . The horizontal asymptote will be y = 0.
2
2
π
3π
The graph of y = cos(x) is below the x-axis for − < x < − and passes through
2
2
1
will also be below the x-axis in this
the point (–π, –1). This means that y =
cos(x)
interval and will pass through the point (–π, –1). It will follow a similar pattern in
π
3π
the region < x < .
2
2
π
π
In the region − < x < , the graph of y = cos(x) is above the x-axis and passes
2
2
1
will also be above the x-axis and
through the point (0, 1). This means that y =
cos(x)
will pass through (0, 1).
1
The graph of y =
(or y = sec(x)) is shown below.
cos(x)
E
y
PR
O
O
FS
In the portion of the graph shown, the x-intercepts occur at −
PA
G
4
3
2
1
TE
D
−π
– –π2
0
–1
π
–
2
π
3π
––
2
x
–2
–3
EC
3π
– ––
2
–4
R
The graph of y = cosec(x )
y
U
N
C
O
R
In a similar fashion, the graph of y = sin(x) can be used to determine the graph
1
(or y = cosec(x)).
of y =
sin(x)
The graph of y = sin(x) is shown below.
3
2
1
−2π
3π
– ––
2
−π
– –π2
0
–1
π
–
2
π
3π
––
2
2π
x
–2
–3
Note that in this instance, the x-intercepts occur at –2π, –π, 0, π and 2π.
Topic 2 Trigonometry c02Trigonometry.indd 107
107
20/08/15 10:52 AM
The graph of y =
1
looks like the following.
sin(x)
y
4
3
2
1
−2π
−π
3π
– ––
2
– –π2
0
–1
π
π
–
2
3π
––
2
–3
Use the graph of y = 2 cos(x) to sketch y =
−2π ≤ x ≤ 2π.
WritE/draW
1 Sketch y = 2 cos(x).
1
over the domain
2 cos(x)
E
tHinK
PR
O
31
O
–4
WorKeD
eXaMPLe
x
FS
–2
2π
PA
G
y
4
Period: 2π
Amplitude: 2
Horizontal shift: 0
Vertical shift: 0
3
TE
D
2
3π
– ––
2
−π
– –π2
0
–1
π
–
2
π
3π
––
2
2π
x
–2
EC
−2π
1
–3
R
R
–4
O
2 Find the x-intercepts
y
4
U
N
C
and hence the vertical
asymptotes for the
reciprocal graph.
3π
π
π
3π
, x = − , x = and x = . These will
2
2
2
2
be the vertical asymptotes for the reciprocal function.
x-intercepts occur at x = −
3
2
1
−2π
3π
– ––
2
−π
– –π2
0
–1
π
–
2
π
3π
––
2
2π
x
–2
–3
–4
108
MaTHs QuesT 12 sPeCIaLIsT MaTHeMaTICs VCe units 3 and 4
c02Trigonometry.indd 108
20/08/15 10:43 AM
3 The graph of y = 2 cos(x)
y
is above the x-axis in the
−3π
regions −2π ≤ x < −
,
2
π
π
− < x < and
2
2
3π
< x ≤ 2π. The graph of
2
1
y=
will also be
2 cos(x)
above the x-axis in these
regions. A maximum value
of y = 2 is reached in the
original graph, meaning
that a minimum of y = 12
will be reached in the
reciprocal graph.
4
3
2
1
−2π
−π
3π
– ––
2
– –π2
0
–1
π
π
–
2
3π
––
2
x
2π
–2
FS
–3
PR
O
O
–4
4 The graph of y = 2 cos(x)
y
is below the x-axis in the
3π
π
regions − < x < −
2
2
π
3π
and < x < .
2
2
1
Therefore, y =
2 cos(x)
is also below the
x-axis in these regions.
The minimum of
y = −2 will become a
maximum of y = −12.
3π
– ––
2
−π
TE
D
−2π
PA
G
E
4
– –π2
3
2
1
0
–1
π
–
2
π
3π
––
2
x
2π
–2
–3
1
−2π ≤ x ≤ 2π.
x
sina b
2
x
Use the graph of y = sina b to sketch y =
2
R
32
U
N
tHinK
C
O
WorKeD
eXaMPLe
R
EC
–4
x
1 Sketch y = sina b.
2
Period: 4π
Amplitude: 1
Horizontal shift: 0
Vertical shift: 0
over the domain
WritE/draW
y
3
2
1
−2π
3π
– ––
2
−π
– –π2
–1
0
π
–
2
π
3π
––
2
2π
x
–2
–3
Topic 2 TrIgonoMeTry
c02Trigonometry.indd 109
109
20/08/15 10:43 AM
x-intercepts occur at x = −2π, x = 0 and x = 2π. These will be the
vertical asymptotes for the reciprocal function.
2 Find the x-intercepts
and hence the vertical
asymptotes for the
reciprocal graph.
y
4
3
2
1
3π
– ––
2
−π
– –π2
–1
0
π
–
2
π
3π
––
2
–2
y
3
E
2
−2π
3π
– ––
2
PA
G
1
−π
– –π2
π
–
2
π
π
–
2
π
3π
––
2
2π
3π
––
2
2π
x
TE
D
–4
R
y
R
4
3
2
1
C
U
N
110 0
–3
O
is below the x-axis in
the region −2π ≤ x < 0.
1
The graph of y =
x
sina b
2
is also below the
x-axis in this region.
The minimum of
y = −1 will become a
maximum of y = −1.
–1
–2
EC
x
2
4 The graph of y = sina b
4
PR
O
–4
x
2
is above the x-axis in the
region 0 < x ≤ 2π. The
1
graph of y =
x
sina b
2
will also be above the
x-axis in this region.
A maximum value of
y = 1 is reached in the
original graph, meaning
that a minimum of y = 1
will be reached in the
reciprocal graph.
x
O
–3
3 The graph of y = sina b
2π
FS
−2π
−2π
3π
– ––
2
−π
– –π2
–1
0
x
–2
–3
–4
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c02Trigonometry.indd 110
20/08/15 10:43 AM
The graph of y = cot(x )
1
The graph of y = tan(x) can be used to find the graph of y =
(or y = cot(x)).
tan(x)
The graph of y = tan(x) is shown below.
y
8
6
4
2
3π
– ––
2
−π
– –π2
–2
0
π
–
2
π
2π
FS
−2π
3π
––
2
O
–4
x
PR
O
–6
–8
E
In the portion of the graph shown, the x-intercepts occur at –2π, –π, 0, π and 2π. This
means that the reciprocal function will have vertical asymptotes at x = −2π, x = −π,
x = 0, x = π and x = 2π.
−3π
3π
−π
π
,x=
, x = and x = . Therefore, the
2
2
2
2
reciprocal function will have x-intercepts at these positions.
PA
G
y = tan(x) has asymptotes at, x =
EC
TE
D
Remembering that sections of the graph that are above the x-axis for y = tan(x) will
1
and similarly for sections below the x-axis, the
also be above the x-axis for y =
tan(x)
1
graph of y =
(or y = cot (x)) looks like this:
tan(x)
y
R
8
6
R
4
U
N
C
O
2
−2π
3π
– ––
2
−π
– –π2
–2
0
π
–
2
π
3π
––
2
2π
x
–4
–6
–8
Topic 2 Trigonometry c02Trigonometry.indd 111
111
20/08/15 10:43 AM
33
1
x
Use the graph of y = tana b to sketch y =
2
−2π ≤ x ≤ 2π.
tHinK
WritE/draW
y
x
1 Sketch y = tana b.
2
Period: 2π
Dilation: 1
Horizontal shift: 0
Vertical shift: 0
8
6
FS
4
–2π
3π
– ––
2
–π
– –π2
–2
0
–6
π
The x-intercepts will be x = −π and x = π.
3 Find the x-intercepts for
x-intercepts occur at x = −2π, x = 0 and x = 2π.
These will be the vertical asymptotes for the
reciprocal function.
TE
D
EC
6
4
R
O
2π
8
2
–2π
3π
– ––
2
U
N
C
3π
––
2
x
y
R
for the reciprocal graph.
2π
PA
G
2 The graph of y = tana b
x
y = tana b and hence
2
the vertical asymptotes
3π
––
2
E
–8
x
2
has asymptotes at x = −π
and x = π. These will be
the x-intercepts of the
reciprocal function.
π
–
2
O
2
–4
112
over the domain
x
tana b
2
PR
O
WorKeD
eXaMPLe
–π
– –π2
–2
0
π
–
2
π
x
–4
–6
–8
MaTHs QuesT 12 sPeCIaLIsT MaTHeMaTICs VCe units 3 and 4
c02Trigonometry.indd 112
20/08/15 10:43 AM
y
4 If we consider the region
8
6
4
2
–2π
–π
3π
– ––
2
– –π2
–2
0
π
π
–
2
x
2π
3π
––
2
–4
–6
FS
between x = −2π and
x = 0, the graph of
x
y = tana b is initially
2
above the x-axis between
x = −2π and x = −π and
is then below the x-axis.
This will also be true for
the reciprocal function.
O
–8
y
5 In a similar fashion, the
graph for x = 0 to x = 2π
can be obtained.
PR
O
8
6
4
–π
π
––
2
PA
G
3π
– ––
2
–2
0
π
π
–
2
x
2π
3π
––
2
–4
–6
TE
D
–2π
E
2
Transformations of reciprocal trigonometric graphs
Sketch the graph of y =
R
34
O
R
WorKeD
eXaMPLe
EC
Transformations can also be applied to the reciprocal trigonometric graphs.
tHinK
U
N
C
π
1 Use the graph of y = sinax + b to find
4
1
.
the graph of y =
π
sinax + b
4
Amplitude: 1
Period: 2π
π
Horizontal shift: left
4
Vertical shift: 0
1
π
sinax + b
4
+ 1 over the domain [−π, 2π] .
WritE/draW
y
4
3
2
1
π
–π – ––
3π – –
– –π4 0
2
4
–1
–2
π
–
4
π
–
2
3π
––
4
π
5π
––
4
3π
––
2
7π
––
4
2π
x
–3
–4
Topic 2 TrIgonoMeTry
c02Trigonometry.indd 113
113
20/08/15 10:43 AM
1
2 Consider the graph of y =
π
sinax + b
4
−π
The asymptotes will occur at x =
,
4
3π
7π
x=
and x = .
4
4
y
.
4
3
2
1
π
–π – ––
3π – –
– –π4 0
2
–1
4
π
–
4
π
–
2
3π
––
4
π
5π
––
4
3π
––
2
7π
––
4
2π
x
–2
FS
–3
y
+ 1,
4
π
sinax + b
4
1
up 1.
move y =
π
sinax + b
4
PR
O
1
3 To graph y =
O
–4
3
2
E
1
PA
G
0
π
–π – ––
3π – –
– –π4
2
–1
4
π
–
4
π
–
2
3π
––
4
π
5π
––
4
3π
––
2
7π
––
4
2π
x
–2
–3
TE
D
–4
Exercise 2.8 Graphs of reciprocal trigonometric functions
WE31
Use the graph of y = 4 cos(x) to sketch y =
EC
1
−2π ≤ x ≤ 2π.
R
PRactise
R
2 Use the graph of y = 2 sin(x) to sketch y =
O
−2π ≤ x ≤ 2π.
U
N
C
3
WE32
−2π ≤ x ≤ 2π.
−2π ≤ x ≤ 2π.
5
WE33
6 Use the graph of y = tan(3x) to sketch y =
−2π ≤ x ≤ 2π.
114 WE34
1
over the domain
sin(2x)
1
over the domain
cos(2x)
Use the graph of y = tan(2x) to sketch y =
−2π ≤ x ≤ 2π.
7
1
over the domain
2 sin(x)
Use the graph of y = sin(2x) to sketch y =
4 Use the graph of y = cos(2x) to sketch y =
1
over the domain
4 cos(x)
1
over the domain
tan(2x)
1
over the domain
tan(3x)
π
Sketch the graph of y = cotax + b + 1 over the domain [−π, 2π].
4
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c02Trigonometry.indd 114
20/08/15 10:43 AM
π
1
secax + b − 1 over the domain [−π, 2π].
2
4
1
9 Use the graph of y = 4 sin(x) to sketch y =
over the domain [−π, π].
4 sin(x)
8 Sketch the graph of y =
x
2
10 Use the graph of y = cosa b to sketch y =
x
3
11 Use the graph of y = tana b to sketch y =
1
x
cosa b
2
1
x
tana b
3
over the domain [−π, π].
over the domain [−3π, 3π].
FS
Consolidate
x
1
coseca b over the domain [0, 2π].
2
2
x
14 Sketch y = cota b − 2 over the domain [−2π, 2π]
4
PR
O
13 Sketch y =
O
12 Sketch y = sec(x) + 1 over the domain [0, 2π].
15 Sketch y = 2 sec(x) − 1 over the domain [−2π, 2π].
π
sinax + b
4
over the domain [−π, π].
E
2
PA
G
16 Sketch y =
π
4
17 Sketch y = 0.25 cosecax − b over the domain [−π, π].
TE
D
π
2
18 Sketch y = 3 seca2x + b − 2 over the domain [−π, π].
1
over the domain
sin(x) + 2
−5π 5π
, d . Sketch both graphs on the same set of axes. Check your
2 2
graphs with CAS.
−3π 3π
1
20aUse the graph of y = cos2 (x) to sketch y =
over
the
domain
, d.
c
2 2
cos2 (x)
Sketch both graphs on the same set of axes. Check your graphs with CAS.
b Hence, determine the graph of y = tan2 (x) for the same domain.
C
O
R
R
c
EC
Master
19 Use the graph of y = sin(x) + 2 to sketch y =
U
N
2.9
AOS 1
Topic 2
Concept 6
Graphs of inverse
circular functions
Concept summary
Practice questions
Graphs of inverse trigonometric functions
There are at least two possible approaches to sketching inverse trigonometric
functions. The first method is to find the inverse of the function (which will be a
trigonometric function) and use your knowledge of trigonometric functions to sketch
the trigonometric function and its inverse.
Alternatively, you could use your knowledge about transforming equations to
transform y = sin−1 (x), y = cos−1 (x) or y = tan−1 (x) as required. In the following
worked examples, we will find the original trigonometric function and then sketch
both functions.
Topic 2 Trigonometry c02Trigonometry.indd 115
115
20/08/15 10:43 AM
WorKeD
eXaMPLe
35
Sketch y = sin −1 (2x).
tHinK
WritE/draW
1 Find the inverse of y = sin−1 (2x).
y = sin−1 (2x)
2x = sin( y)
x = 12 sin( y)
Therefore, the inverse is y = 12 sin(x).
y = 12 sin(x)
Amplitude: 12
Period: 2π
Horizontal shift: 0
Vertical shift: 0
FS
1
2 Sketch y = 2 sin(x).
O
y
2
3π
−2π – ––
2
−π
– –π2
PR
O
1
0
–1
π
–
2
π
3π
––
2
2π
x
3π
––
2
2π
x
3 The domain needs to be restricted so that
π π
Restrict the domain to c − , d .
2 2
R
EC
TE
D
the function is one-to-one. The domain
π π
becomes c − , d .
2 2
PA
G
E
–2
1
R
4 The domain and range of y = 2 sin(x)
U
N
C
O
become the range and domain of
y = sin−1 (2x) respectively.
1
5 Use the graph of y = 2 sin(x) to sketch
y = sin−1 (2x) by reflecting the graph in
the line y = x.
3π
−2π – ––
2
y
2
1
−π
– –π2
0
–1
π
–
2
π
–2
y = 12 sin(x):
π π
1 1
Domain c − , d , range c − , d
2 2
2 2
−1
y = sin (2x):
π π
1 1
Domain c − , d , range c − , d
2 2
2 2
y
π
–
2
π
–
4
–6 –5 –4 –3 –2 –1 0
– –π4
1 2 3 4 5 6 x
– –π2
116
MaTHs QuesT 12 sPeCIaLIsT MaTHeMaTICs VCe units 3 and 4
c02Trigonometry.indd 116
20/08/15 10:43 AM
WorKeD
eXaMPLe
36
Sketch y = cos −1 (x + 5)
WritE/draW
1 Find the inverse of y = cos−1 (x + 5).
y = cos−1 (x + 5)
x + 5 = cos( y)
x = cos( y) − 5
Therefore, y = cos(x) − 5.
2 Sketch y = cos(x) − 5.
y = cos(x) − 5
y
Amplitude: 1
Period: 2π
Horizontal shift: 0
Vertical shift: 5 down
−π
– –π2
0
PR
O
5π −2π
3π
– ––
– ––
2
2
O
1
FS
tHinK
–1
π
–
2
π
π
–
2
π
3π
––
2
x
–2
E
–3
PA
G
−4
TE
D
3 The domain needs to be restricted so that
U
N
C
O
R
R
EC
the function is one-to-one. The domain
becomes [0, π].
4 The domain and range of y = cos(x) − 5
become the range and domain of
y = cos−1 (x + 5) respectively.
−5
−6
Restrict the domain to [0, π].
y
1
5π −2π
3π
– ––
– ––
2
2
−π
– –π2
0
–1
3π
––
2
x
–2
–3
−4
−5
−6
y = cos(x) − 5:
Domain [0, π], range [ −6, −4]
y = cos−1 (x + 5):
Domain [ −6, −4] , range [0, π]
Topic 2 TrIgonoMeTry
c02Trigonometry.indd 117
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20/08/15 10:44 AM
5 Use the graph of y = cos(x) − 5 to sketch
y
π
cos−1 (x
y=
+ 5) by reflecting the graph in
the line y = x.
π
–
2
0
−9 −8 −7 −6 −5 −4 –3 –2 –1
– –π2
1 2 3 4 5 6
x
−π
FS
3π
– ––
2
Sketch y = 3 tan −1 (x).
WritE/draW
y = 3 tan−1 (x)
y
= tan−1 (x)
3
y
x = tana b
3
1 Find the inverse of
y = 3 tan−1 (x).
TE
D
x
Therefore y = tana b.
3
x
3
y
3π
5π
––
2
EC
2 Sketch y = tana b.
2π
3π
––
2
R
R
Period: 3π
Horizontal shift: 0
Vertical shift: 0
E
tHinK
PR
O
37
PA
G
WorKeD
eXaMPLe
O
−2π
U
N
C
O
π
π
–
2
5π −2π
3π
−3π – ––
– ––
2
2
−π
– –π2
0
– –π2
π
–
2
π
3π
––
2
2π
5π
––
2
3π x
−π
3π
– ––
2
−2π
5π
– ––
2
−3π
118
MaTHs QuesT 12 sPeCIaLIsT MaTHeMaTICs VCe units 3 and 4
c02Trigonometry.indd 118
20/08/15 10:44 AM
y
3π
3 The domain needs to be
restricted so that the function
is one-to-one. The domain
3π 3π
becomes a− , b.
2 2
5π
––
2
2π
3π
––
2
π
−π
– –π2
0
– –2π
π
–
2
π
3π
––
2
2π
5π
––
2
2π
5π
––
2
3π x
O
5π −2π
3π
−3π – ––
– ––
2
2
FS
π
–
2
PR
O
−π
3π
– ––
2
−2π
PA
G
E
5π
– ––
2
y = 3 tan−1 (x): Domain R, range a−
x
3
5 Use the graph of y = tana b
5π
––
2
2π
R
to sketch y = 3 tan−1 (x)
by reflecting the graph in
the line y = x.
O
R
3π
––
2
π
C
U
N
3π 3π
, b
2 2
y
EC
x
y = tana b become the range
3
and domain of y = 3 tan−1 (x)
respectively.
3π 3π
x
y = tana b: Domain a− , b, range R
3
2 2
TE
D
4 The domain and range of
−3π
π
–
2
5π −2π
3π
−3π – ––
– ––
2
2
−π
– –π2
0
– –π2
π
–
2
π
3π
––
2
3π x
−π
3π
– ––
2
−2π
5π
– ––
2
Topic 2 Trigonometry c02Trigonometry.indd 119
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20/08/15 10:44 AM
The next worked example is completed by transforming the inverse
trigonometric function.
38
Sketch y = sin −1 (x) +
tHinK
π
4
WritE/draW
y
1 The graph of
π
is the graph
4
π
of y = sin−1 (x) raised by
4
units. Sketch y = sin−1 (x).
y = sin−1 (x) +
π
–
2
FS
WorKeD
eXaMPLe
–2
–1.5
–1
0
–0.5
0.5
1.5
2
x
PA
G
E
– –π4
1
PR
O
O
π
–
4
– –π2
π
4
This means that the
π 3π
range is now c − , d
4 4
EC
TE
D
2 Raise the graph by .
y
π
–
2
–2
–1.5
–1
–0.5
O
R
R
π
–
4
0
0.5
1
1.5
2
x
U
N
C
– –π4
– –π2
ExErcisE 2.9 Graphs of inverse trigonometric functions
PractisE
1
Sketch y = cos−1 (2x).
2 Sketch y = tan−1 (2x).
3
120
WE35
WE36
Sketch y = sin−1 (x + 3).
MaTHs QuesT 12 sPeCIaLIsT MaTHeMaTICs VCe units 3 and 4
c02Trigonometry.indd 120
20/08/15 10:44 AM
4 Sketch y = tan−1 (x − 3).
5
WE37
Sketch y = 2 sin−1 (x).
6 Sketch y = 2 cos−1 (x).
π
Sketch y = cos−1 (x) − .
4
π
8 Sketch y = tan−1 (x) + .
3
x
9 Sketch y = sin−1 a b.
2
7
Consolidate
WE38
FS
x
3
10 Sketch y = tan−1 a b.
Topic 2
O
11 Sketch y = cos−1 (4x).
AOS 1
13 Sketch y = sin−1 (2x + 1).
14 Sketch y = cos−1 (3x − 2).
15 Sketch y = 3 cos−1 (x).
16 Sketch y = 3 sin−1 (2x).
E
Transformations
of inverse circular
functions
Concept summary
Practice questions
PR
O
12 Sketch y = tan−1 (x − 3).
Concept 7
PA
G
π
4
cos−1 (2x − 3)
18 Sketch y =
+ 1.
π
19 a Draw the graph of y = sec(x).
Master
b Identify a suitable domain to make y = sec(x) a one-to-one function.
c Sketch the graph of y = sec−1 (x).
cot−1 (x + 1)
− 2.
π
U
N
C
O
R
R
EC
20 Sketch y =
TE
D
17 Sketch y = 2 tan−1 (x) + .
Topic 2 Trigonometry c02Trigonometry.indd 121
121
20/08/15 10:44 AM
ONLINE ONLY
2.10 Review
the Maths Quest review is available in a customisable
format for you to demonstrate your knowledge of this
topic.
www.jacplus.com.au
• Extended-response questions — providing you with
the opportunity to practise exam-style questions.
a summary of the key points covered in this topic is
also available as a digital document.
the review contains:
• short-answer questions — providing you with the
opportunity to demonstrate the skills you have
developed to efficiently answer questions using the
most appropriate methods
• Multiple-choice questions — providing you with the
opportunity to practise answering questions using
CAS technology
FS
REVIEW QUESTIONS
Units 3 & 4
Trigonometry
Sit topic test
U
N
C
O
R
R
EC
TE
D
studyON is an interactive and highly visual online
tool that helps you to clearly identify strengths
and weaknesses prior to your exams. You can then
confidently target areas of greatest need, enabling
you to achieve your best results.
PA
G
E
PR
O
O
Download the Review questions document from
the links found in the Resources section of your
eBookPLUS.
122
MaTHs QuesT 12 sPeCIaLIsT MaTHeMaTICs VCe units 3 and 4
c02Trigonometry.indd 122
20/08/15 10:44 AM
2 Answers
13a sin( θ) − cos( θ)
Exercise 2.2
b −!2
c −1
d −!3
!3
3
2 !3
3
c −
c −!3 sin(x)
b −
18a
b
5 !21
21
−5 !39
39
!61
6
19a
16
8 !5
21
!29
5
− !15
5
22
q( p2 − q2)
−
a+b
b2
EC
Exercise 2.3
R
Refer to Maths Quest 12 VCE Specialist Maths
Solutions Manual.
!3
2
1
2 2
R
Exercise 2.4
O
1
E
21
( p2 − pq − q2)"p2 − q2
"a2
56
65
−253
325
− !2)
d sin(x)
b 2 − !3
d 2 + !3
63
b −16
204
b −325
19
b −21
ab + "1 − a2 "1 − b2
PA
G
b
1
( !6
4
− !6)
20a −
TE
D
15
b
1
( !2
4
17a
9a
b −!17
b −1
16a !3 cos(x)
c
b
FS
b 2
d !2
O
c −2
d tan(θ)
PR
O
b −!2
b −!15
14a
b sin(θ)
c −tan(θ)
3 !2
4
3 !10
20
−7 !10
20
!15
− 15
!3
12
−6 !35
35
!15
3
13a
d −cos(θ)
15a −cos(θ)
8a −
12a
b sin(θ)
c −sin(θ)
b −
11a
d cos(θ ) − sin(θ)
14a cos(θ)
7a !3
10a
b !3 cos(θ) + sin(θ)
c !3 cos(θ) + sin(θ)
2 !3
1 3
2 !3
2 − 3
!21
3 − 2
!17
4 − 4
2 !3
5a 3
2 !3
6a 3
C
3 2!3 cos( θ) − 2 sin(θ)
U
N
4 sin(θ) + cos(θ)
5 −cos( θ)
6 −sin(θ)
1
4
7 (!6 + !2)
8 !3 − 2
84
9 −85
64
10−1025
11a
!3
1
!2
b 0 c 2 d − 2
2
12a
!3
b !3
3
b"1 − a2 − a"1 − b2
a2 + 2!a + 1 !2a + 1
a a !2a + 1 − 2!a + 1 b
Exercise 2.5
!2
4
!2
2 − 2
4 !2
4 !2
7
3 a 9 b −9 c − 7
56
33
56
4 a 65 b 65 c 33
1 −
π π 2π 3π
π 7π 3π 11π
,
6 , , ,
3 2 3 2
2 6 2 6
7–12 Check with your teacher.
!2
!2
!2
!3
13 a 4 b − 2 c − 2 d 3
5 , ,
3 !55
3 !55
b −23
c − 23
32
32
14 a π
2π 4π
5π
, 2π b 0, , , 2π
3
3 3
3
π 5π 3π
π π 5π 3π
c , , d , , ,
6 6 2
6 2 6 2
π 3π 5π 7π
16a0, , , , , 2π
4 4 4 4
π π 2π 3π
b , , ,
3 2 3 2
π π 5π 7π 3π 11π
c 0, , , , π,
, ,
, 2π
6 2 6
6 2 6
π π 5π 3π 13π 5π 17π 7π
d
, , , ,
, ,
,
12 4 12 4 12 4 12 4
17–24 Check with your teacher.
15a 0, , π,
Topic 2 Trigonometry c02Trigonometry.indd 123
123
20/08/15 10:44 AM
Exercise 2.6
c Domain R, range (–2, 2π)
π
b − c 0.9
3
π
1
b − c 3
6
5π
π
b c 6
6
π
1
b c 4
3
1 a Does not exist 2 a Does not exist 3 a Does not exist 4 a Does not exist 5π 5π
, d
2 2
e Domain [–4, 4], range [0, 6π]
d Domain [–3, 3], range c −
f Domain R, range a−
2
26aDomain c 0, d , range [0, 2π]
3
b Domain [2, 3], range [–π, 2π]
2 !6
5
2 !10
6 7
8 !33
7 49
23
8 32
5
c Domain R, range (–3π, 2π)
π
6
π
5
10 a − b 4
3
PR
O
27aDilation by a factor of c units parallel to the x-axis
(or away from the y-axis), dilation by a factor of b
units parallel to the y-axis (or away from the x-axis),
translation by a units up and parallel to the y-axis (or
away from the x-axis)
bπ
bπ
Domain [–c, c], range c a − , a + d
2
2
1
b Dilation by a factor of units parallel to the x-axis
c
(or away from the y-axis), dilation by a factor of
b units parallel to the y-axis (or away from the x-axis),
translation by a units up and parallel to the y-axis
(or away from the x-axis)
3
15
12 8
33
65
33
14−56
E
13
PA
G
7π π
1 9
,− d
2 2
2 2
b Domain R, range (–1, 5)
15aDomain c , d , range c −
4
16aDomain c −2, −3 d , range [–3, 1]
1 1
Domain c − , d , range [a, a + bπ]
c c
c Dilation by a factor of c units parallel to the x-axis
(or away from the y-axis), dilation by a factor of
b units parallel to the y-axis (or away from the x-axis),
translation by a units up and parallel to the y-axis
(or away from the x-axis)
bπ
bπ
Domain R, range aa − , a + b
2
2
28Check with your teacher.
TE
D
b Domain R, range (–1, 7)
20 a !77
9
5 !89
89
d C
!2
2
e U
N
d −
1
2 !15
16
12
d 13
56
22 a 65
21 a O
R
R
EC
π
17 a b Does not exist
2
π
c − d π
3
2π
e f Does not exist
3
π
π
g h −
3
6
π
π
6
18 a 5
b c 5
6
π
π
π
e f g 10
3
8
b 2
e c −
!21
5
f 5 !11
11
4 !65
65
4
c 5
4 !6
23
63
b 65
f 25
23Check with your teacher.
17
528
697
c −
812
1037
d −
24Check with your teacher.
25a Domain [0, 2], range [–π, π]
b Domain [1, 3], range [0, 3π]
124 Exercise 2.7
c −
b −3!7
e −
π
3
!3
3
f −1
!5
2
1
d 2
h !3
2
b −
range [–4, 1]
f Domain R, range (–1, 7)
114
!3
2
11
d,
3
O
e Domain c −1,
FS
1
d Domain c −2, 2 d , range [0, 4]
9 a b 1.1
19 a 7π 7π
, b
2 2
Note that n ∈ Z.
π
1 (8n ± 1)
4
π
2 (12n ± 5)
12
π
π
3 (12n + 1), (12n + 5)
6
6
π
π
4 (6n − 1), (3n + 2)
6
3
π
5 (4n + 1)
4
π
6 (3n − 1)
6
π
π
7 (6n ± 1), (3n ± 1)
6
3
π
8 (6n ± 1)
12
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c02Trigonometry.indd 124
20/08/15 10:44 AM
b nπ,
26a
π
(6n ± 1)
3
π
(4n ± 1)
4
y
4
3
2
1
3π
−2π – ––
2
−π
– 2π–
−π
– 2π–
PA
G
3π
−2π – ––
2
−π
– 2π–
−π
– 2π–
R
R
0
–1
–2
–3
–4
π
–
2
π
π
–
2
π
π
–
2
π
π
–
2
π
3π
––
2
2π x
0
–1
–2
–3
–4
3π
––
2
2π
3π
––
2
2π
3π
––
2
2π
5π
4π ––
––
3
3
2π
x
y
4
3
2
1
3π
−2π – ––
2
0
–1
–2
–3
–4
x
y
8
6
4
2
5
U
N
2π x
y
4
3
2
1
4
3π
−2π – ––
2
6
3π
––
2
O
PR
O
3π
−2π – ––
2
π
π
–
2
y
4
3
2
1
2
3
0
–1
–2
–3
–4
FS
1
C
O
π
π
(4n ± 1), (3n ± 1)
4
3
Exercise 2.8
EC
TE
D
b
E
2nπ 2π
, (3n + 1)
3 9
π
π
10 (2n + 1), (3n − 1)
2
3
π
π
π
11 (12n + 1),
(12n + 5), (4n + 1)
12
12
4
π
π
12 (6n ± 1), (2n ± 1)
6
2
π
π
13 (4n + 1), (4n − 1)
10
2
π
π
14 (4n + 1), (4n − 1)
14
2
π
15a
(12n ± 1)
18
π
b (3n ± 1)
3
π
π
c (8n − 1), (8n + 5)
8
8
π
π
d
(12n − 1), (12n + 7)
18
18
π
π
16a (6n ± 1), (3n ± 1)
6
3
π
π
π
b (8n ± 1), (8n + 3), (8n + 5)
8
8
8
π
2π
c (6n ± 1),
(3n ± 1)
9
9
π
π
d
(8n ± 1), (8n ± 3)
12
12
π
π
17a (3n − 1)
b
(6n + 1)
3
18
π
π
c (3n ± 1)
d
(6n ± 1)
12
6
nπ π
π
18a
, (12n − 1), (12n + 7)
2 12
12
π
b (4n ± 1), nπ
4
π
π
c (4n ± 1),
(12n ± 5)
4
12
nπ π
π
, (6n + 1), (3n + 1)
d
2 6
3
π
π
π
19a
(12n − 1), (12n + 7), (4n − 1)
12
12
4
π
b (6n ± 1), nπ
6
π
π
π
π
20a (4n + 1), (2n + 1)
b (4n − 1),
(12n + 5)
6
3
4
12
π
π
21a (3n + 1), (2n − 1)
3
2
π
π
b
(24n + 7), (24n − 1)
36
36
nπ
nπ
22a
b
3
2
π
π
π
π
23a (3n − 1), (4n − 1)
b (3n − 1), (4n + 1)
3
4
3
4
π
2nπ
24a nπ, (6n ± 1)
b 2nπ,
3
3
π
π
π
25a (4n ± 1), (12n − 1), (12n + 7)
2
6
6
9
−π
– 2π–
0
–2
–4
–6
–8
x
y
8
6
4
2
−π ––
5π ––
−2π – ––
– π– 0
– 2π
– 4π
3
3
3 3 –2
–4
–6
–8
π
–
3
2π
––
3
π
x
Topic 2 Trigonometry c02Trigonometry.indd 125
125
20/08/15 10:44 AM
π
5π ––
3π ––
7π
––
4 2 4
2π
0
–1
–2
–3
–4
x
y
4
3
2
1
8
−π
π
3π – π
– –– 0
– ––
4
4 2 –1
–2
–3
–4
π 3π
π –
–
4 2 ––
4
π
5π ––
3π ––
7π
––
4 2 4
2π
x
– 2π–
3π
−2π – ––
2
π x
π
–
2
0
–1
–2
–3
π
π
–
2
–π
– –π2
π
–
2
π
–π
–
–3π
4
– –π2
– –π4
3π
––
2
2π
5π
––
2
3π
x
y
4
3
2
1
U
N
12
0
–1
–2
–3
–4
126 π
–
2
π
3π
–
2
–π
–
–3π
4
– –π2
– –π4
π
3π
––
2
2π x
0
–1
–2
–3
–4
π
–
2
π
π
–
4
π
–
2
3π
–
4
π
–
4
π
–
2
3π
–
4
π
–
4
π
–
2
3π
–
4
3π
–
2
2π
x
0
–2
–4
–6
–8
π x
0
–2
–4
–6
–8
π x
y
8
6
4
2
18
2π x
π
–
2
y
8
6
4
2
17
C
O
−π – π– 0
5π −2π ––
−3π – ––
– 3π
2 –2
2
2
–4
–6
–8
x
0
–2
–4
–6
–8
y
8
6
4
2
R
y
8
6
4
2
11
3π
––
2
TE
D
– 2π–
EC
−π
–
–3π
2
16
R
3π
−––
2
– –2π
y
4
3
2
1
–2π
y
5
4
3
2
1
10
−π
15
0
–1
–2
–3
–4
2π x
y
8
6
4
2
PA
G
−π
3π
–
2
14
y
4
3
2
1
9
π
π
–
2
FS
π 3π
π –
–
4 2 ––
4
PR
O
π
3π – π
– –– 0
– ––
4
4 2 –1
–2
–3
–4
y
4
3
2
1
E
−π
13
O
y
4
3
2
1
7
–π
–
–3π
4
– –π2
– –π4
0
–2
–4
–6
–8
π x
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c02Trigonometry.indd 126
20/08/15 10:44 AM
19See figure at foot of page.*
y
14
12
10
8
6
4
2
– –π2
–π
0
–2
–4
π
π
–
2
2π
3π
–
2
x
–2π
–π
–
–3π
2
– –π2
y
5
π
y
14
12
10
8
6
4
2
b
π
–
2
–5 –4 –3 –2 –1 0
– –π2
–π
0
–2
–4
π
π
–
2
2π
3π
–
2
x
Exercise 2.9
3π
–
2
E
y
π
π
–5 –4 –3 –2 –1 0
– –π2
PA
G
π
–
2
1 2 3 4 5 x
TE
D
8
1 2 3 4 5 x
y
π
3π
–
4
π
–
2
1 2 3 4 5 x
C
–5 –4 –3 –2 –1 0
– –π2
–5 –4 –3 –2 –1 0
– –π2
O
π
–
2
y
π
π
–
2
R
y
π
3
1 2 3 4 5 x
3π
–
4
R
1 2 3 4 5 x
7
EC
π
–
2
–5 –4 –3 –2 –1 0
– –π2
π
–
2
–5 –4 –3 –2 –1 0
– –π2
y
π
2
1 2 3 4 5 x
y
2π
6
1
1 2 3 4 5 6 7 8 x
FS
–π
–
–3π
2
–7 –6 –5 –4 –3 –2 –1 0
– –π2
O
–2π
π
–
2
PR
O
20a
y
π
4
U
N
–5 –4 –3 –2 –1 0
– –π2
–π
1 2 3 4 5 x
y
8
6
4
2
*19
–3π
– –2π
–5π
2
–
–3π
2
–π
– –π2
0
–2
–4
–6
–8
π
–
2
π
3π
–
2
2π
5π
–
2
3π
x
Topic 2 Trigonometry c02Trigonometry.indd 127
127
20/08/15 10:44 AM
y
y
14
π
π
π
–
2
π
–
2
1 2 3 4 5 x
–4 –3 –2 –1 0
– –π2
–π
y
3π
15
y
5π
–
2
π
2π
π
–
2
3π
–
2
–7 –6 –5 –4 –3 –2 –1 0
– –π2
π
1 2 3 4 5 6 7 x
π
–
2
–π
–5 –4 –3 –2 –1 0
– –π2
3π
––
2
y
11
π
–
2
0.25 0.5 0.75
x
TE
D
0
– –π2
–π
3π
––
2
EC
y
12
3π
––
2
y
π
–
2
1 2 3 4 5 6 7 x
O
–10 –8 –6 –4 –2 0
– –π2
–π
U
N
3π
––
2
y
3π
–
2
y
2
18
π
1.5
π
–
2
–1.5
–1 –0.5
0
– –π2
2 4 6 8 10 x
–π
3π
––
2
13
1 2 3 4 5 6 7 x
π
C
–7 –6 –5 –4 –3 –2 –1 0
– –π2
π
–
2
–π
17
R
π
–
2
π
–7 –6 –5 –4 –3 –2 –1 0
– –π2
R
π
E
PA
G
π
1 2 3 4 5 x
y
16
–0.75 –0.5 –0.25
FS
10
1 2 3 4 x
O
–5 –4 –3 –2 –1 0
– –π2
PR
O
9
1
0.5
1
1.5 x
–π
0.5
–1
–0.5 0
–0.5
0.5
1
1.5
2
2.5 x
3π
––
2
128 Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c02Trigonometry.indd 128
20/08/15 10:44 AM
y
4
3
2
1
19a
–
–3π
2
–π
– –π2
y
2
1
20
0
–1
–2
–3
–4
π
π
–
2
3π
–
2
–7 –6 –5 –4 –3 –2 –1 0
–1
–2
–3
–4
–5
–6
x
1 2 3 4 5 6 7 x
π
2
b [0, π]/ e f
c
FS
y
2π
3π
–
2
O
π
–8 –7 –6 –5 –4 –3 –2 –1 0
– –π2
PR
O
π
–
2
1 2 3 4 5 6 7 8 x
U
N
C
O
R
R
EC
TE
D
PA
G
E
–π
Topic 2 Trigonometry c02Trigonometry.indd 129
129
20/08/15 10:44 AM