P1: FXS/ABE P2: FXS CHAPTER 11 - Cambridge University Press · 2008. 9. 22. · P1: FXS/ABE P2: FXS...
Transcript of P1: FXS/ABE P2: FXS CHAPTER 11 - Cambridge University Press · 2008. 9. 22. · P1: FXS/ABE P2: FXS...
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P1: FXS/ABE P2: FXS
9780521740494c11.xml CUAU033-EVANS September 9, 2008 13:42
C H A P T E R
11Circular functions II
ObjectivesTo further explore the symmetry properties of circular functions
To further understand and sketch the graphs of circular functions
To solve circular function equations
To evaluate simple trigonometric expressions using trigonometric identities
To prove simple trigonometric identities
To apply addition theorems for circular functions
To apply double angle formulas for circular functions
To simplify expressions of the form a cos x + b sin xTo sketch graphs of functions of the form f (x) = a cos x + b sin xTo solve equations of the form a cos x + b sin x = c
y
x
– θθ
θa
b
a
b
P(θ)π2
Pπ2
– θ
y
x
b
b
a
a
π2
+ θP(θ)
Pπ2
+ θ
θ
θ
11.1 Further symmetry propertiesComplementary relationships
sin(�
2− �
)= a
and since a = cos �sin
(�2
− �)
= cos �Similarly
cos(�
2− �
)= b
and since b = sin �cos
(�2
− �)
= sin �
sin(�
2+ �
)= a = cos �
cos(�
2+ �
)= −b = −sin �
297
SAM
PLE
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298 Essential Advanced General Mathematics
Example 1
If sin � = 0.3 and cos � = 0.8, find the values ofa sin
(�2
− �)
b cos(�
2+ �
)c sin(−�)
Solution
a sin(�
2− �
)= cos �= 0.8
b cos(�
2+ �
)= −sin �= −0.3
c sin (−�) = − sin �= −0.3
Exercise 11A
1 If sin x = 0.3, cos � = 0.6 and tan � = 0.7, find the values ofExample 1a cos(−�) b sin
(�2
+ �)
c tan(−�) d cos(�
2− x
)e sin(−x) f tan
(�2
− �)
g cos(�
2+ x
)h sin
(�2
− �)
i sin
(3�
2+ �
)j cos
(3�
2− x
)
11.2 Addition of ordinatesExample 2
Using the same scale and axes, sketch the graphs of y1 = 2 sin x and y2 = 3 cos 2x for0 ≤ x ≤ 2�.Use addition of ordinates to sketch the graph of y = 2 sin x + 3 cos 2x .
x
y
3210
–1–2–3
–5–4
π2
2ππ
y1 = 2sinx
y2 = 3cos2x
y = 2sinx + 3cos2x
3π2
Solution
The graphs of y1 = 2 sin x and y2 = 3 cos 2x are shown below.To obtain points on the graph of y = 2 sin x + 3 cos 2x the process of addition ofordinates is used.
Let y = y1 + y2 when y1 = 2 sin x and y2 = 3 cos 2xe.g., at
x = 0, y = 0 + 3 = 3x = �
4, y = 2√
2+ 0 = 2√
2= √2
x = �2
, y = 2 − 3 = −1x = �, y = 0 + 3 = 3x = 3�
2,y = −2 − 3 = −5
and so on.
SAM
PLE
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Chapter 11 — Circular functions II 299
Using the TI-NspireCheck that the calculator is in Radian
mode. Open a Graphs & Geometryapplication ( 2) and enter the
functions
f 1 (x) = 2 sin (x)f 2 (x) = 3 cos (2x)f 3 (x) = f 1 (x) + f 2 (x)
The graph of f (3) x is the heavier line.
Add Function Table (b 2 ) and splitthe screen as shown using the Tools menu(/ 5 23) to see that the values
of f 1 (x) and f 2 (x) add to give f 3 (x).
Use the down arrow ( ) to see more xvalues.
Using the Casio ClassPadSet the calculator to Radian mode.
Enter the functions y = sin (x) and y = 12
.
The window settings should be as shown. Tap
$ to produce the graph. Ensure the graph
window is selected (bold border) and tap Analysis,G-solve, intersect to find decimal approximationsfor the solutions. The scroll key moves the cursor
between solutions.SAM
PLE
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300 Essential Advanced General Mathematics
To find exact solutions, the window is
used.
Enter and highlight the equation sin (x) = 12
.
Tap Interactive, Equation/inequality, solve andensure the variable is set to x.
The answer returned is
x = 2�∗constn(9) + �6
,
x = 2�∗constn(10) + 5�6
.
This may be read as
x = 2m� + �6
, 2n� + 5�6
.
It should be clear that there are 4 solutions to the problem. Hence, the values for each of
m and n will be required which produce a solution in the domain. In this case the values are
m = 0, 1 and n = 0, 1. The solutions are
x = �6
,11�
6,
13�
6,
23�
6.
Exercise 11B
1 Use addition of ordinates to sketch the graphs ofExample 2
a y = 2 sin � + cos � b y = 3 cos 2� + 2 sin 2�c y = 1
2sin 2� − cos � d y = 3 sin � + cos 2�
e y = 4 sin � − 2 cos �
11.3 Sketch graphs of the tangent functionA table of values for y = tan x is given below. Use a calculator to check these values.
x −� −3�4
−�2
−�4
0�
4
�
2
3�
4�
5�
4
3�
2
7�
42�
9�
4
5�
2
11�
43�
y 0 1 undefined −1 0 1 undefined −1 0 1 undefined −1 0 1 undefined −1 0SAM
PLE
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Chapter 11 — Circular functions II 301
The graph of y = tan x is given below.
x
y
–π1
10
2
2 3
2π 3π8 94 5 6 7 10
π2
3π2
–π2
5π2
–3–4 –2 –1–1
–2
π
Note: x = −�2
,�
2,
3�
2and
5�
2are asymptotes.
Observations from the graph1 The graph repeats itself every � units, i.e., the period of tan is �.
2 The range of tan is R.
Exercise 11C
1 Sketch the graph of each of the following, showing one complete cycle.
a y = tan 2x b y = 2 tan 3x c y = 2 tan(
x + �4
)d y = 3 tan x + 1 e y = 2 tan
(x + �
2
)+ 1 f y = 3 tan 2
(x − �
4
)− 2
11.4 General solution of circular function equationsThe solution of circular function equations has been discussed in Section 10.9 for functions
over a restricted domain. In this section, we consider the general solutions of such equations
over the maximal domain for each function.
If a circular function equation has one or more solutions in one ‘cycle’, then it will have
corresponding solutions in each ‘cycle’ of its domain, i.e., there will be an infinite number of
solutions.
For example, if cos x = a, then the solution in the interval [0, �] is given by:x = cos−1(a)
By the symmetry properties of the cosine function, other solutions are given by:
−cos−1(a), ±2� + cos−1(a), ±2� − cos−1(a), ±4� + cos−1(a),±4� − cos−1(a), . . . and so on.
In general, if cos (x) = a,x = 2n� ± cos−1(a), where n ∈ Z and a ∈ [−1, 1]
SAM
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302 Essential Advanced General Mathematics
Similarly, if tan (x) = a,x = n� + tan−1(a), where n ∈ Z and a ∈ R
If sin (x) = a,x = 2n� + sin−1(a) or x = (2n + 1)� − sin−1(a), where n ∈ Z and a ∈ [−1, 1]
Note: An alternative and more concise way to express the general solution of sin (x) = a is:x = n� + (−1)n sin−1(a), where n ∈ Z and a ∈ [−1, 1]
Example 3
Find the general solution to each of the following equations.
a cos (x) = 0.5 b √3 tan (3x) = 1 c 2 sin (x) = √2
Solution
a x = 2n� ± cos−1(0.5)= 2n� ± �
3
= (6n ± 1)�3
, n ∈ Z
b tan (3x) = 1√3
3x = n� + tan−1(
1√3
)
= n� + �6
= (6n + 1)�6
x = (6n + 1)�18
, n ∈ Zc sin (x) = 1√
2
x = 2n� + sin−1(
1√2
)or x = (2n + 1)� − sin−1
(1√2
)= 2n� + �
4= (2n + 1)� − �
4
= (8n + 1)�4
, n ∈ Z = (8n + 3)�4
, n ∈ Z
Using the TI-NspireCheck that the calculator is in Radian mode.
a Use Solve( ) from the Algebra menu(b 31) and complete as shown.
Note the use of1
2rather than 0.5 to
ensure that the answer is exact.SA
MPL
E
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Chapter 11 — Circular functions II 303
b Complete as shown.
c Complete as shown.
Using the Casio ClassPada Enter and highlight the equation cos (x) = 0.5,
tap Interactive, Equation/inequation, solveand ensure the variable is set to x.
b Enter and highlight the equation
(3) tan (3x) = 1, tap Interactive,Equation/inequation, solve and ensure thevariable is set to x.
c Enter and highlight the equation (2) sin (x) = 1,tap Interactive, Equation/inequation, solveand ensure the variable is set to x.
Example 4
Find the first three positive solutions to each of the following equations.
a cos (x) = 0.5 b √3 tan (3x) = 1 c 2 sin (x) = √2
SAM
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304 Essential Advanced General Mathematics
Solution
a The general solution (from Example 3) is given by x = (6n ± 1)�3
, n ∈ Z
When n = 0, x = ±�3
, and when n = 1, x = 5�3
or x = 7�3
The first three positive solutions of cos (x) = 0.5 are x = �3
,5�
3,
7�
3
b The general solution (from Example 3) is given by x = (6n + 1)�18
, n ∈ ZWhen n = 0, x = �
18, and when n = 1, x = 7�
18, and when n = 2, x = 13�
18
The first three positive solutions of√
3 tan (3x) = 1 are x = �18
,7�
18,
13�
18
c The general solution (from Example 3) is given by x = (8n + 1)�4
or
x = (8n + 3)�4
, n ∈ Z
When n = 0, x = �4
or3�
4, and when n = 1, x = 9�
4or x = 11�
4
The first three positive solutions of 2 sin (x) = √2 are x = �4
,3�
4,
9�
4
Exercise 11D
1 Find the general solution to each of the following equations.Example 3
a sin (x) = 0.5 b 2 cos (3x) = √3 c √3 tan (x) = −3
2 Find the first two positive solutions to each of the following equations.Example 4
a sin (x) = 0.5 b 2 cos (3x) = √3 c √3 tan (x) = −3
3 Find the general solution to 2 cos(
2x + �4
)= √2, and hence find all the solutions for x in
the interval (−2�, 2�).
4 Find the general solution to√
3 tan(�
6− 3x
)− 1 = 0, and hence find all the solutions for
x in the interval [−�, 0].
5 Find the general solution to 2 sin (4�x) + √3 = 0, and hence find all the solutions for x inthe interval [−1, 1].
11.5 Trigonometric identitiesReciprocal functionsThe functions sin, cos, and tan can be used to form three other functions called the reciprocal
circular functions.
sec � = 1cos �
(cos � �= 0) cosec � = 1sin �
(sin � �= 0)
cot � = cos �sin �
(sin � �= 0)Note: For cos � �= 0 and sin � �= 0, cot � = 1
tan �and tan � = 1
cot �
SAM
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Chapter 11 — Circular functions II 305
Example 5
Find the exact value of each of the following.
a sec2�
3b cot
5�
4c cosec
7�
4
Solution
a sec2�
3= 1
cos2�
3
= 1cos
(� − �
3
)= 1
−cos �3
= 1−1
2= −2
b cot5�
4=
cos5�
4
sin5�
4
=cos
(� + �
4
)sin
(� + �
4
)= −1√
2÷ −1√
2= 1
c cosec7�
4
= 1sin
(2� − �
4
)= 1
−sin(�
4
)= 1
− 1√2
= −√
2
Example 6
Find the value(s) of x between 0 and 2� for which
a sec x = −2 b cot x = −1
Solution
a sec x = −2∴ 1
cos x= −2
∴ cos x = −12
∴ x = � − �3
or x = � + �3
i.e., x = 2�3
or x = 4�3
b cot x = −1implies
tan x = −1∴ x = � − �
4or x = 2� − �
4
i.e. x = 3�4
or x = 7�4
Using the TI-NspireCheck that the calculator is in Radian mode.
Use solve( ) from the Algebra menu (b 31) as shown.SAM
PLE
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306 Essential Advanced General Mathematics
Using the Casio ClassPada Enter and highlight the equation
1
cos(x)= −2,
tap Interactive, Equation/inequation,solve and ensure the variable is set to x.
b Enter and highlight the equation1
tan(x)= −1,
tap Interactive, Equation/inequation,solve and ensure the variable is set to x.
The Pythagorean identityConsider a point, P(�), on the unit circle.
By Pythagoras’ theorem:
OP2 = OM2 + MP2∴ 1 = (cos �)2 + (sin �)2
Now (cos �)2 and (sin �)2 may be written as cos2 � and sin2 �.x
y
–1
–1
1
1
sinθ
cosθ MO
P(θ)
Since this is true for all values of � it is called an identity.
In particular this is called the Pythagorean identity.
cos2 � + sin2 � = 1Other forms of the identity can be derived.
Dividing both sides by cos2 � gives:
cos2 �
cos2 �+ sin
2 �
cos2 �= 1
cos2 �
1 + tan2 � = sec2 �Dividing both sides by sin2 � gives:
cos2 �
sin2 �+ sin
2 �
sin2 �= 1
sin2 �
cot2 � + 1 = cosec2 �
SAM
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Chapter 11 — Circular functions II 307
Example 7
a If cosec x = 74
, find cos x . b If sec x = −32
, find sin x where�
2≤ x ≤ �.
Solution
a Since cosec x = 74, sin x = 4
7Now cos2 x + sin2 x = 1so cos2 x + 16
49= 1
∴ cos2 x = 3349
∴ cos x = ±√
33
7
b Since sec x = −32, cos x = −2
3
cos2 x + sin2 x = 1∴ 4
9+ sin2 x = 1
∴ sin x = ±√
5
3
For P(x) in the 2nd quadrant, sin x is
positive
∴ sin x =√
5
3
Example 8
If sin � = 35
and�
2< � < �, find the value of cos � and tan �.
Solution
Since cos2 � + sin2 � = 1then cos2 � + 3
2
52= 1
∴ cos2 � = 1 − 925
= 1625
∴ cos � = −45
since�
2< � < �
∴ tan � = −34
as tan � = sin �cos �
SAM
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308 Essential Advanced General Mathematics
Example 9
Prove the identity1
1 − cos � +1
1 + cos � = 2 cosec2 �
Solution
LHS = 11 − cos � +
1
1 + cos �= 1 + cos � + 1 − cos �
1 − cos2 �= 2
1 − cos2 �= 2
sin2 �
= 2 cosec2 �= RHS
Exercise 11E
1 Find the exact value of each of the following.Example 5
a cot3�
4b cot
5�
4c sec
5�
6d cosec
�
2
e sec4�
3f cosec
13�
6g cot
7�
3h sec
5�
3
2 Without using a calculator write down the exact value of each of the following.
a cot 135◦ b sec 150◦ c cosec 90◦ d cot 240◦ e cosec 225◦
f sec 330◦ g cot 315◦ h cosec 300◦ i cot 420◦
3 Find the values of x between 0 and 2� for whichExample 6
a cosec x = 2 b cot x = √3 c sec x + √2 = 0 d cosec x = sec x
4 If sec � = −178
and�
2< � < �, findExample 7
Example 8 a cos � b sin � c tan �
5 If tan � = −724
and3�
2< � < 2�, find cos � and sin �.
6 Find the value of sec � if tan � = 0.4 and � is not in the 1st quadrant.
7 If tan � = 43
and � < � <3�
2, evaluate
sin � − 2 cos �cot � − sin � .
8 If cos � = 23
and � is in the 4th quadrant, find the simplest expression in surd form for
tan � − 3 sin �cos � − 2 cot � .
SAM
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Chapter 11 — Circular functions II 309
9 Prove each of the following identities for suitable values of � and �.Example 9Example 9
a (1 − cos2 �)(1 + cot2 �) = 1 b cos2 � tan2 � + sin2 � cot2 � = 1c
tan �
tan �= tan � + cot �
cot � + tan � d (sin � + cos �)2 + (sin � − cos �)2 = 2
e1 + cot2 �
cot � cosec�= sec � f sec � + tan � = cos �
1 − sin �
11.6 Addition formulas and double angle formulasAddition formulasConsider a unit circle.
Let arc length AB = v unitsarc length AC = u units
∴ arc length CB = u − v units
x
y
u – v
C
A
B
(cos u, sin u)(cos v, sin v)
–1 O
1
1
Rotate �OCB so that B is coincident with A.
The point P has coordinates
(cos (u − v), sin (u − v)).Since the triangles CBO and PAO are congruent,
CB = PAu – vP
A
O–1x
1
(1, 0)
(cos(u – v), sin(u – v))
y
Applying the coordinate distance formula
CB2 = (cos u − cos v)2 + (sin u − sin v)2= 2 − 2(cos u cos v + sin u sin v)
PA2 = (cos (u − v) − 1)2 + (sin (u − v) − 0)2= 2 − 2(cos (u − v))
Equating these
2 − 2(cos u cos v + sin u sin v) = 2 − 2(cos (u − v))∴ cos (u − v) = cos u cos v + sin u sin v
Using the TI-NspireUse tExpand( )from the Trigonometrysubmenu of the Algebra menu (b 3
1) as shown.SAM
PLE
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310 Essential Advanced General Mathematics
Replacing v with −vcos (u − (−v)) = cos u cos (−v) + sin u sin (−v)
From symmetry properties
cos (−�) = cos �sin (−�) = − sin �
cos (u + v) = cos u cos v − sin u sin v
Example 10
Evaluate cos 75◦.
Solution
cos 75◦ = cos (45◦ + 30◦)= cos 45◦ cos 30◦ − sin 45◦ sin 30◦
= 1√2
·√
3
2− 1√
2· 1
2
=√
3 − 12√
2
=√
3 − 12√
2×
√2√2
=√
6 − √24
Replacing u with�
2− u in cos (u − v)
∴ cos((�
2− u
)− v
)= cos
(�2
− u)
cos v + sin(�
2− u
)sin v
Applying symmetry properties
sin � = cos(�
2− �
)and cos � = sin
(�2
− �)
∴ cos(�
2− (u + v)
)= sin u cos v + cos u sin v
∴ sin (u + v) = sin u cos v + cos u sin vReplacing v with −v
sin (u − v) = sin u cos (−v) + cos u sin (−v)∴ sin (u − v) = sin u cos v − cos u sin v
Example 11
Evaluate
a sin 75◦ b sin 15◦.
SAM
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Chapter 11 — Circular functions II 311
Solution
a sin 75◦ = sin (30◦ + 45◦)= sin 30◦ cos 45◦ + cos 30◦ sin 45◦
= 12
· 1√2
+√
3
2· 1√
2
= 1 +√
3
2√
2
= 1 +√
3
2√
2×
√2√2
=√
2 + √64
b sin 15◦ = sin (45◦ − 30◦)= sin 45◦ cos 30◦ − cos 45◦ sin 30◦
= 1√2
·√
3
2− 1√
2· 1
2
=√
3 − 12√
2
=√
3 − 12√
2×
√2√2
=√
6 − √24
Addition formula for tangent
Also tan (u + v) = sin (u + v)cos (u + v)
= sin u cos v + cos u sin vcos u cos v − sin u sin v
Divide the numerator and denominator by cos u cos v �= 0
tan (u + v) = tan u + tan v1 − tan u tan v
Similarly it can be shown that
tan (u − v) = tan u − tan v1 + tan u tan v
Example 12
If tan u = 4 and tan v = 35
and u and v are acute angles, show that u − v = �4
.
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312 Essential Advanced General Mathematics
Solution
tan (u − v) = tan u − tan v1 + tan u tan v
=4 − 3
5
1 + 4 × 35
=17
517
5
= 1
∴ u − v = �4
Note: tan � is a one-to-one function for 0 < � <�
2
Derivation of the addition formulas usingrotation about the originThe use of matrices to describe rotations about
the origin has been discussed in Section 10.11.
We can use matrices as an alternative method to
derive the addition formulas. Consider, for
example, the point with coordinates
(cos (u + v), sin (u + v)), which can be regardedas the image of a point with coordinates (cos u, sin u)
under a rotation of vc in an anticlockwise
direction around the origin.
x
y
O
v
u
(cos(u + v), sin(u + v))(cos u, sin u)
The matrix that defines a rotation of v radians anticlockwise about the origin is given by[cos v −sin vsin v cos v
]
i.e.,
[x ′
y′
]=
[cos v −sin vsin v cos v
] [x
y
]
becomes
[cos (u + v)sin (u + v)
]=
[cos v −sin vsin v cos v
] [cos u
sin u
]
i.e., cos (u + v) = cos v cos u − sin v sin u or cos (u + v) = cos u cos v − sin u sin vand sin (u + v) = sin v cos u + cos v sin u or sin (u + v) = sin u cos v + cos u sin v
Similarly, consider a point
(cos (u − v), sin (u − v)), which can beregarded as the image of a point with
coordinates (cos u, sin u) under a rotation
of vc in a clockwise direction around the
origin. O
u
v (cos(u – v), sin(u – v))(cos u, sin u)
x
ySAM
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Chapter 11 — Circular functions II 313
The matrix that defines a rotation of v radians clockwise about the origin is given by[cos (−v) −sin (−v)sin (−v) cos (−v)
]=
[cos v sin v
−sin v cos v
]
i.e.,
[x ′
y′
]=
[cos v sin v
−sin v cos v
] [x
y
]
becomes
[cos (u − v)sin (u − v)
]=
[cos v sin v
−sin v cos v
] [cos u
sin u
]
i.e., cos (u − v) = cos v cos u + sin v sin u or cos (u − v) = cos u cos v + sin u sin vand sin (u − v) = −sin v cos u + cos v sin u or sin (u − v) = sin u cos v − cos u sin v
Double angle formulascos (u + v) = cos u cos v − sin u sin v
Replacing v with ucos (u + u) = cos u cos u −sin u sin ucos 2u = cos2 u −sin2 u
= 2 cos2 u − 1= 1 − 2 sin2 u
since sin2 u = 1 − cos2 usince cos2 u = 1 −sin2 u
Similarly, replacing v with u in sin (u + v) = sin u cos v + cos u sin v∴ sin 2u = sin u cos u + cos u sin u
sin 2u = 2 sin u cos u
Replacing v with u in tan (u + v) = tan u + tan v1 − tan u tan v
∴ tan (u + u) = tan u + tan u1 − tan u tan u
tan 2u = 2 tan u1 − tan2 u
Example 13
If tan � = 43
and 0 < � <�
2, evaluate
a sin 2� b tan 2�
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314 Essential Advanced General Mathematics
5 4
3θ
Solution
a sin � = 45
and cos � = 35
∴ sin 2� = 2 sin � cos �= 2 × 4
5× 3
5= 24
25
b tan 2� = 2 tan �1 − tan2 �
=2 × 4
3
1 − 169
=8
3−79
= −247
Example 14
Prove each of the following identities.
a2 sin � cos �
cos2 � −sin2 � = tan 2� bsin �
sin �+ cos �
cos �= 2 sin (� + �)
sin 2�
c1
cos � + sin � +1
cos � −sin � = tan 2� cosec �
Solution
a LHS = 2 sin � cos �cos2 � −sin2 �
= sin 2�cos 2�
= tan 2�= RHS
Note: Identity holds when cos 2� �= 0
b LHS = sin �sin �
+ cos �cos �
= sin � cos � + cos � sin �sin � cos �
= sin (� + �)1
2sin 2�
= 2 sin (� + �)sin 2�
Note: Identity holds when sin 2� �= 0
c LHS = 1cos � + sin � +
1
cos � −sin �= cos � −sin � + cos � + sin �
cos2 � −sin2 �= 2 cos �
cos 2�
But 2 cos � = 2 sin � cos �sin �
= sin 2�sin �
∴ LHS = sin 2�cos 2� sin �
= tan 2�sin �
= tan 2� cosec �Note: Identity holds when cos 2� �= 0
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Chapter 11 — Circular functions II 315
Sometimes the easiest way to prove two expressions are equal is to simplify each of them. This
is demonstrated in the following example.
Example 15
Prove that (sec A − cos A)(cosecA −sin A) = 1tan A + cot A
Solution
LHS = (sec A − cos A)(cosecA −sin A) RHS = 1tan A + cot A
=(
1
cos A− cos A
) (1
sin A−sin A
)= 1
sin A
cos A+ cos A
sin A
= 1 − cos2 A
cos A× 1 −sin
2 A
sin A= 1
sin2 A + cos2 Acos A sin A
= sin2 A cos2 A
cos A sin A= cos A sin A
sin2 A + cos2 A= cos A sin A = cos A sin A
LHS = RHS
Exercise 11F
1 By making use of the appropriate addition formula find the exact values for each of theExample 10
following.
a cos 15◦ b cos 105◦
2 By making use of the appropriate addition formula find exact values for each of theExample 11
following.
a sin 165◦ b tan 75◦
3 Find exact values of
a cos5�
12b sin
11�
12c tan
−�12
4 If sin u = 1213
and sin v = 35
, evaluate sin (u + v). (Note: There is more than one answer.)Example 12
5 Simplify the following.
a sin(
� + �6
)b cos
(� − �
4
)c tan
(� + �
3
)d sin
(� − �
4
)6 Simplify
a cos (u − v) sin v + sin (u − v) cos v b sin (u + v) sin v + cos (u + v) cos v
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316 Essential Advanced General Mathematics
7 If sin � = −35
and � is in the 3rd quadrant and cos � = −513
and � is in the 2nd quadrant,Example 13
evaluate each of the following without using a calculator.
a cos 2� b sin 2� c tan 2� d sec 2�
e sin (� + �) f cos (� − �) g cosec (� + �) h cot 2�
8 If tan u = 43
and tan v = 512
and both u and v are acute angles evaluate:
a tan (u + v) b tan 2u c cos (u − v) d sin 2u
9 If sin � = 35
and sin � = 2425
and�
2< � < � < � evaluate
a cos 2� b sin (� − �) c tan (� + �) d sin (2�)
10 If sin � = −√
3
2and cos � = 1
2evaluate
a sin 2� b cos 2�
11 Simplify each of the following expressions.
a (sin � − cos �)2 b cos4 � −sin4 �
12 Prove the following identities,Examples 14, 15
a√
2 sin(
� − �4
)= sin � − cos � b cos
(� − �
3
)+ cos
(� + �
3
)= cos �
c tan(
� + �4
)tan
(� − �
4
)= −1 d cos
(� + �
6
)+ sin
(� + �
3
)= √3 cos �
e tan(
� + �4
)= 1 + tan �
1 − tan � fsin (u + v)cos u cos v
= tan v + tan u
gtan u + tan vtan u − tan v =
sin (u + v)sin (u − v) h cos 2� + 2 sin
2 � = 1
i sin 4� = 4 sin � cos3 � − 4 cos � sin3 � j 1 −sin 2�sin � − cos � = sin � − cos �
11.7 a cos x + b sin xIn Section 11.2 the method of addition of ordinates was used in the plotting of the sums of
circular functions. In this section it will be shown how functions with rule of the form
f (x) = a cos x + b sin x may have the rule written in terms of a single circular function.First write
a cos x + b sin x = √a2 + b2(
a√a2 + b2 cos x +
b√a2 + b2 sin x
)
= √a2 + b2 (cos � cos x + sin � sin x)where cos � = a√
a2 + b2 and sin � =b√
a2 + b2Let r = √a2 + b2 and thus
a cos x + b sin x = r cos (x − �)
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Chapter 11 — Circular functions II 317
Similarly it may be shown that
a cos x + b sin x = r sin (x + �)
where r = √a2 + b2, sin � = a√a2 + b2 and cos � =
b√a2 + b2
Example 16
Express cos x − √3 sin x in the form r cos (x − �) and hence find the range of the functionwith rule f (x) = cos x − √3 sin x , and the maximum and minimum values of the function.
Solution
a = 1, b = −√
3 ∴ r = √1 + 3 = 2also cos � = a
r= 1
2and sin � = b
r= −
√3
2∴ � = −�
3∴ cos x −
√3 sin x = 2 cos
(x + �
3
)∴ Range of f is [−2, 2]
The maximum and minimum values of f are 2 and −2 respectively.
Using the TI-NspireUse tCollect( ) from the Trigonometrysubmenu of the Algebra menu (b 3
2) as shown.
Example 17
Solve cos x − √3 sin x = 1 for x ∈ [0, 2�].
Solution
From Example 16, cos x −√
3 sin x = 2 cos(
x + �3
)∴ 2 cos
(x + �
3
)= 1
cos(
x + �3
)= 1
2
x + �3
= �3
,5�
3,
7�
3
x = 0, 4�3
, 2�
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318 Essential Advanced General Mathematics
Using the TI-NspireUse Solve( ) from the Algebra menu (b31) as shown.
The symbol ≤ can be found in thecatalog ( 4), by typing , or by
typing / .
Example 18
Express√
3 sin 2x − cos 2x in the form r sin (2x + �).
Solution
A slightly different technique is used.
Let√
3 sin 2x − cos 2x = r sin (2x + �)Then
√3 sin 2x − cos 2x = r [sin 2x cos � + cos 2x sin �]
This is to hold for all x.
If x = �4
,√
3 = r cos � ... 1
If x = 0, −1 = r sin � ... 2Squaring and adding 1 and 2 gives
r2 cos2 � + r2 sin2 � = 4i.e., r2 = 4
∴ r = ±2The positive solution is taken. Substituting in 1 and 2 gives
∴√
3
2= cos � and − 1
2= sin �
∴ � = −�6
∴√
3 sin 2x − cos 2x = 2 sin(
2x − �6
)Expand the right hand side of the equation to verify.SA
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Chapter 11 — Circular functions II 319
Exercise 11G
1 Find the maximum and minimum values of the following.Example 16
a 4 cos x + 3 sin x b √3 cos x + sin x c cos x −sin xd cos x + sin x e 3 cos x + √3 sin x f sin x − √3 cos xg cos x − √3 sin x + 2 h 5 + 3 sin x − 2 cos x
2 Solve each of the following for x ∈ [0, 2�] or � ◦ ∈ [0, 360].Example 17a sin x − cos x = 1 b √3 sin x + cos x = 1c sin x − √3 cos x = −1 d 3 cos x − √3 sin x = 3e 4 sin �◦ + 3 cos �◦ = 5 f 2√2 sin �◦ − 2 cos �◦ = 3
3 Write√
3 cos 2x −sin 2x in the form r cos (2x + �).
4 Write cos 3x −sin 3x in the form r sin (3x − �).Example 18
5 Sketch the graphs of the following, showing one cycle.
a f (x) = sin x − cos x b f (x) = √3 sin x + cos xc f (x) = sin x + cos x d f (x) = sin x − √3 cos x
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Rev
iew
320 Essential Advanced General Mathematics
Chapter summary
Further symmetry properties: complementary angles
sin(�
2− �
)= cos �
sin(�
2+ �
)= cos �
cos(�
2− �
)= sin �
cos(�
2+ �
)= −sin �
Addition of ordinates
θ
y
3210
y1 = 2 sin x
y2 = 3 cos 2x y = 2 sin x + 3 cos 2x
π2
2ππ
–3–2–1
–4–5
23π
Graph of tangent function
y
π
1
22ππ
4
π2
3π2 2
5π –1
–πθ
y = tan θ
period = π
General solution of circular function equations
If cos (x) = a, x = 2n� ± cos−1(a), where n ∈ Z and a ∈ [−1, 1]If tan (x) = a, x = n� + tan−1(a), where n ∈ Z and a ∈ RIf sin (x) = a, x = 2n� + sin−1(a), or
x = (2n + 1)� −sin−1 (a), where n ∈ Z and a ∈ [−1, 1]Reciprocal circular functions
secant � = sec � = 1cos �
cosecant � = cosec � = 1sin �
cotangent � = cot � = 1tan �
, sin � �= 0 and cos � �= 0
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Review
Chapter 11 — Circular functions II 321
Pythagorean identity
cos2 � + sin2 � = 11 + tan2 � = sec2 �cot2 � + 1 = cosec2 �
Addition formulas
cos (u − v) = cos u cos v + sin u sin vcos (u + v) = cos u cos v −sin u sin vsin (u + v) = sin u cos v + cos u sin vsin (u − v) = sin u cos v − cos u sin vtan (u + v) = tan u + tan v
1 − tan u tan vtan (u − v) = tan u − tan v
1 + tan u tan vDouble angle formulas
cos 2u = cos2 u −sin2 u= 2 cos2 u − 1= 1 − 2 sin2 u
sin 2u = 2 sin u cos utan 2u = 2 tan u
1 − tan2 ua cos x + b sin x can be written as r cos (x − �)
where r =√
a2 + b2 and cos � = a√a2 + b2 and sin � =
b√a2 + b2
It can also be written as r sin (x + �)
where r =√
a2 + b2 and sin � = a√a2 + b2 and cos � =
b√a2 + b2
Multiple-choice questions
1 cosec x −sin x is equal toA cos x cot x B cosec x tan x C 1 −sin2 xD sin x cosec x E
1 −sin xsin x
2 If cos x = −13
, the possible values of sin x are
A−2√2
3,
2√
2
3B
−23
,2
3C
−89
,8
9
D−√2
3,
√2
3E
1
2,−12
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Rev
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322 Essential Advanced General Mathematics
3 If cos � = ab
and 0 < � <�
2, then tan � in terms of a and b is
A
√a2 + b2
bB
√b2 − a2
aC
a√b2 − a2
Da√
b2 + a2 Ea
b√
b2 + a24 The magnitude of ∠ABX is �, AX = 4 cm, XC = x cm
and BC = 2 cm. In terms of x, tan � is equal toA
8
(x + 2)2 B4
xC 8 − x D 8 + xE
8√x2 + 4
2 cm
θ
4 cm
x cm
X
C
A
B
5 For�
2< A < � and � < B <
3�
2, with cos A = t and sin B = t , sin (B + A) is equal to
A 0 B 1 C 2t2 − 1 D 1 − 2t2 E −1
6sin 2A
cos 2A − 1 is equal toA cot 2A − 1 B sin 2A + sec 2A C sin A
cos A − 1D sin 2A − tan 2A E −cot A
7 sin(�
2− x
)is not equal to
A cos (2� − x) B −sin(
3�
2+ x
)C sin x D cos (−x) E sin
(�2
+ x)
8 (1 + cot x)2 + (1 − cot x)2 is equal toA 2 + cot x + 2 cot 2x B 2 C −4 cot x D 2 + cot 2x E 2cosec2x
9 If sin 2A = m and cos A = n, tan A in terms of m and n is equal toA
m
2n2B
n
mC
2n
m2D
2n
mE
2n2
m10 −cos x + sin x , in the form r sin (x + �) where r > 0, is
A√
2 sin(
x + �4
)B −sin
(x + �
4
)C
√2 sin
(x + 5�
4
)
D√
2 sin
(x + 7�
4
)E
√2 sin
(x + 3�
4
)
Short-answer questions (technology-free)
1 Prove each of the following identities.
a sec � + cosec � cot � = sec � cosec2 � b sec � −sin � = tan2 � + cos2 �
sec � + sin �2 Find the maximum and minimum values of each of the following.
a 3 + 2 sin � b 1 − 3 cos � c 4 sin 32
�
d 2 sin21
2� e
1
2 + cos �
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Review
Chapter 11 — Circular functions II 323
3 Find the values of �, � ∈ [0, 2�], for whicha sin2 � = 1
4b sin 2� = 1
2c cos 3� =
√3
2
d sin2 2� = 1 e tan2 � = 13
f tan 2� = −1g sin 3� = −1 h sec 2� = √2
4 Solve the equation tan � = 2 sin � for values of � from 0◦ to 360◦.5 If sin A = 5
13, sin B = 8
17where A and B are acute, find
a cos (A + B) b sin (A − B) c tan (A + B)6 Find
a cos 80◦ cos 20◦ + sin 80◦ sin 20◦ b tan 15◦ + tan 30◦
1 − tan 15◦ tan 30◦7 If A + B = �
2, find the value of
a sin A cos B + cos A sin B b cos A cos B −sin A sin B8 Find the maximum and minimum values of the function with rule
a 3 + 2 sin � b 4 − 5 cos �9 Prove each of the following.
a sin2 A cos2 B − cos2 A sin2 B = sin2 A −sin2 B b sin �1 + cos � +
1 + cos �sin �
= 2sin �
csin � − 2 sin3 �2 cos3 � − cos � = tan �
10 Given that sin A =√
5
3and that A is obtuse, find the value of each of the following:
a cos 2A b sin 2A c sin 4A
11 Prove
a1 − tan2 A1 + tan2 A = cos 2A b
sin A
1 + cos A +1 + cos A
sin A= 2
sin A
12 a Find tan 15◦ in simplest surd form.b Using the identities for sin (u ± v), express 2 sin x cos y as the sum of two sines.
13 Given f : [0, 2�] → R, f (x) = 2√3 cos x − 2 sin x , find the coordinates ofa the y intercept b the x intercepts
c the maximum point d the minimum point.
Hence sketch the graph of f (x) = 2√3 cos x − 2 sin x14 Solve for x, 0 ≤ x ≤ 2�.
a sin x + cos x = 1 b sin 12
x cos1
2x = −1
4c 3 tan 2x = 2 tan x d sin2 x = cos2 x + 1e sin 3x cos x − cos 3x sin x =
√3
2f 2 cos
(2x − �
3
)= −√3
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324 Essential Advanced General Mathematics
15 Sketch graphs of
a y = 2 cos2 x b y = 1 − 2 sin(�
2− x
2
)c f (x) = tan 2x
16 It is given that tan A = 2. Find the exact value of tan �, given that tan (� + A) = 4.17 a Express 2 cos � + 9 sin � in the form r cos (� − �), where r > 0 and 0 < � < �
2b i Give the maximum value of 2 cos � + 9 sin �
ii Give the cosine of � for which this maximum occurs.
iii Find the smallest positive solution of the equation 2 cos � + 9 sin � = 1
Extended-response questions
1 The diagram shows a rectangle ABCD inside a semicircle, centre O and radius 5 cm.
∠BOA = ∠COD = �◦a Show that the perimeter, P cm, of the rectangle
is given by
P = 20 cos � + 10 sin �b Express P in the form r cos (� − �) and hence
find the value of � for which P = 16. θ° θ°A
B C
DO
5 cm 5 cm
c Find the value of k for which the area of the rectangle is k sin 2� cm2.
d Find the value of � for which the area is a maximum.
2 The diagram shows a vertical section through a tent
in which AB = 1 m, BC = 2 m and∠BAD = ∠BCD = �. CD is horizontal.
The diagram is symmetrical about the vertical AD.
a Obtain an expression for AD in terms of �.
b Express AD in the form
r cos (� − �), where r is positive.c State the maximum length of AD
and the corresponding value of �.
A
B
C D
1 m
2 m
θ
θ
d Given that AD = 2.15 m, find the value of � for which � > �.3 a Prove the identity cos 2� = 1 − tan
2 �
1 + tan2 �b i Use the result of a to show 1 + x2 = √2x2 − √2 where x = tan
(67
1
2
)◦ii Hence find the values of integers a and b such that tan
(67
1
2
)◦= a + b√2
c Find the value of tan
(7
1
2
)◦.
SAM
PLE
Cambridge University Press • Uncorrected Sample Pages • 978-0-521-61252-4 2008 © Evans, Lipson, Jones, Avery, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
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P1: FXS/ABE P2: FXS
9780521740494c11.xml CUAU033-EVANS September 9, 2008 13:42
Review
Chapter 11 — Circular functions II 325
4 In the diagram triangle ABC has a right angle at B.
Length of BC = 1 unit.a Find in terms of �
i h1 ii h2 iii h3 iv hn
A
h1
h2h3
B C
θ
b Show that the infinite sum h1 + h2 + h3 + . . . = cos �1 −sin �
c If the infinite sum = √2, find �.5 ABCD is a regular pentagon with side length one unit.
The exterior angles of a regular pentagon each have
magnitude2�
5.
a i Show that the magnitude of ∠BCA is �5
ii Find the length of CA
b i Show the magnitude of ∠DCP is 2�5
ii Use the fact that AC = 2CQ = 2CP + PRto show that 2 cos
�
5= 2 cos 2�
5+ 1
B
CP Q R
A
D E
2π5
iii Use the identity cos 2� = 2 cos2 � − 1 to form a quadratic equation in terms ofcos
�
5iv Find the exact value of cos
�
56 a Prove each of the identities
i cos � =1 − tan2 �
2
1 + tan2 �2
ii sin � =2 tan
�
2
1 + tan2 �2
b Use the result of a to find the value of tan�
2, given 8 cos � − sin � = 4
SAM
PLE
Cambridge University Press • Uncorrected Sample Pages • 978-0-521-61252-4 2008 © Evans, Lipson, Jones, Avery, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
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