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C H A P T E R
11
E
Circular functions II
PL
Objectives
To 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 x
SA
M
To sketch graphs of functions of the form f (x) = a cos x + b sin x
To solve equations of the form a cos x + b sin x = c
11.1
Further symmetry properties
y
Complementary relationships
− =a
2
and since a = cos sin
− = cos 2
sin
Similarly
cos
b
a
π
–θ
2
θ
− =b
P(θ)
b
θ
2
and since b = sin cos
− = sin 2
+ = a = cos sin
2
+ = −b = −sin cos
2
π
–θ
2
P
x
a
y
π
P +θ
2
b
a
π
+θ
2
P(θ)
θ
θ
297
b
a
x
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298
Essential Advanced General Mathematics
Example 1
If sin = 0.3 and cos = 0.8, find the values of
a sin
−
b cos
+
2
2
b cos
+ = −sin 2
= −0.3
Exercise 11A
Example
11.2
1
c sin (−) = − sin = −0.3
E
Solution
a sin
− = cos 2
= 0.8
c sin(−)
1 If sin x = 0.3, cos = 0.6 and tan = 0.7, find the values of
+
c tan(−)
a cos(−)
b sin
2
g cos
+x
−
e sin(−x)
f tan
2
2
3
3
−x
j cos
i sin
+
2
2
−x
2
h sin
−
2
PL
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d cos
Addition of ordinates
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Example 2
Using the same scale and axes, sketch the graphs of y1 = 2 sin x and y2 = 3 cos 2x for
0 ≤ x ≤ 2.
Use addition of ordinates to sketch the graph of y = 2 sin x + 3 cos 2x.
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 of
ordinates is used.
Let y = y1 + y2 when y1 = 2 sin x and y2 = 3 cos 2x
e.g., at
x = 0, y
x= , y
4
x= , y
2
x = , y
3
,y
x=
2
and so on.
y
=0+3=3
√
2
2
= √ +0= √ = 2
2
2
= 2 − 3 = −1
=0+3=3
= −2 − 3 = −5
3
2
1
0
–1
–2
–3
–4
–5
y1 = 2sinx
π
2
π
x
3π
2
2π
y2 = 3cos2x
y = 2sinx + 3cos 2x
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Chapter 11 — Circular functions II
299
Using the TI-Nspire
Check that the calculator is in Radian
mode. Open a Graphs & Geometry
2) and enter the
application (
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.
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M
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) and split
Add Function Table (b 2
the screen as shown using the Tools menu
(/ 5 2 3) 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 x
values.
E
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Using the Casio ClassPad
Set the calculator to Radian mode.
1
Enter the functions y = sin (x) and y = .
2
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 approximations
for the solutions. The scroll key moves the cursor
between solutions.
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300
Essential Advanced General Mathematics
E
To find exact solutions, the
window is
used.
1
Enter and highlight the equation sin (x) = .
2
Tap Interactive, Equation/inequality, solve and
ensure 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 + , 2n + 5 .
6
6
PL
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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=
11 13 23
,
,
,
.
6 6
6
6
SA
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Exercise 11B
Example
2
1 Use addition of ordinates to sketch the graphs of
a y = 2 sin + cos 1
c y = sin 2 − cos 2
e y = 4 sin − 2 cos 11.3
b y = 3 cos 2 + 2 sin 2
d y = 3 sin + cos 2
Sketch graphs of the tangent function
A table of values for y = tan x is given below. Use a calculator to check these values.
x − −
y
0
3
4
1
−
2
−
0
4
4
2
undefined −1 0 1 undefined
3
5
4
4
−1
0
1
3
2
undefined
7
9
2
4
4
−1
0
1
5
2
undefined
11
3
4
−1
0
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301
Chapter 11 — Circular functions II
The graph of y = tan x is given below.
y
π
2
π
–π
–4
–3
–2
0
–1
3π
2
1
2
3
2π
4
–1
–2
3
5
x =− , ,
and
are asymptotes.
2 2 2
2
5
6
3π
7
8
9
x
10
PL
Note:
5π
2
E
–π 2
2 1
Observations from the graph
1 The graph repeats itself every units, i.e., the period of tan is .
2 The range of tan is R.
Exercise 11C
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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 +1
d y = 3 tan x + 1
e y = 2 tan x +
f y = 3 tan 2 x −
−2
2
4
11.4
General solution of circular function equations
The 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]
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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),
Note:
where n ∈ Z and a ∈ [−1, 1]
An alternative and more concise way to express the general solution of sin (x) = a is:
E
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.
√
3 tan (3x) = 1
a cos (x) = 0.5
b
Solution
1
b tan (3x) = √
3
a x = 2n ± cos−1 (0.5)
= 2n ±
3
(6n ± 1)
,n ∈ Z
=
3
3x = n + tan
√
1
sin (x) = √
2
−1
−1
1
√
3
2
6
(6n + 1)
=
6
(6n + 1)
,n ∈ Z
x=
18
= n +
SA
M
c
c 2 sin (x) =
PL
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1
√
2
x = 2n + sin
= 2n +
4
(8n + 1)
,n ∈ Z
=
4
or
−1
x = (2n + 1) − sin
= (2n + 1) −
4
(8n + 3)
=
,n ∈ Z
4
1
√
2
Using the TI-Nspire
Check that the calculator is in Radian mode.
a Use Solve( ) from the Algebra menu
(b 3 1) and complete as shown.
1
Note the use of rather than 0.5 to
2
ensure that the answer is exact.
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Chapter 11 — Circular functions II
303
b Complete as shown.
PL
c Complete as shown.
E
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Using the Casio ClassPad
SA
M
a Enter and highlight the equation cos (x) = 0.5,
tap Interactive, Equation/inequation, solve
and ensure the variable is set to x.
b Enter and highlight the equation
(3) tan (3x) = 1, tap Interactive,
Equation/inequation, solve and ensure the
variable is set to x.
c Enter and highlight the equation (2) sin (x) = 1,
tap Interactive, Equation/inequation, solve
and ensure the variable is set to x.
Example 4
Find the first three positive solutions to each of the following equations.
√
√
3 tan (3x) = 1
c 2 sin (x) = 2
a cos (x) = 0.5
b
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304
Essential Advanced General Mathematics
Solution
(6n ± 1)
,n ∈ Z
3
5
7
or x =
When n = 0, x = ± , and when n = 1, x =
3
3
3
5 7
,
The first three positive solutions of cos (x) = 0.5 are x = ,
3 3 3
(6n + 1)
b The general solution (from Example 3) is given by x =
,n ∈ Z
18
7
13
When n = 0, x =
, and when n = 1, x =
, and when n = 2, x =
18
18
18
√
7 13
,
,
The first three positive solutions of 3 tan (3x) = 1 are x =
18 18 18
(8n + 1)
or
c The general solution (from Example 3) is given by x =
4
(8n + 3)
x=
,n ∈ Z
4
3
9
11
When n = 0, x = or
, and when n = 1, x =
or x =
4
4
4
4
√
3 9
,
The first three positive solutions of 2 sin (x) = 2 are x = ,
4 4 4
E
a The general solution (from Example 3) is given by x =
PL
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Exercise 11D
3
1 Find the general solution to each of the following equations.
√
a sin (x) = 0.5
b 2 cos (3x) = 3
SA
M
Example
Example
11.5
4
c
√
3 tan (x) = −3
2 Find the first two positive solutions to each of the following equations.
√
√
a sin (x) = 0.5
b 2 cos (3x) = 3
c
3 tan (x) = −3
√
3 Find the general solution to 2 cos 2x +
= 2, and hence find all the solutions for x in
4
the interval (−2, 2).
√
− 3x − 1 = 0, and hence find all the solutions for
4 Find the general solution to 3 tan
6
x in the interval [−, 0].
√
5 Find the general solution to 2 sin (4x) + 3 = 0, and hence find all the solutions for x in
the interval [−1, 1].
Trigonometric identities
Reciprocal functions
The functions sin, cos, and tan can be used to form three other functions called the reciprocal
circular functions.
1
1
cosec =
(sin = 0)
(cos = 0)
sec =
sin cos cos (sin = 0)
cot =
sin 1
1
Note: For cos = 0 and sin = 0, cot =
and tan =
tan cot Cambridge University Press • Uncorrected Sample Pages • 978-0-521-61252-4
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Chapter 11 — Circular functions II
305
Example 5
Find the exact value of each of the following.
5
2
b cot
a sec
4
3
1
2
cos
3
1
=
cos −
3
1
=
−cos
3
1
=
1
−
2
= −2
2
a sec
=
3
Example 6
7
4
5
cos
7
5
4
c cosec
b cot
=
5
4
4
sin
1
4
=
sin
2
−
cos +
4
4
=
1
=
sin +
4
−sin
4
−1
−1
= √ ÷√
1
2
2
=
1
=1
−√
2
√
=− 2
E
Solution
c cosec
PL
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M
Find the value(s) of x between 0 and 2 for which
a sec x = −2
b cot x = −1
Solution
a
∴
∴
∴
i.e.,
b cot x = −1
sec x = −2
1
implies
= −2
cos x
−1
tan x = −1
cos x =
2 ∴
x = − or x = 2 −
4
4
x = − or x = +
3
7
3
3
i.e. x =
or x =
4
2
4
4
or x =
x =
3
3
Using the TI-Nspire
Check that the calculator is in Radian mode.
Use solve( ) from the Algebra menu (b 3 1) as shown.
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306
Essential Advanced General Mathematics
Using the Casio ClassPad
1
= −2,
cos(x)
tap Interactive, Equation/inequation,
solve and ensure the variable is set to x.
1
= −1,
b Enter and highlight the equation
tan(x)
tap Interactive, Equation/inequation,
solve and ensure the variable is set to x.
PL
E
a Enter and highlight the equation
The Pythagorean identity
Consider a point, P(), on the unit circle.
By Pythagoras’ theorem:
SA
M
OP2 = OM 2 + MP2
1 = (cos )2 + (sin )2
∴
Now (cos )2 and (sin )2 may be written as cos2 and sin2 . –1
Since this is true for all values of it is called an identity.
In particular this is called the Pythagorean identity.
y
1
P(θ)
sin θ
O cos θ M 1
–1
cos2 + sin2 = 1
Other forms of the identity can be derived.
Dividing both sides by cos2 gives:
sin2 1
cos2 +
=
2
2
cos cos cos2 1 + tan2 = sec2 Dividing both sides by sin2 gives:
1
cos2 sin2 +
=
2
2
sin sin sin2 cot2 + 1 = cosec2 Cambridge University Press • Uncorrected Sample Pages • 978-0-521-61252-4
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x
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Chapter 11 — Circular functions II
307
Example 7
a If cosec x =
7
, find cos x.
4
3
b If sec x = − , find sin x where ≤ x ≤ .
2
2
Solution
Now
so
∴
∴
Example 8
If sin =
2
3
b Since sec x = − , cos x = −
2
3
cos2 x + sin2 x = 1
4
+ sin2 x = 1
9
√
5
sin x = ±
3
E
4
7
, sin x =
4
7
cos2 x + sin2 x = 1
16
=1
cos2 x +
49
33
cos2 x =
49
√
33
cos x = ±
7
a Since cosec x =
∴
∴
For P(x) in the 2nd quadrant, sin x is
positive
√
5
∴
sin x =
3
PL
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3
and < < , find the value of cos and tan .
5
2
SA
M
Solution
Since
then
∴
∴
∴
cos2 + sin2 = 1
32
cos2 + 2 = 1
5
9
2
cos = 1 −
25
16
=
25
4
cos = − since < < 5
2
sin 3
tan = − as tan =
4
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308
Essential Advanced General Mathematics
Example 9
Prove the identity
1
1
+
= 2 cosec2 1 − cos 1 + cos Solution
1
1
+
1 − cos 1 + cos 1 + cos + 1 − cos =
1 − cos2 2
=
1 − cos2 2
=
sin2 = 2 cosec2 = RHS
E
LHS =
PL
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Exercise 11E
5
1 Find the exact value of each of the following.
3
5
5
a cot
b cot
c sec
4
4
6
4
13
7
e sec
f cosec
g cot
3
6
3
SA
M
Example
2
5
h sec
3
d cosec
2 Without using a calculator write down the exact value of each of the following.
a cot 135◦
f sec 330◦
Example
6
Example
7
Example
8
b sec 150◦
g cot 315◦
c cosec 90◦
h cosec 300◦
d cot 240◦
i cot 420◦
3 Find the values of x between 0 and 2 for which
√
√
a cosec x = 2 b cot x = 3
c sec x + 2 = 0
4 If sec =
a cos 5 If tan =
−17
and < < , find
8
2
b sin e cosec 225◦
d cosec x = sec x
c tan −7
3
and
< < 2, find cos and sin .
24
2
6 Find the value of sec if tan = 0.4 and is not in the 1st quadrant.
7 If tan =
3
sin − 2 cos 4
and < <
, evaluate
.
3
2
cot − sin 2
and is in the 4th quadrant, find the simplest expression in surd form for
3
tan − 3 sin .
cos − 2 cot 8 If cos =
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Chapter 11 — Circular functions II
9
9 Prove each of the following identities for suitable values of and .
a (1 − cos2 )(1 + cot2 ) = 1
tan tan + cot c
=
tan cot + tan 1 + cot2 = sec e
cot cosec
11.6
b cos2 tan2 + sin2 cot2 = 1
d (sin + cos )2 + (sin − cos )2 = 2
f sec + tan =
cos 1 − sin Addition formulas and double angle formulas
Addition formulas
E
Example
309
y
Consider a unit circle.
Let arc length AB = v units
u–v
1
(cos u, sin u)
C
PL
arc length AC = u units
∴ arc length CB = u − v units
Rotate OCB so that B is coincident with A.
The point P has coordinates
O
–1
CB = PA
(cos(u – v), sin(u – v))
P
1
SA
M
Applying the coordinate distance formula
CB = (cos u − cos v) + (sin u − sin v)
= 2 − 2(cos u cos v + sin u sin v)
2
PA = (cos (u − v) − 1)2 + (sin (u − v) − 0)2
= 2 − 2(cos (u − v))
2
2
A
1
x
y
(cos (u − v), sin (u − v)).
Since the triangles CBO and PAO are congruent,
(cos v, sin v)
B
2
–1
O
u–v
A
(1, 0)
x
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-Nspire
Use tExpand( )from the Trigonometry
submenu of the Algebra menu (b 3
1) as shown.
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310
Essential Advanced General Mathematics
Replacing v with −v
cos (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
E
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Evaluate cos 75◦ .
Solution
SA
M
PL
cos 75◦ = cos (45◦ + 30◦ )
◦
cos 30◦ − sin 45◦ sin 30◦
= cos 45√
1 1
3
1
−√ ·
=√ ·
2
2 2
√2
3−1
=
√
2
√
√ 2
3−1
2
=
√ ×√
2
√2 2 √
6− 2
=
4
− u in cos (u − v)
2
− u − v = cos
− u cos v + sin
− u sin v
∴
cos
2
2
2
Applying symmetry properties
−
sin = cos
2
−
and cos = sin
2
− (u + v) = sin u cos v + cos u sin v
∴ cos
2
∴
sin (u + v) = sin u cos v + cos u sin v
Replacing u with
Replacing 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◦ .
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Chapter 11 — Circular functions II
311
Solution
E
a sin 75◦ = sin (30◦ + 45◦ )
= sin 30◦ cos 45◦ + cos 30◦ sin 45◦
√
1 1
3 1
·√
= ·√ +
2
2
2
2
√
1+ 3
=
√
2 2
√
√
2
1+ 3
=
√ ×√
2 2
2
√
√
2+ 6
=
4
◦
b sin 15 = sin (45◦ − 30◦ )
SA
M
PL
= sin 45◦ cos 30◦ − cos 45◦ sin 30◦
√
3
1
1 1
=√ ·
−√ ·
2 2
2 2
√
3−1
=
√
2 2
√
√
3−1
2
=
√ ×√
2 2
2
√
√
6− 2
=
4
Addition formula for tangent
Also
sin (u + v)
cos (u + v)
sin u cos v + cos u sin v
=
cos u cos v − sin u sin v
tan (u + v) =
Divide the numerator and denominator by cos u cos v = 0
tan (u + v) =
tan u + tan v
1 − tan u tan v
Similarly it can be shown that
tan (u − v) =
tan u − tan v
1 + tan u tan v
Example 12
If tan u = 4 and tan v =
3
and u and v are acute angles, show that u − v = .
5
4
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312
Essential Advanced General Mathematics
Solution
tan u − tan v
1 + tan u tan v
3
4−
5
=
3
1+4×
5
17
= 5 =1
17
5
∴ u−v =
4
Note: tan is a one-to-one function for 0 < <
2
PL
E
tan (u − v) =
Derivation of the addition formulas using
rotation about the origin
SA
M
y
The use of matrices to describe rotations about
the origin has been discussed in Section 10.11.
(cos(u + v), sin(u + v))
(cos u, sin u)
We can use matrices as an alternative method to
derive the addition formulas. Consider, for
v
example, the point with coordinates
u
x
(cos (u + v), sin (u + v)), which can be regarded
O
as the image of a point with coordinates (cos u, sin u)
under a rotation of v c in an anticlockwise
direction around the origin.
The matrix that
defines a rotation of v radians anticlockwise about the origin is given by
cos v −sin v
sin v
cos v
x
cos v −sin v
x
i.e.,
=
y
sin v
cos v
y
cos (u + v)
cos v −sin v
cos u
becomes
=
sin (u + v)
sin v
cos v
sin u
i.e.,
and
cos (u + v) = cos v cos u − sin v sin u
sin (u + v) = sin v cos u + cos v sin u
or cos (u + v) = cos u cos v − sin u sin v
or sin (u + v) = sin u cos v + cos u sin v
Similarly, consider a point
(cos (u − v), sin (u − v)), which can be
regarded as the image of a point with
(cos u, sin u)
coordinates (cos u, sin u) under a rotation
of v c in a clockwise direction around the
origin.
y
v
u
O
(cos (u – v), sin (u – v))
x
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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)
cos v sin v
=
sin (−v)
cos (−v)
−sin v cos v
x
cos v sin v
x
i.e.,
=
y
−sin v cos v
y
cos (u − v)
cos v sin v
cos u
becomes
=
sin (u − v)
−sin v cos v
sin u
i.e., cos (u − v) = cos v cos u + sin v sin u
and sin (u − v) = −sin v cos u + cos v sin u
E
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or cos (u − v) = cos u cos v + sin u sin v
or sin (u − v) = sin u cos v − cos u sin v
PL
Double angle formulas
cos (u + v) = cos u cos v − sin u sin v
Replacing v with u
cos (u + u) = cos u cos u −sin u sin u
cos 2u = cos2 u −sin2 u
= 2 cos2 u − 1
= 1 − 2 sin2 u
since
since
sin2 u = 1 − cos2 u
cos2 u = 1 −sin2 u
SA
M
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 v
1 − tan u tan v
∴ tan (u + u) =
tan u + tan u
1 − tan u tan u
tan 2u =
2 tan u
1 − tan2 u
Example 13
4
and 0 < < , evaluate
3
2
a sin 2
b tan 2
If tan =
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314
Essential Advanced General Mathematics
Solution
a sin =
2 tan 1 − tan2 4
2×
3
=
16
1−
9
8
= 3
−7
9
24
=−
7
3
4
and cos =
5
5
∴ sin 2 = 2 sin cos 4 3
=2× ×
5 5
24
=
25
b tan 2 =
5
4
θ
E
3
Example 14
PL
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Prove each of the following identities.
2 sin cos cos 2 sin ( + )
sin = tan 2
a
+
=
b
2
2
sin cos sin 2
cos −sin 1
1
c
+
= tan 2 cosec cos + sin cos −sin Solution
2 sin cos cos2 −sin2 sin 2
=
cos 2
= tan 2
= RHS
SA
M
a LHS =
Note:
c
sin cos +
sin cos sin cos + cos sin =
sin cos sin ( + )
=
1
sin 2
2
2 sin ( + )
=
sin 2
b LHS =
Identity holds when cos 2 = 0
Note:
Identity holds when sin 2 = 0
1
1
+
cos + sin cos −sin cos −sin + cos + sin =
cos2 −sin2 2 cos =
cos 2
sin 2
2 sin cos =
But 2 cos =
sin sin sin 2
∴
LHS =
cos 2 sin tan 2
=
sin = tan 2 cosec LHS =
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) =
1
tan A + cot A
Solution
E
1
tan A + cot A
1
=
cos A
sin A
+
cos A
sin A
1
=
sin2 A + cos2 A
cos A sin A
cos A sin A
=
sin2 A + cos2 A
= cos A sin A
= (sec A − cos A)(cosecA −sin A)
1
1
− cos A
−sin A
=
cos A
sin A
=
1 − cos2 A 1 −sin2 A
×
cos A
sin A
RHS =
PL
LHS
sin2 A cos2 A
cos A sin A
= cos A sin A
=
LHS = RHS
SA
M
Exercise 11F
Example
10
1 By making use of the appropriate addition formula find the exact values for each of the
following.
a cos 15◦
Example
11
b cos 105◦
2 By making use of the appropriate addition formula find exact values for each of the
following.
a sin 165◦
b tan 75◦
3 Find exact values of
5
11
a cos
b sin
12
12
Example
12
4 If sin u =
c tan
−
12
3
12
and sin v = , evaluate sin (u + v). (Note: There is more than one answer.)
13
5
5 Simplify the following.
a sin +
b cos −
6
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
Example
Essential Advanced General Mathematics
13
−3
−5
and is in the 3rd quadrant and cos =
and is in the 2nd quadrant,
5
13
evaluate each of the following without using a calculator.
7 If sin =
a cos 2
e sin ( + )
b sin 2
f cos ( − )
c tan 2
g cosec ( + )
d sec 2
h cot 2
5
4
and tan v =
and both u and v are acute angles evaluate:
3
12
a tan (u + v)
b tan 2u
c cos (u − v)
d sin 2u
8 If tan u =
3
24
and sin =
and < < < evaluate
5
25
2
a cos 2
b sin ( − )
c tan ( + )
√
1
3
10 If sin = −
and cos = evaluate
2
2
a sin 2
b cos 2
E
9 If sin =
PL
d sin (2)
11 Simplify each of the following expressions.
a (sin − cos )2
14, 15
12 Prove the following identities,
√
a
2 sin −
= sin − cos 4
c tan +
tan −
= −1
4
4
1 + tan e tan +
=
4
1 − tan tan u + tan v
sin (u + v)
g
=
tan u − tan v
sin (u − v)
i sin 4 = 4 sin cos3 − 4 cos sin3 SA
M
Examples
b cos4 −sin4 11.7
b cos −
+ cos +
= cos 3
3
√
d cos +
+ sin +
= 3 cos 6
3
sin (u + v)
f
= tan v + tan u
cos u cos v
h cos 2 + 2 sin2 = 1
1 −sin 2
j
= sin − cos sin − cos a cos x + b sin x
In 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
√
b
a
cos x + √
sin x
a cos x + b sin x = a 2 + b2
√
a 2 + b2
a 2 + b2
√
= a 2 + b2 (cos cos x + sin sin x)
a
b
where cos = √
and sin = √
a 2 + b2
a 2 + b2
√
Let r = a 2 + 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 =
√
a
b
a 2 + b2 , sin = √
and cos = √
a 2 + b2
a 2 + b2
Example 16
Solution
also
E
√
Express cos x − 3 sin x in the form r cos (x − ) and hence find the range of the function
√
with rule f (x) = cos x − 3 sin x, and the maximum and minimum values of the function.
√
√
a = 1, b = − 3 ∴ r = 1 + 3 = 2√
a
1
b
− 3
cos = = and sin = =
r
2
r
2
∴
=−
3 √
∴ cos x − 3 sin x = 2 cos x +
3
PL
P1: FXS/ABE
∴ Range of f is [−2, 2]
The maximum and minimum values of f are 2 and −2 respectively.
SA
M
Using the TI-Nspire
Use tCollect( ) from the Trigonometry
submenu 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 +
=1
3
1
cos x +
=
3
2
5 7
x+ = ,
,
3
3 3 3
4
x = 0,
, 2
3
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318
Essential Advanced General Mathematics
Using the TI-Nspire
Example 18
Express
E
Use Solve( ) from the Algebra menu (b
3 1) as shown.
The symbol ≤ can be found in the
, or by
catalog ( 4), by typing
typing / .
PL
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√
3 sin 2x − cos 2x in the form r sin (2x + ).
Solution
A slightly different technique is used.
√
3 sin 2x − cos 2x = r sin (2x + )
Let
√
3 sin 2x − cos 2x = r [sin 2x cos + cos 2x sin ]
Then
SA
M
This is to hold for all x.
√
If
x = , 3 = r cos 4
If
x = 0, −1 = r sin ...
1
...
2
Squaring and adding 1 and 2 gives
r 2 cos2 + r 2 sin2 = 4
i.e.,
r2 = 4
∴
r = ±2
The positive solution is taken. Substituting in 1 and 2 gives
√
1
3
= cos and − = sin ∴
2
2
∴
=−
6 √
∴
3 sin 2x − cos 2x = 2 sin 2x −
6
Expand the right hand side of the equation to verify.
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Chapter 11 — Circular functions II
319
Exercise 11G
Example
16
1 Find the maximum and minimum values of the following.
√
a 4 cos x + 3 sin x
b
3 cos x + sin x
√
d cos x + sin x
e 3 cos x + 3 sin x
√
h 5 + 3 sin x − 2 cos x
g cos x − 3 sin x + 2
17
2 Solve each of the following for x ∈ [0, 2] or ◦ ∈ [0, 360].
√
a sin x − cos x = 1
b
3 sin x + cos x = 1
√
√
d 3 cos x − 3 sin x = 3
c sin x − 3 cos x = −1
√
f 2 2 sin ◦ − 2 cos ◦ = 3
e 4 sin ◦ + 3 cos ◦ = 5
√
3 Write 3 cos 2x −sin 2x in the form r cos (2x + ).
Example
18
4 Write cos 3x −sin 3x in the form r sin (3x − ).
PL
E
Example
c cos x −sin x
√
f sin x − 3 cos x
SA
M
5 Sketch the graphs of the following, showing one cycle.
√
a f (x) = sin x − cos x
b f (x) = 3 sin x + cos x
√
c f (x) = sin x + cos x
d f (x) = sin x − 3 cos x
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Essential Advanced General Mathematics
Chapter summary
y1 = 2 sin x
PL
3
2
1
0
–1
–2
–3
–4
–5
E
Further symmetry properties: complementary angles
− = cos sin
2
+ = cos sin
2
− = sin cos
2
+ = −sin cos
2
Addition of ordinates
y
π
π
2
y2 = 3 cos 2x
θ
2π
3π
2
y = 2 sin x + 3 cos 2x
Graph of tangent function
y
SA
M
Review
320
y = tan θ
period = π
1
–π
2
–1
π π
4 2
π
3π
2
2π
5π
2
θ
General solution of circular function equations
If cos (x) = a,
If tan (x) = a,
If sin (x) = a,
where n ∈ Z and a ∈ [−1, 1]
x = 2n ± cos−1 (a),
where n ∈ Z and a ∈ R
x = n + tan−1 (a),
−1
x = 2n + sin (a), or
x = (2n + 1) −sin−1 (a), where n ∈ Z and a ∈ [−1, 1]
Reciprocal circular functions
secant = sec =
1
cos 1
sin 1
cotangent = cot =
, sin = 0 and cos =
0
tan cosecant = cosec =
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Chapter 11 — Circular functions II
321
Review
Pythagorean identity
cos2 + sin2 = 1
1 + tan2 = sec2 cot2 + 1 = cosec2 Addition formulas
PL
cos (u − v) = cos u cos v + sin u sin v
cos (u + v) = cos u cos v −sin u sin v
sin (u + v) = sin u cos v + cos u sin v
sin (u − v) = sin u cos v − cos u sin v
tan u + tan v
tan (u + v) =
1 − tan u tan v
tan u − tan v
tan (u − v) =
1 + tan u tan v
E
P1: FXS/ABE
Double angle formulas
SA
M
cos 2u = cos2 u −sin2 u
= 2 cos2 u − 1
= 1 − 2 sin2 u
sin 2u = 2 sin u cos u
2 tan u
tan 2u =
1 − tan2 u
a cos x + b sin x can be written as r cos (x − )
a
b
where r = a 2 + b2 and cos = √
and sin = √
2
2
2
a +b
a + b2
It can also be written as r sin (x + )
a
b
where r = a 2 + b2 and sin = √
and cos = √
2
2
2
a +b
a + b2
Multiple-choice questions
1 cosec x −sin x is equal to
A cos x cot x
B cosec x tan x
C 1 −sin2 x
1 −sin x
D sin x cosec x
E
sin x
−1
, the possible values of sin x are
2 If cos x =
√ 3√
−2 2 2 2
−2 2
−8 8
,
A
,
B
,
C
3
3
3
3
9 9
√ √
− 2 2
1 −1
D
,
,
E
3
3
2 2
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Essential Advanced General Mathematics
a
3 If cos = and 0 < <
b
√
a 2 + b2
A
B
b
a
E
D √
2
b + a2
, then tan in terms of a and b is
2
√
a
b2 − a 2
C √
a
b2 − a 2
a
√
b b2 + a 2
A
E
4 The magnitude of ∠ABX is , AX = 4 cm, XC = x cm
and BC = 2 cm. In terms of x, tan is equal to
4
8
B
A
2
(x + 2)
x
C 8−x
D 8+x
8
E √
2
x +4
4 cm
X
x cm
θ
PL
B
C
2 cm
3
< A < and < B <
, with cos A = t and sin B = t, sin (B + A) is equal to
2
2
A 0
B 1
C 2t 2 − 1
D 1 − 2t 2
E −1
5 For
6
sin 2A
is equal to
cos 2A − 1
A cot 2A − 1
B sin 2A + sec 2A
C
D sin 2A − tan 2A
E −cot A
7 sin
− x is not equal to 2
3
A cos (2 − x) B −sin
+x
C sin x
2
8 (1 + cot x)2 + (1 − cot x)2 is equal to
A 2 + cot x + 2 cot 2x
B 2
C −4 cot x
SA
M
Review
322
sin A
cos A − 1
D cos (−x)
D 2 + cot 2x
E sin
2
+x
E 2cosec2 x
9 If sin 2A = m and cos A = n, tan A in terms of m and n is equal to
2n
2n
m
n
2n 2
C
D
A
B
E
m2
m
2n 2
m
m
10 −cos x + sin x, in the form r sin (x + ) where r > 0, is
√
√
5
A
2 sin x +
C
2 sin x +
B −sin x +
4 4
4
√
√
3
7
2 sin x +
E
D
2 sin x +
4
4
Short-answer questions (technology-free)
1 Prove each of the following identities.
tan2 + cos2 a sec + cosec cot = sec cosec2 b sec −sin =
sec + sin 2 Find the maximum and minimum values of each of the following.
3
a 3 + 2 sin b 1 − 3 cos c 4 sin 2
1
1
d 2 sin2 e
2
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Chapter 11 — Circular functions II
√
c cos 3 =
3
2
f tan 2 = −1
6 Find
a cos 80◦ cos 20◦ + sin 80◦ sin 20◦
, find the value of
2
a sin A cos B + cos A sin B
7 If A + B =
b
E
4 Solve the equation tan = 2 sin for values of from 0◦ to 360◦ .
8
5
, sin B =
where A and B are acute, find
5 If sin A =
13
17
a cos (A + B)
b sin (A − B)
c tan (A + B)
tan 15◦ + tan 30◦
1 − tan 15◦ tan 30◦
b cos A cos B −sin A sin B
8 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 2
+
=
1 + cos sin sin sin − 2 sin3 = tan 2 cos3 − cos √
5
10 Given that sin A =
and that A is obtuse, find the value of each of the following:
3
a cos 2A
b sin 2A
c sin 4A
SA
M
c
11 Prove
1 − tan2 A
= cos 2A
a
1 + tan2 A
b
sin A
1 + cos A
2
+
=
1 + cos A
sin A
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 of
a 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 x
14 Solve for x, 0 ≤ x ≤ 2.
a sin x + cos x = 1
c 3 tan 2x = 2 tan x
√
3
e sin 3x cos x − cos 3x sin x =
2
1
1
1
b sin x cos x = −
2
2
4
d sin2 x = cos2 x + 1
√
=− 3
f 2 cos 2x −
3
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
Review
3 Find the values of , ∈ [0, 2], for which
1
1
b sin 2 =
a sin2 =
2
4
1
d sin2 2 = 1
e tan2 =
3√
g sin 3 = −1
h sec 2 = 2
323
PL
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Essential Advanced General Mathematics
15 Sketch graphs of
a y = 2 cos2 x
x
c f (x) = tan 2x
2
2
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 < <
2
b 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
−
E
b y = 1 − 2 sin
PL
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
C
is given by
B
P = 20 cos + 10 sin b Express P in the form r cos ( − ) and hence
find the value of for which P = 16.
5 cm
5 cm
θ°
θ°
A
O
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.
SA
M
Review
324
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. B
a Obtain an expression for AD in terms of .
b Express AD in the form
2m
r cos ( − ), where r is positive.
D
A
1m
θ
θ
c State the maximum length of AD
and the corresponding value of .
C
D
d Given that AD = 2.15 m, find the value of for which > .
1 − tan2 3 a Prove the identity cos 2 =
1 + tan2 √
√
1 ◦
2
2
b i Use the result of a to show 1 + x = 2x − 2 where x = tan 67
2
√
1 ◦
ii Hence find the values of integers a and b such that tan 67
=a+b 2
2
◦
1
c Find the value of tan 7
.
2
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Chapter 11 — Circular functions II
A
θ
h2
h1
h3
C
E
B
cos b Show that the infinite sum h 1 + h 2 + h 3 + . . . =
1
−sin √
c If the infinite sum = 2, find .
SA
M
PL
5 ABCD is a regular pentagon with side length one unit.
2π
B
The exterior angles of a regular pentagon each have
5
2
.
magnitude
5
a i Show that the magnitude of ∠BCA is
C
A
5
P
Q
R
ii Find the length of CA
2
b i Show the magnitude of ∠DCP is
5
ii Use the fact that AC = 2CQ = 2CP + PR
2
to show that 2 cos = 2 cos
+1
5
5
E
D
2
iii Use the identity cos 2 = 2 cos − 1 to form a quadratic equation in terms of
cos
5
iv Find the exact value of cos
5
6 a Prove each of the identities
1 − tan2
2 tan
2
2
i cos =
ii sin =
2
1 + tan
1 + tan2
2
2
b Use the result of a to find the value of tan , given 8 cos − sin = 4
2
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
Review
4 In the diagram triangle ABC has a right angle at B.
Length of BC = 1 unit.
a Find in terms of ii h 2
iii h 3
iv h n
i h1
325