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University of Strathclyde
Department of Mathematics & Statistics
Circular Functions — Trigonometry
Summer School Lecture Notes
1
Basic Definitions
OAB is a triangle with a right angle at A.
b =θ
AOB
AB
opposite
=
OB
hypotenuse
B
OA
adjacent
=
OB
hypotenuse
(
)
AB
opposite
sin θ
tan θ :=
=
=
OA
adjacent
cos θ
use
n
ote
hyp
cos θ :=
cot θ :=
O
1
tan θ
REMINDER: SOH CAH TOA
1
θ
adjacent
opposite
sin θ :=
A
2
Summer School: Circular Functions
2
Circular Functions
Y
circle, radius r
From the basic definitions,
sin θ =
opp
y
=
hyp
r
B(x,y)
ϕ
adj
x
cos θ =
=
hyp
r
(
)
opp
y
sin θ
tan θ =
=
=
adj
x
cos θ
r
y
θ
x
O
A
x
y
, cos φ = so we have
r
r
x
sin(90◦ − θ) = sin φ = = cos θ
r
Now φ = 90◦ − θ and sin φ =
cos(90◦ − θ) = cos φ =
y
= sin θ.
r
These are Complementary Angles. Also,
tan(90◦ − θ) = tan φ =
3
x
= cot θ
y
Elementary Angles
When θ = 0◦ : y = 0, x = r
sin(0◦ ) =
y
= 0,
r
cos(0◦ ) =
x
y
= 1, tan(0◦ ) = = 0
r
x
and using complementary angles
sin(90◦ ) = sin(90◦ − 0◦ ) = cos(0◦ ) = 1
cos(90◦ ) = cos(90◦ − 0◦ ) = sin(0◦ ) = 0
tan(90◦ ) = tan(90◦ − 0◦ ) =
1
= undefined
0
X
3
Summer School: Circular Functions
3.1
Isosceles Triangle
B
b + OBA
b = 90◦
AOB
45
b = OBA
b
AOB
o
2a
a
a
A
b = OBA
b = 45◦
⇒ AOB
OB 2 = OA2 + AB 2 = 2a2
√
⇒ OB = 2 a
45
o
O
a
1
cos 45◦ = sin 45◦ = √
=√
2a
2
tan 45◦ = 1
3.2
Equilateral Triangle
B
OB = BC = CO = 2l
o
30
OA = AC = l
Then OB 2 = OA2 + AB 2 ⇒
4l = l + AB ⇒ AB =
2
2
2
√
2l
3l
3l
b = OBC
b = B CO
b = 60◦
C OB
o
b = ABC
b = 30◦
OBA
60
O
l
Hence
1
= cos 60◦
2
√
3
= sin 60◦
cos 30◦ =
2
sin 30◦ =
1
tan 30◦ = √
3
√
tan 60◦ = 3
A
C
4
Summer School: Circular Functions
3.3
Table of Values (IMPORTANT)
0◦ 30◦ 45◦ 60◦ 90◦
√
1
1
3
√
sin θ 0
1
2
2
2
√
3 1
1
√
cos θ 1
0
2
2 2
√
1
tan θ 0 √
1
3 −
3
θ
3.4
3.4.1
Extension to other angles
90◦ < θ ≤ 180◦
Now x < 0, y > 0 and r > 0.
Y
We define
sin θ =
cos θ =
y
>0
r
B
x
<0
r
y
r
θ
tan θ =
y
<0
x
A
x
O
b = 180◦ − θ, so
We also have AOB
sin(180◦ − θ) = sin θ
cos(180◦ − θ) = − cos θ
X
5
Summer School: Circular Functions
3.4.2
180◦ < θ ≤ 270◦
Y
Here x < 0, y < 0, r > 0
sin θ =
y
<0
r
cos θ =
x
<0
r
A
θ
x
X
O
y
tan θ = > 0
x
y
r
B
3.4.3
270◦ < θ ≤ 360◦
Y
Here x > 0, y < 0, r > 0
sin θ =
y
<0
r
cos θ =
x
>0
r
tan θ =
y
<0
x
O
θ
x
r
A
y
B
X
6
Summer School: Circular Functions
3.5
Rule of Signs
Y
What’s positive?
This shows that
st
nd
1 Quad
2 Quad
sin θ > 0 in the 1st and 2nd quadrants
sin
all
cos θ > 0 in the 1st and 4th quadrants
X
tan
tan θ > 0 in the 1st and 3rd quadrants
3.6
cos
rd
th
3 Quad
4 Quad
Negative Angles
Angles are positive when
measured anti-clockwise and
negative
when
Y
measured
clockwise.
y
sin(−θ) = − = − sin θ
r
x
= cos θ
r
y
tan(−θ) = − = − tan θ
x
cos(−θ) =
r
θ
−θ
y
X
x
r
−y
7
Summer School: Circular Functions
4
Degrees and Radians
The angle θ in radians is
θ=
length of arc AB
r
B
(This does not depend on r)
r
θ
The circumference is 2πr. Hence a
A
r
◦
complete rotation (360 ) is
2πr
= 2π radians
r
(
◦
2π radians ≡ 360 ⇒ 1 radian ≡
180
π
)◦
π
radians ≡ 1◦
180
4.1
Table of degrees and radians
Degrees
0◦
30◦
45◦
60◦
90◦
120◦
135◦
150◦
Radians
0
π/6
π/4
π/3
π/2
2π/3 3π/4
5π/6
Degrees
180◦
210◦
225◦
240◦
270◦
300◦
330◦
Radians
π
315◦
7π/6 5π/4 4π/3 3π/2 5π/3 7π/4 11π/6
Now look back at the table of values and make sure you know exact values for radians!!!
NOTE: Be careful to include
◦
when referring to degrees.
8
Summer School: Circular Functions
5
Pythagoras’ Theorem
Pythagoras’ Theorem says
x2 + y 2 = r 2
Dividing by r2 , we get
r
2
y
2
x
y
+ 2 =1
2
r
r
θ
and using the definition of sin θ and cos θ:
x
cos2 θ + sin2 θ = 1
6
Graphs
The graph of sin θ has a maxi-
sin(θ)
mum value of 1 when θ takes the
1
values
···
−3π π 5π
, ,
···
2 2 2
0.5
It takes the minimum value of
−1 when θ takes the values
···
−π 3π
, ,···
2 2
0
−0.5
and is equal to 0 when θ takes
−1
the values
−6
0, ±π, ±2π, · · ·
The graph of sin θ has a period
of 2π.
−4
−2
0
θ
2
4
6
9
Summer School: Circular Functions
The graph of cos θ has a maxicos(θ)
mum value of 1 when θ takes the
values
1
0, ±2π, ±4π · · ·
0.5
It takes the minimum value of
−1 when θ takes the values
±π, ±3π, · · ·
0
−0.5
and is equal to 0 when θ takes
the values
−1
−6.2832
π 3π 5π
± ,± ,± ,···
2
2
2
−3.1416
0
θ
3.1416
The graph of cos θ has a period
of 2π.
Note the similarity between the graphs of sin θ and cos θ.
tan(θ)
6
The graph of tan θ takes the
value 0 when θ takes the values
0, ±π, ±2π, ±3π · · ·
It is undefined when θ takes the
values
π 3π 5π
± ,± ,± ,···
2
2
2
The graph of tan θ has a period
of π.
4
2
0
−2
−4
−6
−4.7124
−1.5708
θ
1.5708
4.7124
6.2832
10
Summer School: Circular Functions
7
Compound Angles
P
α
L
α
N
Q
β
α
O
K
M
b is a right angle, so LN
bQ = α
Triangles OQK, P QN are similar, so QPbN = α. N LP
PK
P L + LK
LK + P L
NM + P L
=
=
=
PO
PO
PO
PO
N M ON
PL PN
=
·
+
·
ON P O P N P O
= sin α cos β + cos α sin β
sin(α + β) =
Hence
sin(α + β) = sin α cos β + cos α sin β
It can also be shown that
cos(α + β) = cos α cos β − sin α sin β
sin(α − β) = sin α cos β − cos α sin β
cos(α − β) = cos α cos β + sin α sin β
tan(α + β) =
tan α + tan β
1 − tan α tan β
tan(α − β) =
tan α − tan β
1 + tan α tan β
Summer School: Circular Functions
Example 7.1 Find the value of sin(75◦ )
sin(75◦ ) = sin(45◦ + 30◦ )
= sin 45◦ cos 30◦ + cos 45◦ sin 30◦
√
√
1 1
1+ 3
1 3
= √
+√
= √
2 2
22
2 2
Example 7.2 Find the value of cos(15◦ )
cos(15◦ ) = cos(45◦ − 30◦ )
= cos(45◦ ) cos(30◦ ) + sin(45◦ ) sin(30◦ )
√
√
1 3
1 1
1+ 3
= √
+√
= √
2 2
22
2 2
Example 7.3 Find the value of tan(105◦ )
tan(105◦ ) = tan(60◦ + 45◦ )
tan(60◦ ) + tan(45◦ )
=
1 − tan(60◦ ) tan(45◦ )
√
√
√
3+1
3+1 1+ 3
√ =
√ ×
√
=
1− 3
1− 3 1+ 3
√
√
3+3+1+ 3
√
√
=
1+ 3− 3−3
√
√
2 3+4
=
=− 3−2
−2
8
Double Angles
From Section 7, we have
sin(2a) = 2 sin a cos a
cos(2a) = cos2 a − sin2 a
We can use cos2 a + sin2 a = 1 to obtain alternative formulae for cos(2a):
cos(2a) = cos2 a − sin2 a = cos2 a − (1 − cos2 a) = 2 cos2 a − 1
cos(2a) = cos2 a − sin2 a = (1 − sin2 a) − sin2 a = 1 − 2 sin2 a
11
12
Summer School: Circular Functions
(
)
1◦
Example 8.1 Find the value of sin 22
.
2
Let a = 22
9
1◦
then consider cos(2a).
2
1
√ = cos(45◦ )
2
(
)
1◦
2
⇒ 2 sin 22
2
(
)
1◦
2
⇒ sin 22
2
(
)
1◦
⇒
sin 22
2
(
)
1◦
= 1 − 2 sin 22
2
√
1
2−1
=1− √ = √
2
2
√
2−1
= √
2 2
√√
2−1
√
=
2 2
2
Secant, Cosecant
We define
sec θ =
1
cos θ
and
csc θ =
1
sin θ
Recall that cos2 θ + sin2 θ = 1.
Divide by cos2 θ to get
1 + tan2 θ = sec2 θ
Divide by sin2 θ to get
cot2 θ + 1 = csc2 θ
10
Identities
Example 10.1 Show that
cos θ
1 − sin θ
=
1 + sin θ
cos θ
cos θ(1 − sin θ)
cos θ(1 − sin θ)
cos θ(1 − sin θ)
1 − sin θ
cos θ
=
=
=
=
2
2
1 + sin θ
(1 + sin θ)(1 − sin θ)
cos θ
cos θ
1 − sin θ
Example 10.2 Show that 1 + sin(2θ) = (cos θ + sin θ)2
1 + sin(2θ) = cos2 θ + sin2 θ + 2 sin θ cos θ
= cos2 θ + 2 cos θ sin θ + sin2 θ
= (cos θ + sin θ)2
13
Summer School: Circular Functions
Example 10.3 Show that
sin(2θ) − cos(2θ) + 1
= tan θ
sin(2θ) + cos(2θ) + 1
sin(2θ) − cos(2θ) + 1
2 sin θ cos θ − (1 − 2 sin2 θ) + 1
=
sin(2θ) + cos(2θ) + 1
2 sin θ cos θ + (2 cos2 θ − 1) + 1
2 sin θ cos θ + 2 sin2 θ
=
2 sin θ cos θ + 2 cos2 θ
sin θ(cos θ + sin θ)
=
cos θ(sin θ + cos θ)
sin θ
=
= tan θ.
cos θ
11
Pythagorean Triples
A set of three positive integers x, y, z such that
x2 + y 2 = z 2
is called a Pythagorean Triple.
Examples are {3, 4, 5}, {5, 12, 13}, {7, 24, 25}, {8, 15, 17}, {9, 40, 41}, {11, 60, 61}, {12, 35, 37},
{13, 84, 85}, {15, 112, 113}, {16, 63, 65}, {17, 144, 145}, {19, 180, 181}, {20, 21, 29}, {20, 99, 101}.
Example 11.1 If sin θ =
15
, what is
17
B
cos θ?
Hint: draw a triangle! From diagram
BC = 15, AB = 17
17
15
Hence
AC 2 = 172 −152 = 289−225 = 64 = 82
AC
8
cos θ =
=
AB
17
A
θ
8
C
14
Summer School: Circular Functions
Example 11.2 If sin α =
5
24
and cos β = , find tan(α − β)
13
25
13
25
5
α
7
β
24
12
tan α − tan β
1 + tan α tan β
7
5
−
12 24 = 5 × 24 − 7 × 12
=
5 7
12 × 24 + 5 × 7
1+
·
12 24
120 − 84
36
=
=
288 + 35
323
tan(α − β) =
12
Product Formulae
Recall that
sin(α + β) = sin α cos β + cos α sin β
sin(α − β) = sin α cos β − cos α sin β
⇒ sin(α + β) + sin(α − β) = 2 sin α cos β
p+q
p−q
Put α + β = p, α − β = q ⇒ α =
,β =
to get
2
2
(
)
(
)
p+q
p−q
sin p + sin q = 2 sin
cos
2
2
(
cos p + cos q = 2 cos
(
cos p − cos q = 2 sin
Example 12.1
p+q
2
p+q
2
(
◦
◦
sin 105 + sin 15
)
(
cos
)
(
sin
p−q
2
q−p
2
)
)
)
( ◦)
120◦
90
= 2 sin
cos
2
2
◦
◦
= 2 sin 60 cos√
45
√
3 1
3
√ =
= 2
2 2
2
15
Summer School: Circular Functions
Example 12.2
sin 105◦ − sin 15◦ = sin 105◦ + sin(−15◦ )
( ◦)
(
)
120◦
90
= 2 sin
cos
2
2
◦
◦
= 2 sin 45 cos 60
1 1
1
= 2√
=√
22
2
Example 12.3
(
◦
◦
cos 15 − cos 105
)
( ◦)
120◦
90
= 2 sin
sin
2
2
◦
◦
= 2 sin 60 sin 45
√
√
3 1
3
√ =√
= 2
2 2
2
Example 12.4
(
)
(
)
5x + x
5x − x
2 sin
cos
sin(5x) + sin x
2
2
(
)
(
)
=
5x + x
5x − x
cos(5x) + cos x
2 cos
cos
2
2
sin(3x)
=
= tan(3x)
cos(3x)
Product formulae can also be re-written in the form
1
sin α cos β = {sin(α + β) + sin(α − β)}
2
1
cos α cos β = {cos(α + β) + cos(α − β)}
2
1
sin α sin β = {cos(α − β) − cos(α + β)}
2
Example 12.5
(
)
( ◦)
)
(
)]
[ (
1◦
1
1
1◦
1◦
1◦
1◦
cos 37
cos 7
=
+ cos 37 − 7
cos 37 + 7
2
2
2
2
2
2
2
1
=
{cos(45◦ ) + cos(30◦ )}
2[
√ ] √
√
1 1
3
2+ 3
√ +
=
=
2
2
4
2
16
Summer School: Circular Functions
Example 12.6
)
)
)
( ◦)
(
(
1◦
1◦
1◦
1
1◦
1◦
2 sin 37
cos 7
= sin 37 + 7
+ sin 37 − 7
2
2
2
2
2
2
◦
◦
= sin(45 ) + sin(30 )
√
1
1
2+1
= √ + =
2
2 2
(
13
Trigonometric Equations
Question: Given a value c of sin(θ), what is θ?
1
Example 13.1 Suppose sin(θ) = :
2
sin(θ)
1
0.5
0
−0.5
−1
−6.2832
−3.1416
0
3.1416
θ
6.2832
9.4248
12.5664
1
There are infinitely many values of θ for which sin(θ) = , for example
2
θ=
π 5π
π
5π
,
, 2π + , 2π +
,...
6 6
6
6
In general
θ=
π
+ 2kπ,
6
θ=
5π
+ 2kπ,
6
k = 0, ±1, ±2, ±3, . . .
Because of these endless answers we are usually asked to find values of θ within a certain
range.
17
Summer School: Circular Functions
Solving for θ ∈ [0, 2π]
13.1
Given we want to solve, for θ ∈ [0, 2π],
trig f unction(θ) = c
we first solve
trig f unction(θ̂) = |c|
i.e. for positive c.
Here trig f unction is sin, cos or tan.
Y
nd
2
st
Quad
1 Quad
sin
all
π−θ
rd
3
π
). We
2
then refer to our ’Rules of Signs’ from
This gives an acute angle (θ̂ <
section 3.5. So all of our functions are
θ
π+θ
2π − θ
tan
cos
positive in two quadrants and negative in
X
two quadrants. This gives, at most, two
values for θ ∈ [0, 2π]. This process is
th
Quad
4 Quad
best seen in an example.
cos(θ)
Example 13.2 Find all the solutions for
x ∈ [0, 2π] of
1
2 cos2 x − 3 cos x + 1 = 0
0.5
Factorise: (2 cos x − 1)(cos x − 1) = 0
0
1st
1
cos x = ,
2
π
x̂ = ,
3
π
x= ,
3
cos x = 1
−0.5
x̂ = 0
1st
x=0
−1
0
4th
x = 2π−
5π
π
=
,
3
3
4th
x = 2π−0 = 2π
1.5708
3.1416
θ
4.7124
6.2832
18
Summer School: Circular Functions
Example 13.3 Find x ∈ [0, 2π] for which
cos(2x) − 3 cos x + 2 = 0
Put cos(2x) = 2 cos2 x − 1 to obtain
2 cos2 x − 3 cos x + 1 = 0
See previous example.
Example 13.4 Find x ∈ [0, 2π] for which
sin(θ)
1
sin x − 2 cos2 x + 2 = 0
0.5
Put cos2 x = 1 − sin2 x:
sin x − 2(1 − sin2 x) + 2 = 0
2 sin2 x + sin x = 0
1
sin x(sin x + ) = 0
2
sin x = 0,
x̂ = 0,
1st
x = 0 or 2π,
2nd
x = π−0 = π,
sin x = −
1
2
π
x̂ =
6
3rd
4th
π
7π
=
6
6
π
11π
x = 2π− =
6
6
x = π+
0
−0.5
−1
0
1.5708
3.1416
θ
4.7124
6.2832
4
x 10
19
Summer School: Circular Functions
Example 13.5 Solve, for x ∈ [0, 2π]
cos2 x − sin2 x − cos x + 1 = 0
cos(θ)
1
Put sin2 x = 1 − cos2 x to obtain
cos2 x − (1 − cos2 x) − cos x + 1 = 0
0.5
2 cos2 x − cos x = 0
0
cos x(2 cos x − 1) = 0
cos x = 0,
x̂ =
1st
4th
14
x=
π
,
2
cos x =
x̂ =
π
,
2
x = 2π−
π
3π
=
,
2
2
1
2
−0.5
π
3
−1
π
3
1st
x=
4th
x = 2π−
0
1.5708
3.1416
θ
4.7124
π
5π
=
3
3
Inverse Functions
We have seen in the previous section that there are an infinite number of solutions, x, to
equations such as
sin x = c
where x is an angle and c is a constant.
[ π π]
with sin(x) = c: This unique
If we restrict c to c ∈ [−1, 1] there is a unique x ∈ − ,
2 2
value is the inverse sine of c. We write
x = sin−1 c or x = arcsin c
Similarly, for each value of c ∈ [−1, 1] there is a unique x ∈ [0, π] with cos(x) = c: This x is
the inverse cosine function
x = cos−1 c or x = arccos c
( π π)
And again, for each c ∈ (−∞, ∞) there is a unique x ∈ − ,
with tan(x) = c: This x
2 2
is the inverse tangent function
x = tan−1 c or x = arctan c
6.2832
20
Summer School: Circular Functions
Note that sin−1 (sin α) = α and sin(sin−1 α) = α etc.
We can use these inverse functions to help find values of some circular functions.
( )
( )
4
5
−1
Example 14.1 If θ = sin
+ sin
find sin θ.
5
13
( )
( )
4
5
−1
−1
Let sin
= α, sin
=β ⇒θ =α+β
5
13
−1
sin θ = sin(α + β) = sin α cos β + cos α sin β
α
13
5
5
3
β
4
12
3
4
cos α = , sin α =
5
5
cos β =
12
,
13
sin β =
5
13
48 + 15
63
4 12 3 5
·
+ ·
=
=
5 13 5 13
65
65
( )
( )
12
24
Example 14.2 If θ = sin−1
+ tan−1
find cos θ.
13
7
sin θ =
(
Let sin
−1
12
13
)
(
−1
= α, tan
24
7
)
=β ⇒θ =α+β
cos θ = cos α cos β − sin α sin β
β
α
13
25
12
cos α =
5
,
13
7
5
24
sin α =
12
13
cos θ =
cos β =
7
,
25
sin β =
5 7
12 24
35 − 288
253
·
−
·
=
=−
13 25 13 25
325
325
24
25
21
Summer School: Circular Functions
15
Polar Coordinates
Y
Let P have Cartesian coordinates (x, y).
OP has length r ≥ 0.
P(x,y)
b = θ.
X OP
Then P has polar coordinates (r, θ).
We have
r
x = r cos θ,
x2 + y 2 = r 2 ,
y
y = r sin θ
r=
√
x2 + y 2
θ
y
tan θ =
x
X
x
O
θ is usually restricted to −π < θ ≤ π or
0 ≤ θ < 2π. (The later in our case).
We use a similar method to find θ as we did for solving trigonometric equations. That is,
firstly solve
y
tan θ̂ = x
Then we check which quadrant we are in and use the same relations as before to find θ.
Example 15.1 (x, y) = (1,
r=
√
x2 + y 2 =
tan(θ̂) =
√
√
√
3)
Y
(1, 3)
1+3=2
3 ⇒ θ̂ =
π
3
π
The point is in the 1st quad so θ = is
( π )3
fine and Polar coordinates are 2,
.
3
θ
X
22
Summer School: Circular Functions
√
Example 15.2 (x, y) = (−1, − 3)
r=
√
Y
1+3=2
θ
√
− 3 √
π
= 3 ⇒ θ̂ =
tan(θ̂) =
−1
3
X
The point is in the 3rd quad so
π
4π
=
3
3
(
)
4π
and Polar coordinates are 2,
.
3
θ=π+
(−1,− 3)
Example 15.3 (x, y) = (4, −4)
r=
√
x2 + y 2 =
√
16 + 16 =
√
√
32 = 4 2
Y
θ
4
π
tan(θ̂) = = 1 ⇒ θ̂ =
4
4
X
The point is in the 4th quad so
π
7π
=
4
4
(
)
√ 7π
and Polar coordinates are 4 2,
.
4
√
Example 15.4 (x, y) = (− 3, 3)
θ = 2π −
r=
√
3+9=
√
(4,−4)
√
12 = 2 3
Y
√
3
π
tan(θ̂) = √ = 3 ⇒ θ̂ =
3
3
(− 3, 3)
The point is in the 2nd quad so
2π
π
=
3
3
(
)
√ 2π
and Polar coordinates 2 3,
.
3
θ=π−
θ
THE END
X