1
Angles
In school you learned about angles and measured them in degrees. In this course
we will use a different unit for angles called radians (which we will write as rad).
The relationship between radians and degrees is given by the equation
180◦ = πrad.
So:
π
180
180
1 rad =
π
1◦ =
π
It follows that an angle θ in degrees corresponds to θ 180
rad in radians while
◦
180
an angle φ in radians corresponds to φ πrad in degrees. Note that when we write
an angle in radians we usually leave out the unit. The following table shows the
conversion of some common angles
Degrees
Radians
2
30◦
45◦
60◦
90◦
π
6
π
4
π
3
π
2
180◦
π
270◦
3π
2
360◦
2π
Trigonometric functions
In school you were taught the definition of (at least some of the) trig functions.
Given a triangle
Hypotenuse
Opposite
θ
Adjacent
the six trig functions were defined by
sin(θ) =
cos(θ) =
tan(θ) =
sec(θ) =
csc(θ) =
cot(θ) =
Opposite
Hypotenuse
Adjacent
Hypotenuse
Opposite
Adjacent
Hypotenuse
Adjacent
Hypotenuse
Opposite
Adjacent
Opposite
1
where it is supposed that Adjacent, Opposite and Hypotenuse represent the
lengths of those sides of the triangle. We would like to extend these definitions
to any angle θ not only acute angles. We note that each positive real number r
and each angle θ determine a point as shown in the diagram
x
(x, y)
r
θ
y
We now define the six trigonometric functions as follows
y
r
y
tan(θ) =
x
r
csc(θ) =
y
x
r
r
sec(θ) =
x
x
cot(θ) = .
y
sin(θ) =
3
cos(θ) =
Trigonometric identities
These identities follow directly from the definition of the trig functions
3.1
Basic identities
1. csc(θ) =
1
;
sin(θ)
2. sec(θ) =
1
;
cos(θ)
3. cot(θ) =
1
;
tan(θ)
4. tan(θ) =
sin(θ)
;
cos(θ)
5. cot(θ) =
cos(θ)
.
sin(θ)
2
Using the definition of sin(θ) and cos(θ) we find that
y2
x2
+
r2
r2
2
x + y2
=
r2
2
r
= 2
r
= 1,
sin2 (θ) + cos2 (θ) =
from which we obtain the following identities:
3.2
Pythagorean identities
1. sin2 (θ) + cos2 (θ) = 1;
2. tan2 (θ) + 1 = sec2 (θ);
3. 1 + cot2 (θ) = csc2 (θ).
Exercise 3.2.1. Show that:
(a) Identity 3.2.2 follows from 3.2.1 and some of the basic identities;
(b) Identity 3.2.3 follows from 3.2.1 and some of the basic identities.
Exercise 3.2.2. Show, using the definitions of the various trig functions that:
(a) 3.2.2 holds;
(b) 3.2.3 holds.
Recall that given an angle θ and a radius r we are able to obtain a point
P (x, y) as depicted below
P (x, y)
O
θ
−θ
Q(x, −y)
3
It is easy to see that the point Q obtained from −θ and r will be the point
obtained by reflecting P (x, y) in the x-axis and hence will have coordinates
Q(x, −y). Using the definitions of sin(θ) and cos(θ) we obtain:
3.3
Even and odd identities
1. sin(−θ) = − sin(θ);
2. cos(−θ) = cos(θ).
We have:
3.4
Periodic identities
1. sin(θ + 2π) = sin(θ);
2. cos(θ + 2π) = cos(θ).
3.5
Addition and subtraction formulas
1. sin(θ + φ) = sin(θ) cos(φ) + sin(φ) cos(θ);
2. sin(θ − φ) = sin(θ) cos(φ) − sin(φ) cos(θ);
3. cos(θ + φ) = cos(θ) cos(φ) − sin(φ) sin(θ);
4. cos(θ − φ) = cos(θ) cos(φ) + sin(φ) sin(θ);
5. tan(θ + φ) =
tan(θ) + tan(φ)
;
1 − tan(θ) tan(φ)
6. tan(θ − φ) =
tan(θ) − tan(φ)
;
1 + tan(θ) tan(φ)
Exercise 3.5.1.
(a) Prove the rule of cosines. Show that in a triangle
B
a
c
θ
C
A
b
4
c2 = a2 + b2 − 2ab cos(θ)
[Hint: In order to do so construct a line segment perpendicular to CA from
B to CA. Then use pythagorus theorem on each right angled sub-triangle.]
(b) Prove the subtraction formula
cos(α − β) = cos(α) cos(β) − sin(α) sin(β)
using the figure
A(cos(α), sin(α))
c
B(cos(β), sin(β))
α β
O
Using the rule of cosines we have that
c2 = a2 + b2 − 2ab cos(α − β)
= 1 + 1 − 2 cos(α − β)
= 2 − 2 cos(α − β).
While using the distance formula we find that
c2 = (sin(α) − sin(β))2 + (cos(α) − cos(β))2
= (sin2 (α) − 2 sin(α) sin(β) + sin2 (β)) + (cos2 (α) − 2 cos(α) cos(β) + cos2 (β))
= ((sin2 (α) + cos2 (α)) + (sin2 (β) + cos2 (β))) − 2 sin(α) sin(β) − 2 cos(α) cos(β)
= (1 + 1) − 2(sin(α) sin(β) + cos(α) cos(β))
= 2 − 2(sin(α) sin(β) + cos(α) cos(β)).
Equating these two equations for c2 we obtain the above formula.
(c) Show how Formula 5 can be obtained from Formulas 1 and 3
5
We have
sin(θ + φ)
cos(θ + φ)
sin(θ) cos(φ) + sin(φ) cos(θ)
=
cos(θ) cos(φ) − sin(θ) sin(φ)
tan(θ + φ) =
=
sin(θ) cos(φ)+sin(φ) cos(θ)
cos(θ) cos(φ)
cos(θ) cos(φ)−sin(θ) sin(φ)
cos(θ) cos(φ)
=
sin(θ) cos(φ)
cos(θ) cos(φ)
cos(θ) cos(φ)
cos(θ) cos(φ)
=
sin(θ)
sin(φ)
cos(θ) + cos(φ)
sin(θ) sin(φ)
1 − cos(θ)
cos(φ)
=
tan(θ) + tan(φ)
.
1 − tan(θ) tan(φ)
+
−
sin(φ) cos(θ)
cos(θ) cos(φ)
sin(θ) sin(φ)
cos(θ) cos(φ)
(d) Show how Formula 6 can be obtained from Formulas 2 and 4;
3.6
Double-angle formulas
1. sin(2θ) = 2 sin(θ) cos(θ);
2. cos(2θ) = cos2 (θ) − sin2 (θ);
3. cos(2θ) = 2 cos2 (θ) − 1;
4. cos(2θ) = 1 − 2 sin2 (θ).
Exercise 3.6.1. Find all values of x in the interval [0, 2π] such that
(a) sin(x) = sin(2x)
We have:
sin(x) = sin(2x)
⇔ sin(x) = 2 sin(x) cos(x)
⇔ sin(x) − 2 sin(x) cos(x) = 0
⇔ sin(x)(1 − 2 cos(x)) = 0
⇔ sin(x) = 0 or 1 − 2 cos(x) = 0
1
⇔ sin(x) = 0 or cos(x) =
2
n
π
πo
⇔x ∈ 0, π, 2π, , 2π −
3
3
6
(b) sin(x) = cos(2x)
We have:
sin(x) = cos(2x)
⇔ sin(x) = 1 − 2 sin2 (x)
⇔2 sin2 (x) + sin(x) − 1 = 0
⇔(2 sin(x) − 1)(sin(x) + 1) = 0
1
⇔ sin(x) = or sin(x) = −1
2
nπ
3π
πo
or x ∈
,π −
⇔x ∈
6
6
2
π
π 3π
⇔x ∈
,π − ,
6
6 2
(c) sin(2x) = cos(2x).
Using the two last identities (from the previous list) and solving from cos2 (θ)
and sin2 (θ) in each we obtain
3.7
Half-angle formulas
1. cos2 (θ) =
1 + cos(2θ)
;
2
2. sin2 (θ) =
1 − cos(2θ)
.
2
3.8
Product formulas
1. sin(θ) cos(φ) =
1
2
[sin(θ + φ) + sin(θ − φ)];
2. cos(θ) cos(φ) =
1
2
[cos(θ + φ) + cos(θ − φ)];
3. sin(θ) sin(φ) =
1
2
[cos(θ − φ) − cos(θ + φ)].
7
4
Graphs of trigonometric functions
The graph of y = sin(x):
y
1
0.5
−2π − 3π −π
2
π
2
π
2
π
3π
2
2π
π
2
π
3π
2
2π
x
−0.5
−1
sin(x)
The graph of y = cos(x):
y
1
0.5
−2π − 3π −π
2
π
2
−0.5
−1
cos(x)
The graph of y = tan(x):
8
x
y
10
5
π
2
−2π − 3π −π
2
π
π
2
3π
2
2π
−5
−10
tan(x)
The graph of y = cot(x):
y
10
5
−2π − 3π −π
2
π
2
π
2
π
3π
2
−5
−10
cot(x)
The graph of y = sec(x):
9
2π
x
x
y
10
5
π
2
−2π − 3π −π
2
π
π
2
3π
2
2π
−5
−10
sec(x)
The graph of y = csc(x):
y
10
5
−2π − 3π −π
2
π
2
π
2
π
3π
2
−5
−10
csc(x)
10
2π
x
x