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MATH 201 - Week 9
Ferenc Balogh
Concordia University
2008 Winter
Based on the textbook
J. Stuart, L. Redlin, S. Watson, Precalculus - Mathematics for Calculus, 5th Edition, Thomson
All figures and videos are made using MAPLE 11 and ImageMagick-convert.
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The Unit Circle (Section 5.1)
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Trigonometric Functions of Real Numbers (Section 5.2)
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The Unit Circle
Terminal Points on the Unit Circle
The Reference Number
The Trigonometric Functions
Values of Trigonometric Functions
Fundamental Identities
Trigonometric Graphs (Section 5.3)
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Graphs of the Sine and Cosine Functions
Graphs of Transformations of Sine and Cosine
The unit circle is the circle of radius 1 centered at the origin.
The equation of the unit circle is
x 2 + y 2 = 1.
Example. The points
√
P(1, 0),
Q
3 1
,
2 2
!
are on the unit circle because
12 + 02
√ !2 2
3
1
+
2
2
2 2
1
1
√
+ −√
2
2
=
1
=
1
=
1
,
R
1
1
√ , −√
2
2
Example. Find the x-coordinate of the point P on the unit circle
located in the second quadrant whose y -coordinate is equal to 12 .
Solution. The point P(x0 , y0 ) is on the unit circle therefore its
coordinates satisfy the equation x 2 + y 2 = 1.
Since y0 = 12 , x = x0 solves the equation
2
1
x +
= 1.
2
2
2
1
1
3
x =1−
=1− = .
2
4
4
2
Since P is in the second quadrant, we
have to take the negative solution, so
r
√
3
3
x0 = −
=−
.
4
2
Let t be an arbitrary real number.
The terminal point corresponding to the number t on the unit
circle is given by the following procedure:
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If t = 0 then the corresponding terminal point is (1, 0).
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If t > 0 then we take the endpoint of the (possibly
self-overlapping) arc starting at (1, 0) of length t along the
circumference of the unit circle in positive (counterclockwise)
direction.
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If t < 0 then we take the endpoint of the (possibly
self-overlapping) arc starting at (1, 0) of length |t| along the
circumference of the unit circle in negative (clockwise)
direction.
Remark. The notion of terminal point is similar to the notion of
terminal direction for angles.
Example. Find the terminal points for
t = π,
t=
π
,
2
t = −3π.
Solution. First of all, the unit circle has a perimeter of 2π.
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The terminal point for t = π is (−1, 0) since π is exactly the
half of the perimeter.
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The terminal point for t =
the total length.
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The terminal point for t = −3π is (−1, 0) since we have to
measure a distance of one and a half of the total perimeter in
the negative direction on the circumference.
π
2
is (0, 1) since
π
2
is one-fourth of
Here is a list of some special terminal points:
t terminal point
0
0) (1,
√
3 1
π
,
6
2 2 π
√1 , √1
4
2 √ 2
3
π
1
3
2, 2
π
2
π
2π
(0, 1)
(−1, 0)
(1, 0)
Remark. Notice that different values of t can give the same
terminal point on the unit circle (one could call these coterminal
numbers).
The reference number t̄ associated with a real number t is the
shortest distance along the circumference of the unit circle between
the terminal point determined by t and the x-axis.
This is the analogue of the reference angle.
Examples.
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The reference number of π is 0 since the terminal point
(−1, 0) is on the x-axis.
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The reference number of π2 is π2 since the distance of its
terminal point (0, 1) from the x-axis is π2 along the unit circle.
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π
The reference number of 2π
3 is 3 since the distance of its
terminal point from the x-axis is 2π
3 along the unit circle.
Example. Find the reference numbers for
3
t = − π,
2
t=
7π
,
8
t=−
5π
.
6
Solution.
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For t = − 32 π the terminal point is (0, 1). The corresponding
reference number is t̄ = π2 .
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If t = 7π
8 the terminal point is in the second quadrant and it
is closer to the negative half of the x-axis. The reference
π
number is π − 7π
8 = 8.
√
3
1
For t = − 5π
,
the
terminal
point
is
given
by
−
,
−
6
2
2 .
π
Therefore the reference number is 6 .
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How to find terminal points using reference numbers?
Suppose that t is a given real number.
1. Find the reference number corresponding to t̄.
2. Find the terminal point determined by the reference number t̄
of the form (a, b).
3. The terminal point determined by t is of the form (±a, ±b)
where the signs are fixed by the quadrant of the terminal
point of t.
Example. Find the terminal points using reference numbers for
t=
2π
,
3
3
t = π,
4
t=
7π
,
6
π
t=− .
4
Solution.
π
For t = 2π
3 , the reference number
√ is t̄ = 3 .
The terminal point of t̄ is 12 , 23 .
The terminal point of t lies in the second quadrant therefore the
signs of the x-coordinate and the y -coordinate are negative and
positive respectively.
√ 3
1
So the terminal point of t = 2π
is
−
,
3
2 2 .
the reference number
is t̄ = π4 .
The terminal point of t̄ is √12 , √12 .
The terminal point of t lies in the second quadrant therefore the
signs of the x-coordinate and the y -coordinate are negative and
positive respectively.
3π
1 √1
√
So the terminal point of t = 4 is − 2 , 2 .
For t =
3π
4 ,
Solution (Continued).
π
For t = 7π
6 , the reference number
√
is t̄ = 6 .
The terminal point of t̄ is 23 , 12 .
The terminal point of t lies in the third
(−, −).
√ quadrant:
3
1
7π
So the terminal point of t = 6 is − 2 , − 2 .
For t = − π4 , the reference number is t̄ = π4 .
The terminal point of t̄ is √12 , √12 .
The terminal point of t lies in the fourth
quadrant:
(+, −).
π
1
1
So the terminal point of t = − 4 is √2 , − √2 .
Definition of Trigonometric Functions of Real Numbers
Let t be a real number and let P(x, y ) denote the terminal point
of t. The trigonometric functions of t are defined to be
sin t = y
csc t =
1
y
cos t = x
tan t =
y
x
1
x
cot t =
x
y
sec t =
Remark. Note the similarities and differences between the
expressions defining the trigonometric functions for angles and for
real numbers.
Example. Find the trigonometric functions of t = 5π
6 .
5π
Solution. The reference number of t is t̄ = π − 6 = π6 therefore
√
the terminal point is (−
3 1
2 , 2 ).
5π
6
5π
cos
6
5π
tan
6
5π
csc
6
5π
sec
6
5π
cot
6
sin
√
So x = −
= y=
1
2√
3
2
y
1
= −√
x
3
1
=2
y
1
2
= −√
x
3
√
x
=− 3
y
= x =−
=
=
=
=
3
2
and y = 21 .
The domains of Trigonometric Functions
Function
sin
cos
tan
csc
sec
cot
Domain
R
R R \ π2 + kπ | k is an integer
R \ {kπ
| k is an integer} R \ π2 + kπ | k is an integer
R \ {kπ | k is an integer}
Range
[−1, 1]
[−1, 1]
R
(−∞, −1] ∪ [1, ∞)
(−∞, −1] ∪ [1, ∞)
R
Special values and signs of the trigonometric functions
t
0
π
6
π
4
π
3
π
2
sin t cos t tan t csc t sec t cot t
0
1
0
−
1
−
√
√
3
1
1
2
√
√
2
3
2
2
3
√
√3
1
1
√
√
1
2
2
1
2
√2
√
3
1
2
√
√1
3
2
2
2
3
3
1
0
−
1
−
0
Quad. sin t cos t tan t csc t sec t cot t
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+
+
+
+
+
+
I
+
−
−
+
−
−
III
−
−
+
−
−
+
IV
−
+
−
−
+
−
Example. Use the known special values to find the exact values of
π
4π
20π
sin , cot −
.
, csc
3
6
3
Solution.
I t̄ = 4π − π = π . The terminal point of 4π is in the third
3
3
3
quadrant so
√
π
3
4π
= − sin = −
.
sin
3
3
2
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t̄ = π6 . The terminal point of − π6 is in the fourth quadrant so
π
√
π
= − cot = − 3.
cot −
6
6
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t̄ = π3 . The terminal point of
csc
20π
3
is in the second quadrant so
20π
π
2
= csc = √ .
3
3
3
Fundamental Identities
Reciprocial Identities
csc t =
1
sin t
sec t =
tan t =
sin t
cos t
1
cos t
cot t =
cot t =
1
tan t
cos t
sin t
Pythagorean Identities
sin2 t + cos2 t = 1
tan2 t + 1 = sec2 t
1 + cot2 t = csc2 t
Example. Use the calculator to find the approximate values of
sin 22.4,
cot (−1.3) ,
csc 31.
Solution. First of all, make sure that the calculator is in rad
mode.
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sin 22.4 ≈ −0.3976.
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cot (−1.3) =
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csc 31 =
1
≈ −0.2776
tan (−1.3)
1
≈ −2.475
sin 31
Reminder.
A function f is said to be even if
f (−x) = f (x)
for all x in its domain. A function f is said to be odd if
f (−x) = −f (x)
for all x in its domain.
Even-Odd Properties
sin(−t) = − sin t
cos(−t) = cos t
tan(−t) = − tan t
csc(−t) = − csc t
sec(−t) = sec t
cot(−t) = − cot t
Example. Express sin t in terms of cos t where t is chosen from
the fourth quadrant.
Solution. In the fourth quadrant, the sine function is negative.
Using the basic Pythagorean identity we get
sin2 t + cos2 t = 1
sin2 t = 1 − cos2 t
p
sin t = ± 1 − cos2 t
Since sin t is negative in the fourth quadrant and the square root is
non-negative we have
p
sin t = − 1 − cos2 t.
Remark. If t is in the third quadrant, for example, then the + sign
is valid.
Example. If sec t = 3 and t is from the fourth quadrant, find the
values of the other trigonometric functions.
Solution. It is convenient to find the sine and cosine first:
1
1
cos t =
= .
sec t
3
Using the expression from the previous page we have
r
r
√
p
8
8
1
=−
.
sin t = − 1 − cos2 t = − 1 − = −
9
9
3
From these, it is easy to generate the other trigonometric
functions:
√
sin t
tan t =
=− 8
cos t
1
3
csc t =
= −√
sin t
8
cos t
1
cot t =
= −√
sin t
8
A function f is said to be periodic if there is a number p such that
f (t + p) = f (t)
for all t. The least such positive p (if this exists) is called the
period of f .
Periodicity of sin and cos
The sine and cosine functions are periodic and their period is 2π:
sin(t + 2π) = sin t
cos(t + 2π) = cos t.
This fact is very important: it is enough to find the shape of their
graphs in the interval between 0 and 2π and then ’make copies’.
We can use the special values of the trigonometric values to sketch
the graphs of the sine and cosine functions:
The graph of the sine function
The graph of the cosine function
Vertical and Horizontal Strechings and Shrinkings
The graph of a transformed sine function
y = A sin(kx)
is obtained from the original graph y = sin x using
1. horizontal stretching/shrinking depending on the parameter k
2. vertical stretching/shrinking depending on the parameter A
The period of A sin(kx) is 2π
k :
2π
A sin k x +
= A sin (kx + 2π) = A sin (kx) .
k
The magnitude of A sin(kx) is given by the number |A|: this is
called the amplitude.
Additional Horizontal Shifts
The graph of a transformed sine function
y = A sin(k(x − b))
is obtained from the original graph y = sin x using
1. horizontal stretching/shrinking depending on the parameter k
2. horizontal shifting given by b
3. vertical stretching/shrinking depending on the parameter A
The number b is called the phase shift.
Example. Consider the function
f (x) = −3 sin(2(x − 5)).
What is the amplitude, period and phase shift of f (x)?
Sketch the graph of f .
Solution. The amplitude is 3, the period is 2π
2 = π and the phase
shift is 5.
The graph is obtained from the graph of
y = sin x using the following sequence
of elementary transformations:
1. horizontal shrinking of factor
1
2
2. horizontal shifting to the right of 5
units
3. vertical stretching of factor 3
4. reflection to the x-axis.