MATH 201 - Week 9 - math.u-szeged.hubaloghf/homepage/teaching/math201/math201_week... · MATH 201 - Week 9 Overview The Unit Circle (Section 5.1) The Unit Circle Terminal Points on

MATH 201 - Week 9

MATH 201 - Week 9

Ferenc Balogh

Concordia University

2008 Winter

Based on the textbookJ. Stuart, L. Redlin, S. Watson, Precalculus - Mathematics for Calculus, 5th Edition, Thomson

All figures and videos are made using MAPLE 11 and ImageMagick-convert.

MATH 201 - Week 9

Overview

The Unit Circle (Section 5.1)

The Unit CircleTerminal Points on the Unit CircleThe Reference Number

Trigonometric Functions of Real Numbers (Section 5.2)

The Trigonometric FunctionsValues of Trigonometric FunctionsFundamental Identities

Trigonometric Graphs (Section 5.3)

Graphs of the Sine and Cosine FunctionsGraphs of Transformations of Sine and Cosine

MATH 201 - Week 9

Overview







MATH 201 - Week 9

Overview







MATH 201 - Week 9

The Unit Circle

The unit circle is the circle of radius 1 centered at the origin.The equation of the unit circle is

x2 + y2 = 1.

Example. The points

P(1, 0), Q

(√3

2,

1

2

), R

(1√2,− 1√

2

)are on the unit circle because

12 + 02 = 1(√3

2

)2

+

(1

2

)2

= 1

(1√2

)2

+

(− 1√

2

)2

= 1

MATH 201 - Week 9

The Unit Circle

The unit circle is the circle of radius 1 centered at the origin.The equation of the unit circle is

x2 + y2 = 1.

Example. The points

P(1, 0), Q

(√3

2,

1

2

), R

(1√2,− 1√

2

)are on the unit circle because

12 + 02 = 1(√3

2

)2

+

(1

2

)2

= 1

(1√2

)2

+

(− 1√

2

)2

= 1

MATH 201 - Week 9

The Unit Circle

Example. Find the x-coordinate of the point P on the unit circlelocated in the second quadrant whose y -coordinate is equal to 1

2 .

Solution. The point P(x0, y0) is on the unit circle therefore itscoordinates satisfy the equation x2 + y2 = 1.

Since y0 = 12 , x = x0 solves the equation

x2 +

(1

2

)2

= 1.

x2 = 1−(

1

2

)2

= 1− 1

4=

3

4.

Since P is in the second quadrant, wehave to take the negative solution, so

x0 = −√

3

4= −√

3

2.

MATH 201 - Week 9

The Unit Circle

Example. Find the x-coordinate of the point P on the unit circlelocated in the second quadrant whose y -coordinate is equal to 1

2 .Solution. The point P(x0, y0) is on the unit circle therefore itscoordinates satisfy the equation x2 + y2 = 1.

Since y0 = 12 , x = x0 solves the equation

x2 +

(1

2

)2

= 1.

x2 = 1−(

1

2

)2

= 1− 1

4=

3

4.

Since P is in the second quadrant, wehave to take the negative solution, so

x0 = −√

3

4= −√

3

2.

MATH 201 - Week 9

The Unit Circle

Terminal Points on the Unit Circle

Let t be an arbitrary real number.

The terminal point corresponding to the number t on the unitcircle is given by the following procedure:

If t = 0 then the corresponding terminal point is (1, 0).

If t > 0 then we take the endpoint of the (possiblyself-overlapping) arc starting at (1, 0) of length t along thecircumference of the unit circle in positive (counterclockwise)direction.

If t < 0 then we take the endpoint of the (possiblyself-overlapping) arc starting at (1, 0) of length |t| along thecircumference of the unit circle in negative (clockwise)direction.

Remark. The notion of terminal point is similar to the notion ofterminal direction for angles.

MATH 201 - Week 9

The Unit Circle








MATH 201 - Week 9

The Unit Circle








MATH 201 - Week 9

The Unit Circle








MATH 201 - Week 9

The Unit Circle








MATH 201 - Week 9

The Unit Circle


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.

The terminal point for t = π is (−1, 0) since π is exactly thehalf of the perimeter.

The terminal point for t = π2 is (0, 1) since π

2 is one-fourth ofthe total length.

The terminal point for t = −3π is (−1, 0) since we have tomeasure a distance of one and a half of the total perimeter inthe negative direction on the circumference.

MATH 201 - Week 9

The Unit Circle



t = π, t =π

2, t = −3π.






MATH 201 - Week 9

The Unit Circle



t = π, t =π

2, t = −3π.






MATH 201 - Week 9

The Unit Circle



t = π, t =π

2, t = −3π.






MATH 201 - Week 9

The Unit Circle



t = π, t =π

2, t = −3π.






MATH 201 - Week 9

The Unit Circle


Here is a list of some special terminal points:

t terminal point0 (1, 0)π6

(√3

2 ,12

)π4

(1√2, 1√

2

)π3

(12 ,√

32

)π2 (0, 1)

π (−1, 0)

2π (1, 0)

Remark. Notice that different values of t can give the sameterminal point on the unit circle (one could call these coterminalnumbers).

MATH 201 - Week 9

The Unit Circle


Here is a list of some special terminal points:

t terminal point0 (1, 0)π6

(√3

2 ,12

)π4

(1√2, 1√

2

)π3

(12 ,√

32

)π2 (0, 1)

π (−1, 0)

2π (1, 0)

Remark. Notice that different values of t can give the sameterminal point on the unit circle (one could call these coterminalnumbers).

MATH 201 - Week 9

The Unit Circle

The Reference Number

The reference number t̄ associated with a real number t is theshortest distance along the circumference of the unit circle betweenthe terminal point determined by t and the x-axis.

This is the analogue of the reference angle.

Examples.

The reference number of π is 0 since the terminal point(−1, 0) is on the x-axis.

The reference number of π2 is π

2 since the distance of itsterminal point (0, 1) from the x-axis is π

2 along the unit circle.

The reference number of 2π3 is π

3 since the distance of itsterminal point from the x-axis is 2π


MATH 201 - Week 9

The Unit Circle


The reference number t̄ associated with a real number t is theshortest distance along the circumference of the unit circle betweenthe terminal point determined by t and the x-axis.

This is the analogue of the reference angle.

Examples.

The reference number of π is 0 since the terminal point(−1, 0) is on the x-axis.

The reference number of π2 is π

2 since the distance of itsterminal point (0, 1) from the x-axis is π


The reference number of 2π3 is π

3 since the distance of itsterminal point from the x-axis is 2π


MATH 201 - Week 9

The Unit Circle


Example. Find the reference numbers for

t = −3

2π, t =

7π

8, t = −5π

6.

Solution.

For t = −32π the terminal point is (0, 1). The corresponding

reference number is t̄ = π2 .

If t = 7π8 the terminal point is in the second quadrant and it

is closer to the negative half of the x-axis. The referencenumber is π − 7π

8 = π8 .

For t = −5π6 , the terminal point is given by

(−√

32 ,−

12

).

Therefore the reference number is π6 .

MATH 201 - Week 9

The Unit Circle



t = −3

2π, t =

7π

8, t = −5π

6.

Solution.





8 = π8 .


(−√

32 ,−

12

).


MATH 201 - Week 9

The Unit Circle



t = −3

2π, t =

7π

8, t = −5π

6.

Solution.





8 = π8 .


(−√

32 ,−

12

).


MATH 201 - Week 9

The Unit Circle



t = −3

2π, t =

7π

8, t = −5π

6.

Solution.





8 = π8 .


(−√

32 ,−

12

).


MATH 201 - Week 9

The Unit Circle


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 t̄of 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 terminalpoint of t.

MATH 201 - Week 9

The Unit Circle


Example. Find the terminal points using reference numbers for

t =2π

3, t =

3

4π, t =

7π

6, t = −π

4.

Solution.For t = 2π

3 , the reference number is t̄ = π3 .

The terminal point of t̄ is(

12 ,√

32

).

The terminal point of t lies in the second quadrant therefore thesigns of the x-coordinate and the y -coordinate are negative andpositive respectively.

So the terminal point of t = 2π3 is

(−1

2 ,√

32

).

For t = 3π4 , the reference number is t̄ = π

4 .


1√2, 1√

2

).



(− 1√

2, 1√

2

).

MATH 201 - Week 9

The Unit Circle



t =2π

3, t =

3

4π, t =

7π

6, t = −π

4.




12 ,√

32

).



(−1

2 ,√

32

).


4 .


1√2, 1√

2

).



(− 1√

2, 1√

2

).

MATH 201 - Week 9

The Unit Circle



t =2π

3, t =

3

4π, t =

7π

6, t = −π

4.




12 ,√

32

).



(−1

2 ,√

32

).


4 .


1√2, 1√

2

).



(− 1√

2, 1√

2

).

MATH 201 - Week 9

The Unit Circle



t =2π

3, t =

3

4π, t =

7π

6, t = −π

4.




12 ,√

32

).



(−1

2 ,√

32

).


4 .


1√2, 1√

2

).



(− 1√

2, 1√

2

).

MATH 201 - Week 9

The Unit Circle


Solution (Continued).For t = 7π


The terminal point of t̄ is(√

32 ,

12

).

The terminal point of t lies in the third quadrant: (−,−).


(−√

32 ,−

12

).

For t = −π4 , the reference number is t̄ = π

4 .


1√2, 1√

2

).

The terminal point of t lies in the fourth quadrant: (+,−).

So the terminal point of t = −π4 is

(1√2,− 1√

2

).

MATH 201 - Week 9

The Unit Circle





32 ,

12

).



(−√

32 ,−

12

).


4 .


1√2, 1√

2

).



(1√2,− 1√

2

).

MATH 201 - Week 9

The Unit Circle





32 ,

12

).



(−√

32 ,−

12

).


4 .


1√2, 1√

2

).



(1√2,− 1√

2

).

MATH 201 - Week 9

Trigonometric Functions of Real Numbers

Definition of Trigonometric Functions of Real Numbers

Let t be a real number and let P(x , y) denote the terminal pointof t. The trigonometric functions of t are defined to be

sin t = y cos t = x tan t =y

x

csc t =1

ysec t =

1

xcot t =

x

y

Remark. Note the similarities and differences between theexpressions defining the trigonometric functions for angles and forreal numbers.

MATH 201 - Week 9


Example. Find the trigonometric functions of t = 5π6 .

Solution. The reference number of t is t̄ = π − 5π6 = π

6 therefore

the terminal point is (−√

32 ,

12 ). So x = −

√3

2 and y = 12 .

sin5π

6= y =

1

2

cos5π

6= x = −

√3

2

tan5π

6=

y

x= − 1√

3

csc5π

6=

1

y= 2

sec5π

6=

1

x= − 2√

3

cot5π

6=

x

y= −√

3

MATH 201 - Week 9


Example. Find the trigonometric functions of t = 5π6 .

Solution. The reference number of t is t̄ = π − 5π6 = π

6 therefore

the terminal point is (−√

32 ,

12 ). So x = −

√3

2 and y = 12 .

sin5π

6= y =

1

2

cos5π

6= x = −

√3

2

tan5π

6=

y

x= − 1√

3

csc5π

6=

1

y= 2

sec5π

6=

1

x= − 2√

3

cot5π

6=

x

y= −√

3

MATH 201 - Week 9


The domains of Trigonometric Functions

Function Domain Range

sin R [−1, 1]cos R [−1, 1]tan R \

{π2 + kπ | k is an integer

}R

csc R \ {kπ | k is an integer} (−∞,−1] ∪ [1,∞)sec R \

{π2 + kπ | k is an integer

}(−∞,−1] ∪ [1,∞)

cot R \ {kπ | k is an integer} R

MATH 201 - Week 9


Values of Trigonometric Functions

Special values and signs of the trigonometric functions

t sin t cos t tan t csc t sec t cot t

0 0 1 0 − 1 −π6

12

√3

21√3

2 2√3

√3

π4

1√2

1√2

1√

2√

2 1

π3

√3

212

√3 2√

32 1√

3π2 1 0 − 1 − 0

Quad . sin t cos t tan t csc t sec t cot t

I + + + + + +

I + − − + − −III − − + − − +

IV − + − − + −

MATH 201 - Week 9



Special values and signs of the trigonometric functions

t sin t cos t tan t csc t sec t cot t

0 0 1 0 − 1 −π6

12

√3

21√3

2 2√3

√3

π4

1√2

1√2

1√

2√

2 1

π3

√3

212

√3 2√

32 1√

3π2 1 0 − 1 − 0

Quad . sin t cos t tan t csc t sec t cot t

I + + + + + +

I + − − + − −III − − + − − +

IV − + − − + −

MATH 201 - Week 9



Example. Use the known special values to find the exact values of

sin4π

3, cot

(−π

6

), csc

20π

3.

Solution.

t̄ = 4π3 − π = π

3 . The terminal point of 4π3 is in the third

quadrant so

sin4π

3= − sin

π

3= −√

3

2.

t̄ = π6 . The terminal point of −π

6 is in the fourth quadrant so

cot(−π

6

)= − cot

π

6= −√

3.

t̄ = π3 . The terminal point of 20π

3 is in the second quadrant so

csc20π

3= csc

π

3=

2√3.

MATH 201 - Week 9




sin4π

3, cot

(−π

6

), csc

20π

3.

Solution.

t̄ = 4π3 − π = π


quadrant so

sin4π

3= − sin

π

3= −√

3

2.



cot(−π

6

)= − cot

π

6= −√

3.



csc20π

3= csc

π

3=

2√3.

MATH 201 - Week 9




sin4π

3, cot

(−π

6

), csc

20π

3.

Solution.

t̄ = 4π3 − π = π


quadrant so

sin4π

3= − sin

π

3= −√

3

2.



cot(−π

6

)= − cot

π

6= −√

3.



csc20π

3= csc

π

3=

2√3.

MATH 201 - Week 9




sin4π

3, cot

(−π

6

), csc

20π

3.

Solution.

t̄ = 4π3 − π = π


quadrant so

sin4π

3= − sin

π

3= −√

3

2.



cot(−π

6

)= − cot

π

6= −√

3.



csc20π

3= csc

π

3=

2√3.

MATH 201 - Week 9



Fundamental Identities

Reciprocial Identities

csc t =1

sin tsec t =

1

cos tcot t =

1

tan t

tan t =sin t

cos tcot t =

cos t

sin t

Pythagorean Identities

sin2 t + cos2 t = 1 tan2 t + 1 = sec2 t 1 + cot2 t = csc2 t

MATH 201 - Week 9



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 radmode.

sin 22.4 ≈ −0.3976.

cot (−1.3) =1

tan (−1.3)≈ −0.2776

csc 31 =1

sin 31≈ −2.475

MATH 201 - Week 9




sin 22.4, cot (−1.3) , csc 31.


sin 22.4 ≈ −0.3976.

cot (−1.3) =1

tan (−1.3)≈ −0.2776

csc 31 =1

sin 31≈ −2.475

MATH 201 - Week 9




sin 22.4, cot (−1.3) , csc 31.


sin 22.4 ≈ −0.3976.

cot (−1.3) =1

tan (−1.3)≈ −0.2776

csc 31 =1

sin 31≈ −2.475

MATH 201 - Week 9




sin 22.4, cot (−1.3) , csc 31.


sin 22.4 ≈ −0.3976.

cot (−1.3) =1

tan (−1.3)≈ −0.2776

csc 31 =1

sin 31≈ −2.475

MATH 201 - Week 9



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

MATH 201 - Week 9



Example. Express sin t in terms of cos t where t is chosen fromthe 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

sin t = ±√

1− cos2 t

Since sin t is negative in the fourth quadrant and the square root isnon-negative we have

sin t = −√

1− cos2 t.

Remark. If t is in the third quadrant, for example, then the + signis valid.

MATH 201 - Week 9



Example. Express sin t in terms of cos t where t is chosen fromthe fourth quadrant.Solution. In the fourth quadrant, the sine function is negative.Using the basic Pythagorean identity we get

sin2 t + cos2 t = 1


sin t = ±√

1− cos2 t


sin t = −√

1− cos2 t.


MATH 201 - Week 9



Example. Express sin t in terms of cos t where t is chosen fromthe fourth quadrant.Solution. In the fourth quadrant, the sine function is negative.Using the basic Pythagorean identity we get

sin2 t + cos2 t = 1


sin t = ±√

1− cos2 t


sin t = −√

1− cos2 t.


MATH 201 - Week 9



Example. If sec t = 3 and t is from the fourth quadrant, find thevalues of the other trigonometric functions.

Solution. It is convenient to find the sine and cosine first:

cos t =1

sec t=

1

3.

Using the expression from the previous page we have

sin t = −√

1− cos2 t = −√

1− 1

9= −

√8

9= −√

8

3.

From these, it is easy to generate the other trigonometricfunctions:

tan t =sin t

cos t= −√

8

csc t =1

sin t= − 3√

8

cot t =cos t

sin t= − 1√

8

MATH 201 - Week 9



Example. If sec t = 3 and t is from the fourth quadrant, find thevalues of the other trigonometric functions.Solution. It is convenient to find the sine and cosine first:

cos t =1

sec t=

1

3.

Using the expression from the previous page we have

sin t = −√

1− cos2 t = −√

1− 1

9= −

√8

9= −√

8

3.

From these, it is easy to generate the other trigonometricfunctions:

tan t =sin t

cos t= −√

8

csc t =1

sin t= − 3√

8

cot t =cos t

sin t= − 1√

8

MATH 201 - Week 9

Trigonometric Graphs

Graphs of the Sine and Cosine Functions

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 theperiod 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 theirgraphs in the interval between 0 and 2π and then ’make copies’.

MATH 201 - Week 9




f (t + p) = f (t)






MATH 201 - Week 9




f (t + p) = f (t)






MATH 201 - Week 9



We can use the special values of the trigonometric values to sketchthe graphs of the sine and cosine functions:

The graph of the sine function

The graph of the cosine function

MATH 201 - Week 9



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 2πk :

A sin

(k

(x +

2π

k

))= A sin (kx + 2π) = A sin (kx) .

The magnitude of A sin(kx) is given by the number |A|: this iscalled the amplitude.

MATH 201 - Week 9





y = A sin(kx)





A sin

(k

(x +

2π

k

))= A sin (kx + 2π) = A sin (kx) .


MATH 201 - Week 9





y = A sin(kx)





A sin

(k

(x +

2π

k

))= A sin (kx + 2π) = A sin (kx) .


MATH 201 - Week 9



Additional Horizontal Shifts


y = A sin(k(x − b))



2 horizontal shifting given by b


The number b is called the phase shift.

MATH 201 - Week 9



Additional Horizontal Shifts


y = A sin(k(x − b))



2 horizontal shifting given by b


The number b is called the phase shift.

MATH 201 - Week 9



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 2π2 = π and the phase

shift is 5.The graph is obtained from the graph ofy = sin x using the following sequenceof elementary transformations:

1 horizontal shrinking of factor 12

2 horizontal shifting to the right of 5units

3 vertical stretching of factor 3

4 reflection to the x-axis.

MATH 201 - Week 9



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 phaseshift is 5.The graph is obtained from the graph ofy = sin x using the following sequenceof elementary transformations:

1 horizontal shrinking of factor 12

2 horizontal shifting to the right of 5units

3 vertical stretching of factor 3

4 reflection to the x-axis.

MATH 201 - Week 9



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MATH 201 - Week 9 - math.u-szeged.hubaloghf/homepage/teaching/math201/math201_week... · MATH 201 - Week 9 Overview The Unit Circle (Section 5.1) The Unit Circle Terminal Points on