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CHAPTER 4
Trigonometric Functions
4.1 Angles & Radian Measure
• Objectives
– Recognize & use the vocabulary of angles
– Use degree measure
– Use radian measure
– Convert between degrees & radians
– Draw angles in standard position
– Find coterminal angles
– Find the length of a circular arc
– Use linear & angular speed to describe motion on a
circular path
Angles
• An angle is formed when two rays have a
common endpt.
• Standard position: one ray lies along the xaxis extending toward the right
• Positive angles measure counterclockwise
from the x-axis
• Negative angles measure clockwise from
the x-axis
Angle Measure
• Degrees: full circle = 360 degrees
– Half-circle = 180 degrees
– Right angle = 90 degrees
Radians: one radian is the measure of the central
angle that intercepts an arc equal in length to the
length of the radius (we can construct an angle
of measure = 1 radian!)
Full circle = 2  radians
Half circle =
radians


Right angle = 2 radians
Radian Measure
• The measure of the angle in radians is the ratio of the
arc length to the radius
s

r
• Recall half circle = 180 degrees= radians
• This provides a conversion factor. If they are equal,
their ratio=1, so we can convert from radians to
degrees (or vice versa) by multiplying by this “wellchosen one.”
• Example: convert 270 degrees to radians

3
270 

180 2
Convert 145 degrees to radians.
Coterminal angles
• Angles that have rays at the same spot.
• Angle may be positive or negative (move
counterclockwise or clockwise) (i.e. 70
degree angle coterminal to -290 degree
angle)
• Angle may go around the circle more than
once (i.e. 30 degree angle coterminal to
390 degree angle)
Find the complement and supplement
of these angles in degrees and radians.
a ) 24

b ) 132
answers
11
a) complement 66;
30
13
Supplement 156;
15
b) complement none
4
Supplement 48;
15
Converting degree to decimals.
• Write down the angle value in the degreeminute-second form. For example, the
angle is 27 degrees, 12 minutes and 45
seconds.
• Divide the minutes by 60
• Divide the seconds by 3,600
• Then add them
• You will get 27.2125
Converting decimal to degrees
• The whole units of degrees will remain the same
(i.e. in 121.135° longitude, start with 121°).
• Multiply the decimal by 60 (i.e. .135 * 60 = 8.1).
• The whole number becomes the minutes (8').
• Take the remaining decimal and multiply by 60.
(i.e. .1 * 60 = 6).
• The resulting number becomes the seconds (6").
Seconds can remain as a decimal.Take your
three sets of numbers and put them together,
using the symbols for degrees (°), minutes (‘),
and seconds (") (i.e. 121°8'6" longitude)
Convert the angle measures to
DM ' S "
a )  12430 '
b )  115.8
c ) .7865
answers
a )  124.5
b)  11548'
c) 453' 47"
Arc length
• Since radians are defined as the central
angle created when the arc length = radius
length for any given circle, it makes sense to
consider arc length when angle is measured
in radians
• Recall theta (in radians) is the ratio of arc
length to radius
• Arc length = radius x theta (in radians)
s  r
Linear speed & Angular speed
• Speed a particle moves along an arc of the circle
(v) is the linear speed (distance, s, per unit time, t)
s
v
t
• Speed which the angle is changing as a particle
moves along an arc of the circle is the angular
speed.(angle measure in radians, per unit time, t)


t
Relationship between linear speed
& angular speed
• Linear speed is the product of radius and angular
speed.
v  r 
• Example: The minute hand of a clock is 6 inches
long. How fast is the tip of the hand moving?
• We know angular speed = 2 pi per 60 minutes
 2
v  6in
 60 min
2in
 in
in


 .6

min
 10 min 5 min
4.2 Trigonometric Functions: The
Unit Circle
• Objectives
– Use a unit circle to define trigonometric
functions of real numbers
– Recognize the domain & range of sine &
cosine
– Find exact values of the trig. functions at pi/4
– Use even & odd trigonometric functions
– Recognize & use fundamental identities
– Use periodic properties
– Evaluate trig. functions with a calculator
What is the unit circle?
• A circle with radius = 1 unit
• Why are we interested in this circle? It provides
convenient (x,y) values as we work our way around
the circle.
• (1,0), theta = 0
• (0,1), theta = pi/2
• (-1,0), theta = pi
• (0,-1), theta = 3 pi/2
• ALSO, any (x,y) point on the circle would be at the
end of the hypotenuse of a right triangle that
extends from the origin, such that x 2  y 2  1
sin t and cos t
• For any point (x,y) found on the unit circle,
x=cos t and y=sin t
• t = any real number, corresponding to the
arc length of the unit circle
• Example: at the point (1,0), the cos t = 1
and sin t = 0. What is t? t is the arc length
at that point AND since it’s a unit circle, we
know the arc length = central angle, in
radians. THUS, cos (0) = 1 and sin (0)=0
Relating all trigonometric functions
to sin t and cos t
sin( t ) y
tan( t ) 

cos(t ) x
1
1
csc(t ) 

sin( t ) y
1
1
sec(t ) 

cos(t ) x
cos(t ) x
cot(t ) 

sin( t ) y
Pythagorean Identities
• Every point (x,y) on the unit circle corresponds to a
real number, t, that represents the arc length at that
point
2
2
x

y
 1 and x = cos(t) and y=sin(t),
• Since
then cos 2 t  sin 2 t  1
• If each term is divided by cos 2 t , the result is
sin 2 t
1
2
2
1

,
1

tan
t

sec
t
2
2
cos t cos t
• If each term is divided by
2
sin t , the result is
cos 2 t
1
2
2

1

,
cot
t

1

csc
t
2
2
sin t
sin t
Given csc t = 13/12, find the values
of the other 6 trig. functions of t
•
•
•
•
•
sin t = 12/13 (reciprocal)
cos t = 5/13 (Pythagorean)
sec t = 13/5 (reciprocal)
tan t = 12/5 (sin(t)/cos(t))
cot t = 5/12 (reciprocal)
Trig. functions are periodic
• sin(t) and cos(t) are the (x,y) coordinates
around the unit circle and the values repeat
every time a full circle is completed
• Thus the period of both sin(t) and cos(t) = 2
pi
• sin(t)=sin(2pi + t)
cos(t)=cos(2pi + t)
• Since tan(t) = sin(t)/cos(t), we find the
values repeat (become periodic) after pi,
thus tan(t)=tan(pi + t)
4.3 Right Triangle Trigonometry
• Objectives
– Use right triangles to evaluate trig.
Functions
– Find function values for 30 degrees, 45
   
degrees & 60 degrees  , , 
6 4 3
– Use equal cofunctions of complements
– Use right triangle trig. to solve applied
problems
Within a unit circle, and right
triangle can be sketched
• The point on the circle is (x,y) and the
hypotenuse = 1. Therefore, the x-value is
the horizontal leg and the y-value is the
vertical leg of the right triangle formed.
• cos(t)=x which equals x/1, therefore the
cos (t)=horizontal leg/hypotenuse =
adjacent leg/hypotense
• sin(t)=y which equals y/1, therefore the
sin(t) = vertical leg/hypotenuse = opposite
leg/hypotenuse
The relationships holds true for ALL
right triangles (other 3 trig.
functions are found as reciprocals)
opposite
sin  
hypotenuse
adjacent
cos  
hypotenuse
sin  opposite
tan  

cos  adjacent
Find the value of 6 trig. functions of
the angles in a right triangle.
• Given 2 sides, the value of the 3rd side can
be found, using Pythagorean theorem
• After side lengths of all 3 sides is known,
find sin as opposite/hypotenuse
• cos = adjacent/hypotenuse
• tan = opposite/adjacent
• csc = 1/sin
• sec = 1/cos
• cot= 1/tan
Given a right triangle with
hypotenuse =5 and side adjacent
angle B of length=2, find tan B
1) 21
21
2)
2
2
3)
21
2
4)
5
Special Triangles
• 30-60 right triangle, ratio of sides of the triangle is
1:2: 3, 2 (longest) is the length of the
hypotenuse, the shortest side (opposite the 30
degree angle) is 1 and the remaining side
(opposite the 60 degree angle) is 3
• 45-45 right triangle: The 2 legs are the same
length since the angles opposite them are equal,
thus 1:1. Using pythagorean theorem, the
remaining side, the hypotenuse, is 2
Cofunction Identities
• Cofunctions are those that are the reciprocal
functions (cofunction of tan is cot, cofunction of
sin is cos, cofunction of sec is csc)
• For an acute angle, A, of a right triangle, the side
opposite A would be the side adjacent to the other
acute angle, B
• Therefore sin A = cos B
• Since A & B are the acute angles of a right
triangle, their sum = 90 degrees, thus B= 90  A
• function(A)=cofunction (90  A)
4.4 Trigonometric Functions of Any
Angle
• Objectives
– Use the definitions of trigonometric
functions of any angle
– Use the signs of the trigonometric
functions
– Find reference angles
– Use reference angles to evaluate
trigonometric functions
Trigonometric functions of Any
Angle
• Previously, we looked at the 6 trig.
functions of angles in a right triangle.
These angles are all acute. What about
negative angles? What about obtuse
angles?
• These angles exist, particularly as we
consider moving around a circle
• At any point on the circle, we can drop a
vertical line to the x-axis and create a
triangle. Horizontal side = x, vertical
side=y, hypotenuse=r.
Trigonometric Functions of Any
Angle (continued)
• If, for example, you have an angle whose
terminal side is in the 3rd quadrant, then the x
& y values are both negative. The radius, r, is
always a positive value.
• Given a point (-3,-4), find the 6 trig. functions
associated with the angle formed by the ray
containing this point.
2
2
(

3
)

(

4
)
 25  5
• x=-3, y=-4, r =
• (continued next slide)
Example continued
• sin A = -4/5, cos A = -3/5, tan A = 4/3
• csc A = -5/4, sec A = -5/3, cot A = ¾
• Notice that the same values of the trig.
functions for angle A would be true for the
angles 360+A, A-360 (negative values)
Examining the 4 quadrants
• Quadrant I: x & y are positive
– all 6 trig. functions are positive
• Quadrant II: x negative, y positive
– positive: sin, csc
negative: cos, sec, tan, cot
• Quadrant III: x negative, y negative
– positive: tan, cot
negative: sin, csc, cos, sec
• Quadrant IV: x positive, y negative
– positive: cos, sec
negative: sin, csc, cot, tan
Reference angles
• Angles in all quadrants can be related to a
“reference” angle in the 1st quadrant
• If angle A is in quadrant II, it’s related
angle in quad I is 180-A. The numerical
values of the 6 trig. functions will be the
same, except the x (cos, sec, tan, cot) will
all be negative
• If angle A is in quad III, it’s related angle in
quad I is 180+A. Now x & y are both neg,
so sin, csc, cos, sec are all negative.
Reference angles cont.
• If angle A is in quad IV, the reference
angle is 360-A. The y value is negative,
so the sin, csc, tan & cot are all negative.
Special angles
• We often work with the “special angles” of the
“special triangles.” It’s good to remember them
both in radians & degrees
30 

6
,60 

3
,45 

4
,90 

2
• If you know the trig. functions of the special angles
in quad I, you know them in every quadrant, by
determining whether the x or y is positive or
negative
4.5 Graphs of Sine & Cosine
• Objectives
– Understand the graph of y = sin x
– Graph variations of y = sin x
– Understand the graph of y = cos x
– Graph variations of y = cos x
– Use vertical shifts of sin & cosine curves
– Model periodic behavior
Graphing y = sin x
• If we take all the
values of sin x from
the unit circle and plot
them on a coordinate
axis with x = angles
and y = sin x, the
graph is a curve
• Range: [-1,1]
• Domain: (all reals)
Graphing y = cos x
• Unwrap the unit
circle, and plot all x
values from the circle
(the cos values) and
plot on the coordinate
axes, x = angle
measures (in radians)
and y = cos x
• Range: [-1,1]
• Domain: (all reals)
Comparisons between y=cos x and
y=sin x
• Range & Domain: SAME
– range: [-1,1], domain: (all reals)
• Period: SAME (2 pi)
• Intercepts: Different
– sin x : crosses through origin and intercepts the
x-axis at all multiples of  , (....  3 ,2 , ,0,  ,2 ,3 ,...)
– cos x: intercepts y-axis at (0,1) and intercepts xaxis at all odd multiples of  ,  ...  3 ,  ,  , 3 ,...
2 
2
2 2
2

Amplitude & Period
• The amplitude of sin x & cos
x is 1. The greatest distance
the curves rise & fall from the
axis is 1.
• The period of both functions
is 2 pi. This is the distance
around the unit circle.
• Can we change amplitude?
Yes, if the function value (y)
is multiplied by a constant,
that is the NEW amplitude,
example: y = 3 sin x
Amplitude & Period (cont)
• Can we change the
period? Yes, the
length of the period is
a function of the xvalue.
• Example: y = sin(3x)
– The amplitude is still 1.
(Range: [-1,1])
– Period is 2
3
Phase Shift
• The graph of y=sin x is
“shifted” left or right of
the original graph
• Change is made to the
x-values, so it’s
addition/subtraction to
x-values.

• Example: y = sin(x- 3 ),
the graph of y=sin x is
shifted right 
3
Vertical Shift
• The graph y=sin x
can be shifted up or
down on the
coordinate axis by
adding to the y-value.
• Example:
• y = sin x + 3 moves
the graph of sin x up
3 units.
Graph y =

2cos(x- 4
• Amplitude = 2

• Phase shift =
right
4
• Vertical shift = down 2
)-2
4.6 Graphs of Other Trigonometric
Functions
• Objectives
– Understand the graph of y = tan x
– Graph variations of y = tan x
– Understand the graph of y = cot x
– Graph variations of y = cot x
– Understand the graphs of y = csc x and y = sec x
y = tan x
• Going around the unit
circle, where the y
value is 0, (sin x = 0),
the tangent is
undefined.
 3    3
(...
• At x = 2 , 2 , 2 , 2 ,...)
the graph of y = tan x
has vertical
asymptotes
• x-intercepts where
cos x = 0,
x = (...  2 , ,0,  ,2 ,...)
Characteristics of y = tan x
•
•
•
•
•
•
Period = 
Domain: (all reals except odd multiples of
Range: (all reals)
Vertical asymptotes: odd multiples of 
2
x – intercepts: all multiples of 
Odd function (symmetric through the origin, quad
I mirrors to quad III)
Transformations of y = tan x
• Shifts (vertical & phase) are done as the
shifts to y = sin x
• Period change (same as to y=sin x, except
the original period of tan x is pi, not 2 pi)
Graph y = -3 tan (2x) + 1
•
•
•
•
Period is now pi/2
Vertical shift is up 1
-3 impacts the “amplitude”
Since tan x has no amplitude, we consider
the point ½ way between intercept &
asymptote, where the y-value=1. Now the
y-value at that point is -3.
• See graph next slide.
Graph y = -3 tan (2x) + 1
Graphing y = cot x
• Vertical asymptotes
are where sin x = 0,
(multiples of pi)
• x-intercepts are
where cos x = 0 (odd
multiples of pi/2)
y = csc x
• Reciprocal of y = sin x
• Vertical tangents where sin x = 0 (x = integer
multiples of pi)
• Range: (,1]  [1, )
• Domain: all reals except integer multiples of pi
• Graph on next slide
Graph of y = csc x
y = sec x
• Reciprocal of y = cos x
• Vertical tangents where cos x = 0 (odd multiples
of pi/2)
• Range: (,1]  [1, )
• Domain: all reals except odd multiples of pi/2
• Graph next page
Graph of y = sec x
4.7 Inverse Trigonometric
Functions
• Objectives
– Understand the use the inverse sine function
– Understand and use the inverse cosine function
– Understand and use the inverse tangent function
– Use a calculator to evaluate inverse trig. functions
– Find exact values of composite functions with
inverse trigonometric functions
What is the inverse sin of x?
•
•
•
•
It is the ANGLE (or real #) that has a sin value of x.
Example: the inverse sin of ½ is pi/6 (arcsin ½ = pi/6)
Why? Because the sin(pi/6)= ½
Shorthand notation for inverse sin of x is arcsin x or
sin
1
x
• Recall that there are MANY angles that would have a
sin value of ½. We want to be consistent and specific
about WHICH angle we’re referring to, so we limit the
range to    ,  
(quad I & IV)
 2
2 
Find the domain of y = sin
1
x
• The domain of any function becomes the range of
its inverse, and the range of a function becomes
the domain of its inverse.
• Range of y = sin x is [-1,1], therefore the domain of
the inverse sin (arcsin x) function is [-1,1]
Trigonometric values for special
angles
• If you know sin(pi/2) = 1, you know the
inverse sin(1) = pi/2
• KNOW TRIG VALUES FOR ALL SPECIAL
ANGLES (once you do, you know the
inverse trigs as well!)


2
1

Find sin 
 2 



1)
4
7
2)
4
 3
3)
4

4)
4
Graph y = arcsin (x)
The inverse cosine function
• The inverse cosine of x refers to the angle (or
number) that has a cosine of x
• Inverse cosine of x is represented as arccos(x)
or cos 1 x
• Example: arccos(1/2) = pi/3 because the
cos(pi/3) = ½
• Domain: [-1,1]
• Range: [0,pi] (quadrants I & II)
Graph y = arccos (x)
The inverse tangent function
• The inverse tangent of x refers to the angle (or
number) that has a tangent of x
• Inverse tangent of x is represented as arctan(x)
1
tan
x
or
• Example: arctan(1) = pi/4 because the
tan(pi/4)=1
• Domain: (all reals)
• Range: [-pi/2,pi/2] (quadrants I & IV)
Graph y = arctan(x)
Evaluating compositions of
functions & their inverses
• Recall: The composition of a function and its
inverse = x. (what the function does, its
inverse undoes)
• This is true for trig. functions & their
inverses, as well ( PROVIDED x is in the
range of the inverse trig. function)
• Example: arcsin(sin pi/6) = pi/6, BUT
arcsin(sin 5pi/6) = pi/6
• WHY? 5pi/6 is NOT in the range of arcsin x,
but the angle that has the same sin in the
appropriate range is pi/6
4.8 Applications of Trigonometric
Functions
• Objectives
– Solve a right triangle.
– Solve problems involving bearings.
– Model simple harmonic motion.
Solving a Right Triangle
• This means find the values of all angles and all side
lengths.
• Sum of angles = 180 degrees, and if one is a right
angle, the sum of the remaining angles is 90
degrees.
• All sides are related by the Pythagorean Theorem:
a b  c
2
2
2
• Using ratio definition of trig functions (sin x =
opposite/hypotenuse, tan x = opposite/adjacent, cos
x = adjacent/hypotenuse), one can find remaining
sides if only one side is given
Example: A right triangle has an
hypotenuse = 6 cm with an angle =
35 degrees. Solve the triangle.
•
•
•
•
•
cos(35 degrees) = .819 (using calculator)
cos(35 degrees) = adjacent/6 cm
Thus, .819 = adjacent/6 cm, adjacent = 4.9 cm
Remaining angle = 55 degrees
Remaining side:
a 2  (4.9) 2  6 2
a 2  36  24  12
a  12cm
Trigonometry & Bearings
• Bearings are used to describe position in
navigation and surveying. Positions are
described relative to a NORTH or SOUTH axis (yaxis). (Different than measuring from the
standard position, the positive x-axis.)
• N 55E means the direction is 55 degrees from
the north toward the east (in quadrant I)
• S 35W means the direction is 35 degrees from
the south toward the west (in quadrant III)