Transcript Thinking Mathematically by Robert Blitzer

Chapter 4
Angles and Their
Measure
Angles
An angle is formed by two rays that have a
common endpoint called the vertex. One ray is
called the initial side and the other the terminal
side. The arrow near the vertex shows the
direction and the amount of rotation from the initial
side to the terminal side.
C
A
è
Initial Side
Terminal Side
B
Vertex
Angles of the Rectangular Coordinate System
An angle is in standard position if
• its vertex is at the origin of a rectangular coordinate system and
• its initial side lies along the positive x-axis.
y
y
 is positive
Terminal
Side

Vertex
Initial Side
x
Vertex
Initial Side
x
Terminal
Side

 is negative
Positive angles rotate counterclockwise.
Negative angles rotate clockwise.
Measuring Angles Using Degrees
The figures below show angles classified by
their degree measurement. An acute angle measures
less than 90º. A right angle, one quarter of a complete
rotation, measures 90º and can be identified by a small
square at the vertex. An obtuse angle measures more
than 90º but less than 180º. A straight angle has
measure 180º.

Acute angle
0º <  < 90º
90º
Right angle
1/4 rotation

Obtuse angle
90º <  < 180º
180º
Straight angle
1/2 rotation
Coterminal Angles
An angle of xº is coterminal with angles of
xº + k · 360º
where k is an integer.
Text Example
Assume the following angles are in standard position. Find a positive angle
less than 360º that is coterminal with:
a. a 420º angle
b. a –120º angle.
Solution We obtain the coterminal angle by adding or subtracting 360º.
Our need to obtain a positive angle less than 360º determines whether we
should add or subtract.
a. For a 420º angle, subtract 360º to find a positive coterminal angle.
420º – 360º = 60º
A 60º angle is coterminal with a 420º angle. These angles, shown on the
next slide, have the same initial and terminal sides.
Text Example cont.
y
y
240º
60º
x
420º
x
-120º
Solution
b. For a –120º angle, add 360º to find a positive coterminal angle.
-120º + 360º = 240º
A 240º angle is coterminal with a –120º angle. These angles have the
same initial and terminal sides.
Finding Complements and
Supplements
• For an xº angle, the complement is a 90º –
xº angle. Thus, the complement’s measure
is found by subtracting the angle’s measure
from 90º.
• For an xº angle, the supplement is a 180º –
xº angle. Thus, the supplement’s measure is
found by subtracting the angle’s measure
from 180º.
Definition of a Radian
• One radian is the measure of the central
angle of a circle that intercepts an arc equal
in length to the radius of the circle.
Radian Measure
Consider an arc of length s on a circle or radius r.
The measure of the central angle that intercepts
the arc is
s
 = s/r radians.

r
O
r
Conversion between Degrees and
Radians
• Using the basic relationship  radians =
180º,
• To convert degrees to radians, multiply
degrees by ( radians) / 180
• To convert radians to degrees, multiply
radians by 180 / ( radians)
Example
Convert each angle in degrees to radians
40º
75º
-160º
Example cont.
Solution:
• 40º = 40*/180 = 2  /9
• 75º = 75*  /180 = 5  /12
• -160º = -160*  /180 = -8  /9
Length of a Circular Arc
Let r be the radius of a circle and  the nonnegative radian measure of a central angle
of the circle. The length of the arc
s
intercepted by the central angle is

s=r
O
r
Example
A circle has a radius of 7 inches. Find the
length of the arc intercepted by a central
angle of 2/3
Solution:
s = (7 inches)*(2  /3)
=14  /3 inches
Definitions of Linear and
Angular Speed
If a point is in motion on a circle of radius r
through an angle of  radians in time t, then
its linear speed is
v = s/t
where s is the arc length given by s = r , and
its angular speed is
 = /t
Linear Speed in Terms of
Angular Speed
• The linear speed, v, of a point a distance r
from the center of rotation is given by v=r
where  is the angular speed in radians per
unit of time.
Trigonometric
Functions: The Unit
Circle
Definitions of the Trigonometric
Functions in Terms of a Unit Circle
If t is a real number and P = (x, y) is a point
on the unit circle that corresponds to t, then
sint  y
cost  x
1
csc t  , y  0
y
1
sec t  ,x  0
x
y
t ant  , x  0
x
x
cott  , y  0
y
Text Example
Use the Figure to find the values of the
trigonometric functions at t=/2.
Solution:
(0,1)
The point P on the unit circle that
Corresponds to t= /2 has coordinates
(0,1). We use x=0 and y=1 to find the
Values of the trigonometric functions


sin
2
cos
2
x0
1 1
 x 0
  1 cot    0
2 y 1
2 y 1
y
1 1
t ant  andsec t    undefined
x
x 0
csc

 y 1

/2
/2
(1,0)
x2  y2  1
The Domain and Range of the
Sine and Cosine Functions
• The domain of the sine function and the
cosine function is the set of all real numbers
The range of these functions is the set of all
real numbers from -1 to 1, inclusive.
Trigonometric Functions at /4.

2
sin 
4
2
csc

4
 2

2
cos 
4
2
sec

4
 2
t an
cot

4

4
1
1
Even and Odd Trigonometric
Functions
The cosine and secant functions are even.
cos(-t) = cos t
sec(-t) = sec t
The sine, cosecant, tangent, and cotangent
functions are odd.
sin(-t) = -sin t
csc(-t) = -csc t
tan(-t) = -tan t
cot(-t) = -cot t
Example
• Use the value of the trigonometric function
at t = /4 to find sin (- /4 ).
sin
csc

4

4
2
2
cos
 2
sec


4

4

2
2
 2
t an
cot

4

4
1
1
Solution:
sin(-t) = -sin t, so sin(- /4 ) = -sin(/4 ) = -2/2
Reciprocal Identities
1
sint 
csct
1
csc t 
sint
1
cost 
sect
1
sec t 
cost
1
t ant 
cott
1
cott 
t ant
Quotient Identities
sint
t ant 
cost
cost
cott 
sin t
Pythagorean Identities
2
2
sin t  cos t  1
2
2
1  t an t  sec t
1  cot t  csc t
2
2
Definition of a Periodic Function
A function f is periodic if there exists a
positive number p such that
f(t + p) = f(t)
For all t in the domain of f. The smallest
number p for which f is periodic is called
the period of f.
Periodic Properties of the Sine
and Cosine Functions
sin(t + 2) = sin t and cos(t + 2) = cos t
The sine and cosine functions are periodic
functions and have period 2.
Periodic Properties of the
Tangent and Cotangent Functions
tan(t + ) = tan t and cot(t + ) = cot t
The tangent and cotangent functions are
periodic functions and have period .
Repetitive Behavior of the Sine,
Cosine, and Tangent Functions
For any integer n and real number t,
sin(t + 2n) = sin t, cos(t + 2n) = cos t, and
tan(t + n) = tan t
Right Triangle
Trigonometry
The Six Trigonometric Functions
The figure below shows a right triangle with one of its acute
angles labeled . The side opposite the right angle is known as the
hypotenuse. The other sides of triangle are described by the position
relative to the acute angle . One side is opposite  and one is adjacent
to .
Hypotenuse
Side opposite .

Side adjacent to .
Right Triangle Definitions of
Trigonometric Functions
sin  
Opp
Hyp
csc  
Hyp
Opp
cos  
Adj
Hyp
sec  
Hyp
Adj
tan  
Opp
Adj
cot  
Adj
Opp
Hyp
Opp

Adj
Example
adj.
cos A 
hyp.
x
cos55  .5735
9
x  9(0.5735)

x  5.16
Sines, Cosines, and Tangents of
Special Angles

1

3

3
sin30  sin  , cos30  cos 
, t an30  t an 
6 2
6
2
6
3

2

2

sin 45  sin 
, cos45  cos 
, t an45  t an  1
4
2
4
2
4
sin60  sin

3

3
 1

, cos60  cos  , t an60  t an  3
2
3 2
3
Cofunction Identities
The value of a trigonometric function of  is
equal to the cofunction of the complement
of .
sin  = cos (90º – )cos  = sin (90º – )
tan  = cot (90º – ) cot  = tan (90º – )
sec  = csc (90º – )csc  =sec (90º – )
If  is in radians, replace 90º with /2.
Text Example
Sighting the top of a building, a surveyor measured the angle of elevation to
be 22º. The transit is 5 feet above the ground and 300 feet from the building.
Find the building’s height.
Solution Let a be the height of the portion of the building that lies above
the transit in the figure shown. The height of the building is the transit’s
height, 5 feet, plus a. Thus, we need to identify a trigonometric function that
will make it possible to find a. In terms of the 22º angle, we are looking for
the side opposite the angle.
a
Transit
5 feet
22º
300 feet
h
Text Example cont.
Solution
The transit is 300 feet from the building, so the side adjacent to the 22º angle
is 300 feet. Because we have a known angle, an unknown opposite side, and
a known adjacent side, we select the tangent function.
a
tan 22º = 300
Length of side opposite the 22º angle
Length of side adjacent to the 22º angle
a = 300 tan 22º 300(0.4040) 121
Multiply both sides of the equation by 300.
The height of the part of the building above the transit is approximately 121
feet. If we add the height of the transit, 5 feet, the building’s height is
approximately 126 feet.
a
Transit
5 feet
22º
300 feet
h
Trigonometric
Functions of Any
Angle
Definitions of Trigonometric
Functions of Any Angle
• Let  is be any angle in standard position, and let P = (x,
y) be a point on the terminal side of . If r = x2 + y2 is the
distance from (0, 0) to (x, y), the six trigonometric
functions of  are defined by the following ratios.
y
x
y
sin   ,
cos  ,
t an  , x  0
r
r
x
r
r
x
csc   ,y  0 sec   , x  0 cot   ,y  0
y
x
y
Text Example
Let P = (-3, -4) be a point on the terminal side of . Find each of the six
trigonometric functions of .
Solution The situation is shown below. We need values for x, y, and r to
evaluate all six trigonometric functions. We are given the values of x and y.
Because P = (-3, -4) is a point on the terminal side of , x = -3 and y = -4.
Furthermore,
y
5

-5
5
r
-5
P = (-3, -4)
x = -3
y = -4
x
Text Example Cont.
Solution
Now that we know x, y, and r, we can find the
six trigonometric functions of .
y 4
4
sin   
  , cos 
r
5
5
r
5
5
csc   
  , sec  
y 4
4
x 3
3
y 4 4

  , t an  

r
5
5
x 3 3
r
5
5
x 3 3

  , cot  

x 3
3
y 4 4
The bottom row shows the
reciprocals of the row above.
The Signs of the Trigonometric
Functions
y
Quadrant II
Sine and
cosecant
positive
Quadrant I
All functions
positive
x
Quadrant III
tangent and
cotangent
positive
Quadrant IV
cosine and
secant
positive
Definition of a Reference Angle
• Let  be a nonacute angle in standard
position that lies in a quadrant. Its reference
angle is the positive acute angle ´ prime
formed by the terminal side or  and the xaxis.
Example
Find the reference angle , for the following
b
angle:
 =315º
Solution:
315
a
 =360º - 315º = 45º
a
45
b
P(a, b)
Using Reference Angles to Evaluate
Trigonometric Functions
• The values of a trigonometric functions of a
given angle, , are the same as the values
for the trigonometric functions of the
reference angle, ´, except possibly for the
sign. A function value of the acute angle, ´,
is always positive. However, the same
functions value for  may be positive or
negative.
A Procedure for Using Reference Angles
to Evaluate Trigonometric Functions
• The value of a trigonometric function of any
angle  is found as follows:
• Find the associated reference angle, ´, and
the function value for ´.
• Use the quadrant in which  lies to prefix
the appropriate sign to the function value in
step 1.
Text Example
Use reference angles to find the exact value of
the following trigonometric functions.
a. sin 135°
Solution
a. We use our two-step procedure to find sin 135°.
Step 1 Find the reference angle,  ´, and sin  ´.
135º terminates in quadrant II with a
reference angle  ´ = 180º – 135º = 45º.
y
135°
45°
x
Solution
Text Example cont.
The function value for the reference angle is sin 45º = 2 / 2.
Step 2 Use the quadrant in which è lies to prefix the
appropriate sign to the function value in step 1. The angle
135º lies in quadrant II. Because the sine is positive in
quadrant II, we put a + sign before the function value of the
reference angle. Thus, sin135= +sin45=2 / 2
Graphs of Sine and
Cosine Functions
The Graph of y=sinx
The trigonometric functions can be graphed in a rectangular coordinate system
by plotting points whose coordinates belong to the function. Thus, we graph
y = sin x by listing some points on the graph. Because the period of the sine
function is 2 we will graph the function on the interval [0, 2]. The rest of the
graph is made up of repetitions of this portion.
y
y = sin x, 0 < x < 2
1

6

3

2
˝
-1
Period: 2
3
2
2 x
Graphing Variations of y=sinx
• Identify the amplitude and the period.
• Find the values of x for the five key points – the three xintercepts, the maximum point, and the minimum point.
Start with the value of x where the cycle begins and add
quarter-periods – that is, period/4 – to find successive
values of x.
• Find the values of y for the five key points by evaluating
the function at each value of x from step 2.
• Connect the five key points with a smooth curve and graph
one complete cycle of the given function.
• Extend the graph in step 4 to the left or right as desired.
Text Example
Determine the amplitude of y = 1/2 sin x. Then graph y = sin x
and y = 1/2 sin x for 0 < x < 2.
Solution
Step 1 Identify the amplitude and the period. The equation
y = 1/2 sin x is of the form y = A sin x with A = 1/2. Thus, the
amplitude |A| = 1/2. This means that the maximum value of y is
1/2 and the minimum value of y is -1/2. The period for both y =
1/2 sin x and y = sin x is 2.
Text Example cont.
Solution
Step 2 Find the values of x for the five key points. We
need to find the three x-intercepts, the maximum point, and
the minimum point on the interval [0, 2]. To do so, we
begin by dividing the period, 2, by 4.
Period/4 = 2 /4 = /2
We start with the value of x where the cycle begins: x = 0.
Now we add quarter periods,  /2, to generate x-values for
each of the key points. The five x-values are
x = 0, x = 0 + /2 = /2, x = /2 + /2 = ,
x =  + /2 = 3 /2, x= 3 /2 + /2 = 2 
Text Example cont.
Solution
Step 3 Find the values of y for the five key points. We
evaluate the function at each value of x from step 2.
y = 1/2 sin0 = 0
y = 1/2sin/2 = 1/2*1 = 1/2
y = 1/2sin = 1/2*0 = 0
y = 1/2 sin 3/2 = 1/2(-1) = -1/2
y = 1/2sin2 = 1/2*0 = 0
(0,0)
(/2, 1/2)
(,0)
(3 /2, -1/2)
(2, 0)
There are x-intercepts at 0,  and 2 . The maximum and
minimum points are (/2, 1/2) and (3 /2, -1/2)
Text Example cont.
Solution
Step 4 Connect the five key points with a smooth curve and graph one
complete cycle of the given function. The five key points for y = 1/2sin x
are shown below. By connecting the points with a smooth curve, the figure
shows one complete cycle of y = 1/2sin x. Also shown is graph of y = sin x.
The graph of y = 1/2sin x shrinks the graph of y = sin x.
y
y = sin x
1
y = 1/2sinx
2˝
˝
-1
x
Amplitudes and Periods
The graph of y = A sin Bx has
amplitude = | A|
period = 2 /B.
y
y = A sin Bx
Amplitude: | A|
Period: 2 /B
x
The Graph of y = Asin(Bx - C)
The graph of y = A sin (Bx – C) is obtained by horizontally shifting the graph
of y = A sin Bx so that the starting point of the cycle is shifted from x = 0 to
x = C/B. The number C/B is called the phase shift.
y
amplitude = | A|
period = 2 /B.
y = A sin Bx
Amplitude: | A|
x
Starting point: x = C/B
Period: 2/B
Example
Determine the amplitude, period, and phase
shift of y = 2sin(3x-)
Solution:
Amplitude = |A| = 2
period = 2/B = 2/3
phase shift = C/B = /3
Example cont.
• y = 2sin(3x- )
3
2
1
-6 -5 -4 -3 -2 -1
-1
-2
-3
1 2 3 4 5 6
The Graph of y = AcosBx
We graph y = cos x by listing some points on the graph. Because the period
of the cosine function is 2, we will concentrate on the graph of the basic
cosine curve on the interval [0, 2 ]. The rest of the graph is made up of
repetitions
of this portion.
.
y
1
-3 /2
-
/2
-/2

-1
Period: 2
3/2
2
5/2
x
The Graph of y = AcosBx
The graph of y = A cos Bx has
amplitude = | A|
period = 2/B.
y
y = A cos Bx
Amplitude: | A|
x
2/B
Period:
2/B
The Graph of y = Acos(Bx - C)
The graph of y = A cos (Bx – C) is obtained by horizontally shifting the graph
of y = A cos Bx so that the starting point of the cycle is shifted from x = 0 to
x = C/B. The number C/B is called the phase shift.
y = A cos( Bx-C)
y
amplitude = | A|
period = 2 /B.
Amplitude: | A|
x
Period:
2/B
Graphs of Other
Trigonometric
Functions
The Tangent Curve: The Graph of
y=tanx and Its Characteristics
y
Period: 
Domain: All real numbers
except  /2 + k ,
k an integer
1
–2
–
5
2
–
–
3
2
–

2
–1

2
Range: All real numbers
2

0
3
2
5
2
x
Symmetric with respect to
the origin
Vertical asymptotes at
odd multiples of  /2
Graphing y = A tan(Bx – C)
y = A tan (Bx – C)
Bx – C = - /2
x-intercept
between
asymptotes
1. Find two consecutive asymptotes by setting the
variable expression in the tangent equal to -/2
and /2 and solving
Bx – C =  /2
Bx – C = -/2 and Bx – C = /2
2. Identify an x-intercept, midway between
consecutive asymptotes.
x
3. Find the points on the graph 1/4 and 3/4 of the
way between and x-intercept and the asymptotes.
These points have y-coordinates of –A and A.
4. Use steps 1-3 to graph one full period of the
function. Add additional cycles to the left or right
as needed.
Text Example
Graph y = 2 tan x/2 for – < x < 3 
Solution
Step 1 Find two consecutive asymptotes.
Thus, two consecutive asymptotes occur at x = -  and x = .
Step 2 Identify any x-intercepts, midway between consecutive
asymptotes. Midway between x = - and x =  is x = 0. An x-intercept is 0
and the graph passes through (0, 0).
Text Example cont.
Solution
Step 3 Find points on the graph 1/4 and 1/4 of the way between an xintercept and the asymptotes. These points have y-coordinates of –A and A.
Because A, the coefficient of the tangent, is 2, these points have y-coordinates of
-2 and 2.
Step 4 Use steps 1-3 to graph one
full period of the function. We use the
two consecutive asymptotes, x = - and
x = , an x-intercept of 0, and points
midway between the x-intercept and
asymptotes with y-coordinates of –2
and 2. We graph one full period of
y = 2 tan x/2 from – to . In order to
graph for – < x < 3 , we continue the
pattern and extend the graph another
full period on the right.
y = 2 tan x/2
y
4
2
˝
-˝
-2
-4
3˝
x
The Cotangent Curve: The Graph
of y = cotx and Its Characteristics
The Graph of y = cot x and Its Characteristics
Characteristics
y
4
2
-
-
 /2
 /2
-2
-4
˝
3
 /2
2
x
Period: 
Domain: All real numbers except
integral multiples of 
Range: All real numbers
Vertical asymptotes: at integral
multiples of 
n x-intercept occurs midway between
each pair of consecutive asymptotes.
Odd function with origin symmetry
Points on the graph 1/4 and 3/4 of the way
between consecutive asymptotes have ycoordinates of –1 and 1.
Graphing y=Acot(Bx-C)
1. Find two consecutive asymptotes by setting the
variable expression in the cotangent equal to 0
and ˝ and solving
Bx – C = 0 and Bx – C = 
Bx – C = 
2. Identify an x-intercept, midway between
consecutive asymptotes.
x
3. Find the points on the graph 1/4 and 3/4 of the
y-coordway between an x-intercept and the asymptotes.
inate is -A.
These points have y-coordinates of –A and A.
4. Use steps 1-3 to graph one full period of the
function. Add additional cycles to the left or right
as needed.
y = A cot (Bx – C)
y-coordinate is A.
x-intercept
between
asymptotes
Bx – C
=0
Example
Graph y = 2 cot 3x
Solution:
3x=0 and 3x=
x=0 and x = /3 are vertical asymptotes
An x-intercepts occurs between 0 and /3 so an xintercepts is at (/6,0)
The point on the graph midway between the
asymptotes and intercept are /12 and 3/12.
These points have y-coordinates of -A and A or -2
and 2
Graph one period and extend as needed
Example cont
• Graph y = 2 cot 3x
10
8
6
4
2
-3
-2
-1
1
-2
-4
-6
-8
-10
2
3
The Cosecant Curve: The Graph
of y = cscx and Its Characteristics
y
-  /2 1
-2
-3  /2
-˝
-1
3  /2
 /2
˝
2
x
Characteristics
Period: 2
Domain: All real numbers except
integral multiples of 
Range: All real numbers y such that
y < -1 or y > 1
Vertical asymptotes: at integral
multiples of 
Odd function with origin symmetry
The Secant Curve: The Graph of
y=secx and Its Characteristics
y
1
-2
-3  /2
-
-˝ /2
-1
˝  /2

3  /2
2
x
Characteristics
Period: 2
Domain: All real numbers except odd
multiples of  /2
Range: All real numbers y such that
y < -1 or y > 1
Vertical asymptotes: at odd multiples
of / 2
Even function with origin symmetry
Text Example
Use the graph of y = 2 sin 2x to obtain the graph of y = 2 csc 2x.
y
y
2
2
˝
-˝
-2
x
˝
x
-2
Solution The x-intercepts of y = 2 sin 2x correspond to the vertical
asymptotes of y = 2 csc 2x. Thus, we draw vertical asymptotes through the xintercepts. Using the asymptotes as guides, we sketch the graph of y = 2 csc 2x.
Graphs of Other
Trigonometric
Functions
Inverse
Trigonometric
Functions
The Inverse Sine Function
The inverse sine function, denoted by sin-1, is the inverse of the restricted sine
function y = sin x, -  /2 < x <  / 2. Thus,
y = sin-1 x means sin y = x,
where -  /2 < y <  /2 and –1 < x < 1. We read y = sin-1 x as “ y equals the
inverse sine at x.”
y
y = sin x
-  /2 < x < /2
1
-  /2
 /2
-1
x
Domain: [-  /2,  /2]
Range: [-1, 1]
Finding Exact Values of
-1
sin x
• Let  = sin-1 x.
• Rewrite step 1 as sin  = x.
• Use the exact values in the table to find the
value of  in [-/2 , /2] that satisfies sin 
= x.
Example
• Find the exact value of sin-1(1/2)
1 1
  sin
2
1
sin  
2
 1
sin 
6 2


6
The Inverse Cosine Function
The inverse cosine function, denoted by cos1, is the inverse of the restricted cosine
function
y = cos x, 0 < x < . Thus,
y = cos-1 x means cos y = x,
where 0 < y <  and –1 < x < 1.
Text Example
Find the exact value of cos-1 (-3 /2)
Solution
Step 1 Let  = cos-1 x. Thus =cos-1 (-3 /2)
We must find the angle , 0 <  < , whose
cosine equals -3 /2
Step 2 Rewrite  = cos-1 x as cos  = x. We
obtain cos  = (-3 /2)
Text Example cont.
Find the exact value of cos-1 (-3 /2)
Solution
Step 3 Use the exact values in the table to
find the value of  in [0, ] that satisfies cos 
= x. The table on the previous slide shows that
the only angle in the interval [0, ] that satisfies
cos  = (-3 /2) is 5/6. Thus,  = 5/6
The Inverse Tangent Function
The inverse tangent function, denoted by
tan-1, is the inverse of the restricted tangent
function
y = tan x, -/2 < x < /2. Thus,
y = tan-1 x means tan y = x,
where -  /2 < y <  /2 and –  < x < .
Inverse Properties
The Sine Function and Its Inverse
sin (sin-1 x) = x for every x in the interval [-1, 1].
sin-1(sin x) = x for every x in the interval [-/2,/2].
The Cosine Function and Its Inverse
cos (cos-1 x) = x for every x in the interval [-1, 1].
cos-1(cos x) = x for every x in the interval [0, ].
The Tangent Function and Its Inverse
tan (tan-1 x) = x for every real number x
tan-1(tan x) = x for every x in the interval (-/2,/2).
Applications of
Trigonometric
Functions
Solving Right Triangles
Solving a right triangle means finding the missing lengths of its
sides and the measurements of its angles. We will label right
triangles so that side a is opposite angle A, side b is opposite angle
B, and side c is the hypotenuse opposite right angle C.
B
c
a
A
b
C
Text Example
B
Solve the right triangle shown.
c
a
34.5º
A
b = 10.5
C
Solution We begin by finding the the measure of angle B. We do not need
a trigonometric function to do so. Because C = 90º and the sum of a
triangle’s angles is 180, we see that A + B = 90º. Thus,
B = 90º – A = 90º – 34.5º = 55.5º.
Now we need to find a. Because we have a known angle, and unknown
opposite side, and a known adjacent side, we use the tangent function.
tan34.5º = a/10.5
Now we solve for a.
A = 10.5tan34.5=7.22
Text Example cont.
B
Solve the right triangle shown.
c
55.5º
7.22
34.5º
A
b = 10.5
Solution
Finally, we need to find c. Because we have a known angle, a known
adjacent side, and an unknown hypotenuse, we use the cosine function.
os34.5 = 10.5/c
c=10.5/cos34.5 = 12.74
In summary, B = 55.5º, a = 7.22, and c = 12.74.
C
Text Example
Use the figure to find: a. the bearing from O to B. b. the bearing from O to A.
N
B
40º
A
20º
W
75º
O
C
Solution
E
D
25º
S
a. To find the bearing from O to B, we need the acute angle between the
ray OB and the north-south line through O. The measurement of this
angle is given to be 40º. The figure shows that the angle is measured
from the north side of the north-south line and lies west of the northsouth line. Thus, the bearing from O to B is N 40º W.
Text Example cont.
Use the figure to find: a. the bearing from O to B. b. the bearing from O to A.
N
B
40º
A
20º
W
75º
O
C
E
D
25º
Solution
S
b. To find the bearing from O to A, we need the acute angle between the
ray OA and the north-south line through O. This angle is specified by
the voice balloon in the figure. The figure shows that this angle
measures 90º – 20º, or 70º. This angle is measured from the north side
of the north-south line. This angle is also east of the north-south line.
This angle is also east of the north-south line. Thus the bearing from O
to A is N 70º E.
Simple Harmonic Motion
An object that moves on a coordinate axis is in
simple harmonic motion if its distance from the
origin, d, at time t is given by either
d = a cos  t or d = a sin  t.
The motion has amplitude |a|, the maximum
displacement of the object from its rest position.
The period of the motion is 2/ , where  > 0.
The period gives the time it takes for the motion to
go through one complete cycle.
Example
• An object in simple harmonic motion has a
frequency of 1/4 oscillation per minute and
an amplitude of 8 ft. Write an equation in
the form d  a sin t for the objects
simple harmonic motion.
Example cont.
• Solution:
• a = 8 and the period is 4 minutes since it
travels 1/4 oscillation per minute
2
4

2  4


2
d  8 sin

2
t
Frequency of an Object in Simple
Harmonic Motion
• An object in simple harmonic motion given
by d=acost or d = asint has frequency f
given by f = /2, >0.
• Equivalently, f = 1/period.
Example
• A mass moves in simple harmonic motion
described by the following equation, with t
measured in seconds and d in centimeters.
Find the maximum displacement, the
frequency, and the time required for one
cycle.
d  8 cos

3
t
Example cont.
• Since a=8, the maximum displacement is 8
cm.

d  8 cos t
3
a 8
Example cont.
• The frequency is 1/6 cm per second.

d  8 cos t
3


3


1
3
f 


2 2 6
Example cont.
• The time required for one cycle is 6
seconds.

d  8 cos t
3


3
period 
2


2

3
6