Chapter 7

Chapter 7 -...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter
7:

Trigonometric
Functions
of
Real
Numbers:
 
 Radian
Measure
(7.1)
 
 Definition:

Place
the
vertex
of
the
angel
at
the
center
of
a
circle
with
radius
r
as
 shown
below.

Let
s
denote
the
length
of
the
arc
intercepted
by
the
angel
θ .

The
 s arc length of s radian
measure
of
the
angel
θ 
=
 = 
 r radius r 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Ex:

Find
θ 
for
s
=
200cm,
r
=
1m
 
 
 
 
 
 
 
 
 
 
 
 Note:

1)

Radian
measure
is
a
“dimensionless”
quantity.

It
is
the
ratio
of
lengths
 with
same
units.
 
 
 
 
 
 
 
 
 2)

Radian
measure
of
angel
θ 
is
independent
of
the
size
of
the
circle
in
which
it
is
 placed.

See
below:
 
 
 
 
 
 
 
 3)


For
a
full
circle
(θ =3600)
 
 s 2πr 
 θ= = = 2π r r θ = 3600 in degree = 2π in radian π in radian = 360 = 1800 2 
 
 
 
 
 
 
 
 
 
 
 The
conversion
formula
between
radian
and
degree
can
be
derived
with
the
 information
above.
 
 1800 



(to
convert
angel
to
degree
measure
multiply
by
this
factor)
 1 radian = π 
 
 
 π Similarly

10 = 

(to
convert
angel
to
radian
measure
multiply
by
this
factor)
 1800 
 
 
 
 Here
is
a
table
with
the
Unit
Circle
special
angels
expressed
in
both
radian
and
 degree
measure.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Ex:

Express
50
in
radian
measure
 
 
 
 
 
 
 
 
 
 11π Ex:

Express
 
radian
to
degree
measure
 6 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Ex:

Compute
the
value
of
the
following
trigonometric
functions:
 
 π cos( ) = 3 π sin( ) = 3 cos( −2π )= 3 sin( −2π ) 3 tan( 7π ) 4 
 cot( −7π ) 4 cos 4 = sin 4 = cos( −1) = sin( −1) = 
 Radian
Measure
and
Geometry
(7.2)
 
 
 Remember:

In
a
circle
of
radius
r,
the
arc
length
s
determined
by
a
central
angel
θ 
is

 s θ = 






(s
and
r
have
the
same
units)






(see
picture
below)
 r 
 
 
 
 
 
 
 
 
 
 
 
 s
=
rθ 





 
 
 The
formula
to
find
the
arc
length
of
a
circular

 sector
with
radius
r
and
intercepted
angel
θ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 The
area
of
a
sector
with
central
angel
θ 
is
 1 A = r 2θ 
 2 
 
 Proof:
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Ex:

Given
r=6cm,





θ = 2π 


Find
s
and
A
 3 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Ex:

Given
r
=
4
inches,
θ =750






Find
s
and
A.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Ex:

In
a
circle
of
radius
3
m,
the
area
of
a
certain
sector
is
20
m2.

Find
the
degree
 measure
of
the
central
angel.
 
 
 
 
 
 
 
 
 
 Definition:

Angular
speed
and
Linear
speed
 
 
 Suppose
a
wheel
rotates
about
its
axis
at
a
constant
rate
.

If
a
radial
line
turns
 through
an
angel
θ 
in
time
t
(see
picture)
the
angular
speed

 
 θ angle travelled 
 ω= = t time taken 
 
 
 
 
 
 
 
 
 
 
 Suppose
a
wheel
turns
about
its
axis
at
a
constant
rate.

It
the
point
P
on
the
rotating
 wheel
turns
a
distance
s
in
time
t,
the
linear
speed
of
P,
 
 s arc length travelled 
 v= = t time taken Note:

In
both
definitions,
θ 
will
be
in
radian
measure.


 
 Claim:

 v = rω 




































Linear
speed=(radius)(angular
speed)
 
 Proof:
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Ex:

A
circular
gear
in
a
motor
rotates
at
the
rate
of
100
rpm
(revolutions
per
 minute).
 
 a) What
is
the
angular
speed
of
the
gear,
in
radians
per
minute?
 
 
 
 
 
 
 
 
 
 b) Find
the
linear
speed
of
a
point
on
the
gear
4
cm
from
the
center.
 
 
 
 
 
 
 
 
 
 
 Ex:

The
rotation
of
a
wheel
is
10800/sec,


r
=
25
cm.
 
 a) Find
the
angular
speed
in
radian/sec
 
 
 
 
 
 
 
 
 
 
 
 
 
 b) Find
the
linear
speed
in
cm/sec
of
a
point
on
the
circumference
of
the
wheel
 
 
 Ex:

It
is
roughly
true
that
the
earth
revolves
once
every
24
hours
on
an
axis
through
 the
north
and
south
poles.
 
 a) What
is
the
angular
speed
that
the
earth
travels
in
radians
per
hour?
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 b) Given
that
the
radius
of
the
earth
is
approximately
3960
miles,
what
is
the
 linear
speed
in
miles
per
hour
with
which
a
person
at
the
equator
is
 travelling
through
space
at
any
given
moment?

Using
the
calculator
,
 approximate
it
to
the
nearest
mile
per
hour.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Trigonometric
Functions
of
Real
Numbers

(7.3)
 
 
 Let
P
=
(x,
y)
denote
the
point
on
the
unit
circle
that
has
angel
measure
t
from
(1,
0).

 Let
P
denote
the
point
where
the
terminal
side
of
the
angel
with
the
radian
measure
 t
intersects
the
unit
circle
(see
the
picture
below).

The
six
trig
functions
of
real
 number
t
are
defines
as
follows:


(Remember
SOHCAHTOA)
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 y 1 sin t = csct = ( y ≠ 0) 1 y cos t = x 1 sect = 1 x (x ≠ 0) 
 tant = y (x ≠ 0) x cott = x y (y ≠ 0) 
 
 Note:

You
can
view
the
input
angel
t
as
a
real
number
(since
it
is
in
radian
 measure).
 
 
 Evaluate:


 
 2π cos( ) 3
 
 
 
 
 sin( 
 
 
 
 
 
 
 
 
 
 
 
 tan( −5π ) =
 4 7π ) =
 4 
 
 
 
 
 
 
 
 
 
 
 
 Review
of
basic
identities:
 
 
 sin t tan t = cos t cott = cost sint sect = 1 cost 
 csc t = 1 sin t 
 
 Pythagorian
Identites:
 
 
 sin 2 t + cos 2 t = 1 
 cott = 1 tant 1 + tan 2 t = sec2 t 1 + cot 2 t = csc2 t 
 Opposite
–
Angel
Identities:
 
 
 sin( −t ) = − sin t cos(-t) = cost 
 
 
 
 
 
 3 π Ex:

 If sint = and < t < π , compute tant 
 4 2 
 
 
 
 
 
 
 
 
 
 3 3π If sin(-s) = and π < s < , compute 5 2 sins = coss = cos(-s) = tans + tan(-s) = 
 
 tan(-t) = -tant Graphs
of
sine
and
cosine
functions

(7.4):
 
 The
focus
of
this
section
will
be
on
studying
sine
and
cosine
functions.

They
are
 periodic
functions.

To
understand
what
it
means
to
be
periodic,
let’s
define
some
 vocabulary.
 
 Definition:

A
non‐constant
function
f
is
said
to
be
periodic
if
there
is
a
number
p>0
 such
that

f(x+p)
=
f(x)
for
all
x
in
the
domain
of
f.

The
smallest
such
number
p
is
 called
the
period.
 
 
 Ex:

Graph
f(x)
=
sinx
with
period
 2π 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Definition:
Let
f

be
a
periodic
function
and
let
m
and
M
denote
the
minimum
and
 maximum
values
of
the
function
respectively.


 
 M −m Then
the
amplitude
of
f
=
 
 2 
 The
example
shows
the
amplitude
of
f(x)
=
sinx
as
 
 
 Ex:

Find
the
amplitude
of
the
following
graph:
 
 
 
 
 
 
 
 
 
 
 
 
 Property
Summary:
 f ( t ) = sin t 
 
 
 
 Basic
graph:
 
 
 
 
 
 
 
 
 
 
 
 
 Domain:
 
 Range:
 
 Period:
 
 Amplitude:
 
 Since
sin(‐t)
=
‐sint,
it
is
an
odd
function
(symmetric
to
the
origin).

The
behavior
of
 the
basic
sine
function
is
it
crosses
the
x‐axis

at
the
beginning,
middle
and
end
of
 the
cycle.

The
curve
reaches
its
highest
point
one
quarter
of
the
way
through
the
 cycle
and
its
lowest
point
three
quarters
of
the
way
through
the
cycle.

See
the
 picture
above
 
 
 
 
 
 Property
Summary:

f(t)
=
cost
 
 Basic
Graph:
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Domain:
 
 
 Range:
 
 
 Period:
 
 
 Amplitude:
 
 
 Since
cos(‐t)=cost,
it
is
an
even
function
(symmetric
to
y‐axis).

The
basic
cosine
 function
crosses
the
x‐axis
one
quarter
of
the
way
and
again
three
quarter
of
the
 way.

The
curve
reaches
its
highest
point
at
the
beginning
and
end
of
the
basic
cycle.

 It
reaches
its
lowest
point
half
way
through
the
cycle.

See
the
picture
above.
 
 
 There
are
very
many
examples
of
periodic
function
generated
by
electricity,
sound
 wave
etc.

See
page
522‐523
in
your
textbook
for
more
details.
 
 
 
 
 Let’s
construct
the
basic
graphs
of
sine
and
cosine
functions
by
using
our
knowledge
 of
special
angels.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 π Note:

 sin( x + ) = cos x or sin(x + 900 ) = cos x 
as

observed
in
the
above
graph
and
 2 the
table.

 Graphs
of
 y = A sin( Bx − C ) and y = Acos(Bx - C)


(7.5)
 
 
 In
order
to
investigate
the
role
of
A

 
 Graph
 y = sin x , y = 3sinx, y = -3sinx 

on
the
same
set
of
axis.
 
 
 Table:
 
 
 
 
 
 
 
 
 
 Graph:
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Conclusion:

 A 
=
Amplitude
=
How
high
or
low
the
graph
will
attain
 Investigating
the
role
of
B
 
 Graph
 y = cos x , y = cos2x, y = cos4x on
the
same
set
of
axis.


 
 



 
 
 
 
 
 
 
 Table:
 
 
 
 
 
 
 
 Graph:
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Conclusion:

The
period
of
y
=
cosx:
 
 
 










The
period
of
y
=
cos2x
 
 
 











The
period
of
y
=
cos4x
 
 
 In
general,
the
period
of
y
=
cos(Bx)
 
 Let’s
combine
the
ideas
of
amplitude
and
period
and
graph
 
 Ex:

Graph
y
=
2sin(3x)
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Given
the
following
graph,
figure
out
the
function
by
figuring
out
the
values
of
A
and
 B
in
the
form
y
=
Asin(Bx)
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Now
let’s
cover
the
concept
of
Horizontal
Phase
shift
and
Vertical
shift.
 
 Horizontal
Phase
shift:

In
order
to
understand
the
role
of
C,
let’s
graph

 
 2π π y = 4 sin( 2x − ) = 4 sin( 2( x − )) 

 3 3 
 Note:

This
equation
is
obtained
from
y
=
4sin(2x)
by
replacing
x
with
 ( x − Thus,
the
graph
of
 y = 4 sin( 2( x − right
by
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 π 
 3 π )
 3 π )) 
is
translating
the
graph
of
y
=
4sin(2x)
to
the
 3 Conclusion:


Horizontal
Phase
shift
of
a
sine
or
cosine
graph
can
be
obtained
by
 C calculating
 − 
from
the
original
form
of
the
 B equations y = A sin( Bx − C ) and y = Acos(Bx - C).

Another
word,
the
graphs
of
 y = A sin( Bx − C ) and y = Acos(Bx - C)
are
obtained
by
horizontally
translating
the
 graphs
of
y
=
AsinBx
and
y
=
AcosBx
respectively
so
that
the
starting
point
of
the
 basic
cycle
is
shifted
from
x=0
to
x=‐C/B
 
 
 
 
 π Ex:

Graph
 y = −4 cos( 3x + ) 
by
figuring
out
the
amplitude,
period,
horizontal
 4 phase
shift
and
reflection
across
the
x‐axis.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 π π⎤ ⎡ ) = −4 cos⎢3( x + )⎥ 

then
it
makes
sense
to
say
that
 4 12 ⎦ ⎣ π the
graph
of
 y = −4 cos( 3x + ) 
is
obtained
by
horizontally
shifting
the
graph
of
 4 π 
units
to
the
left.

So
it
shows
that
the
formula
for
the
horizontal
 y = −4 cos 3x , 12 phase
shift
written
above
works.
 
 
 
 
 
 The
other
kind
of
shift
that
you
will
have
to
perform
on
the
basic
graphs
is
vertical
 shift.

This
shift
is
very
easy
to
recognize.

For
example
let’s
add
on
the
following
 feature
to
the
graph
that
we
just
graphed.
 
 π y = −4 cos( 3x + ) + 1 
 4 
 This
means
that
the
previous
graph
is
shifted
1
unit
up
vertically
 
 In
general
for
equations
of
the
form
 y = A sin( Bx − C ) + D and y = Acos(Bx - C) + D ,
 if
the
constant
D
is
added,
the
graph
shifts
vertically
up
D
unit
and
if
the
constant
D
 is
subtracted,
the
graph
shifts
down
D
units.
 
 π Now
let’s
look
at
the
final
graph
of
 y = −4 cos( 3x + ) + 1 
 4 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Check:

Since
 y = −4 cos( 3x + Graphs
of
the
Tangent
Function
(7.7)
 
 This
section
will
focus
on
the
graphs
of
the
y
=
tanx,
y
=
cscx,
y
=
secx
and
y
=
cotx.
 
 Like

y=
sinx
and
y
=
cosx,
these
new
trig
functions
are
also
periodic.

The
difference

 
 is
that
these
new
four
functions
posses
vertical
asymptotes.

Let’s
focus
at
first
on

 
 the

graph
of
y
=
tanx
 
 sin x Recall:

 y = tan x = 
 cos x 
 Table:
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Looking
at
the
denominator,
as
x
varies
from
 π π 0 to cosx will vary from cos0 = 1 to cos = 0.

So
the
value
of
y
=
tanx
will
get
 2 2 larger
and
larger
as
the
denominator
gets
smaller
and
smaller
between
 π x = 0 to x = .

Notice
that
the
value
of
tanx
increases
slowly
at
first
and
more
 2 π rapidly
as
cosx
approaches
the
value
of
0
at
 x = .

 2 
 Let’s
construct
the
graph
of
y
=
tanx
by
using
the
table
above:
 
 
 
 
 
 
 
 
 
 
 
 We
can
finish
up
this
graph
without
plugging
in
any
additional
values
remembering
 sin(-x) − sin x the
fact
that
 tan( − x ) = − tan x because tan(-x) = = = − tan x 
 cos(-x) cos x 
 Therefore,
we
can
conclude
that
tanx
is
an
odd
function.

Thus,
it
is
symmetric
about
 the
origin.

The
graph
with
a
full
cycle
looks
as
follows:
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 You
can
observe
from
the
graph
above
that
the
period
for
 sin(x + π ) − sin x y = tan x is π because tan(x + π ) = = = tan x 
 cos(x + π ) − cos x 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Ex:

Graph
the
following
tangent
functions
for
one
period:
 
 y = tan x y = tan( x + π ) 4 y = -tan(x + 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 π ) 4 translation of y = tanx, π units to the left 4 translation of the graph above across the x - axis 
 Ex:

Graph
the
following
functions
for
one
period:
 
 1 y = tan x stretch the graph of y = tanx by a factor of 2 2 y = 2tanx compress the graph of y = tanx by a factor of 1 
 2 y = tan 2x change the period of y = tanx 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Note:

The
period
or
a
full
cycle
for
the
graph
of
 y = tan x is π .


Therefore,
the
period
 π of
 y = A tan Bx is 
by
the
following
justification:
 B 
 
 
 
 
 
 
 
 Property
Summary
of
 y = tan x , origin.

The
lines
 x = ± -π π < x < .

The
graph
is
symmetric
about
the
 2 2 π 
are
vertical
asymptotes.
 2 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Domain:
 
 Range:
 
 Period:
 
 Asymptotes:
 
 x‐intercepts:
 
 y‐intercepts:
 
 
 
 
 Note:

The
x‐intercept
takes
place
mid
way
between
consecutive
vertical
asymptotes
 due
to
the
symmetry
of
the
graph.
 
 To
graph
the
cotangent
function
 y = cot x remember the fact cotx = -tan(x 
 Proof:
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Therefore,
we
can
graph
 y = cot x 
by
shifting
the
 y = tan x graph and
then
reflecting
it
across
the
x‐axis.
 
 Graph
of
y
=
cotx
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 π )
 2 π 
units
to
the
right
 2 Property
Summary
for
 y = cot x (cotx = -tan(x - π ))
 2 
 Domain:
 
 Range:
 
 Period:
 
 Asymptotes:
 
 x‐intercepts:
 
 Finally
we
will
construct
the
graphs
of
 y = csc x and y = secx .

The
graph
of
 y = csc x 
 1 is
closely
related
to
the
graph
of
 y = sin x 
since
 csc x = .

So
the
vertical
 sin x asymptotes
for
cscx
is
when
 sin x = 0, which implies x = o, ± π ,±2π ,.......................
 
 First
make
a
table
y
=
cscx
keeping
in
mind
the
vertical
asymptotes.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Graph:
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 The
graph
of
 y = csc x 
is
symmetric
about
the
origin
because

 
 1 1 csc( − x ) = = = − csc x 
 sin( −x ) − sin x 
 Let’s
draw
one
period
of
the
graph
of
 y = csc x 
 
 Notice
how
it
is
related
to
the
y
=
sinx
graph.
 
 Property
Summary:
 
 Domain:
 
 Range:
 
 Period:
 
 Asymptotes:
 
 Intercepts:
 
 
 
 
 
 
 Finally,
to
graph
 y = sec x ,
remember
the
following
identities:
 
 π 1 sin( x + ) = cos x and = sec x 2 cosx 
 Therefore we can conclude secx = 1 = cosx 1 sin( x + 
 
 Take
 y = csc x 
graph
and
horizontally
move
it
 y
=
secx
 
 
 
 
 π ) 2 = csc( x + π ) 2 π 
to
the
left
to
obtain
the
graph
of

 2 Graph
of
y
=
secx
from
the
graph
of
y
=
cosx:
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Property
Summary:
 
 Domain:
 
 
 Range:
 
 
 Period:
 
 Asymptotes:
 
 y‐intercept
 
 x‐intercept
 
 
 The
same
rules
of
shifting,
reflecting
and
stretching
of
graphing
applies
to
 each
and
every
trigonometric
function.
 
 
 
 
 
 
 
 
 
 ...
View Full Document

Ask a homework question - tutors are online