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Remote Learning Packet Please submit scans of written work in Google Classroom at the end of the week.
May 4-8, 2020 Course: 10 Precalculus Teacher(s): Mr. Simmons Weekly Plan: Monday, May 4 ⬜ Story time! ⬜ Problems 11-14 and 16 from “Relationship between Trig Functions” Tuesday, May 5 ⬜ Read “Radian measure.” Wednesday, May 6 ⬜ Problems 1-14 Thursday, May 7 ⬜ Read “Introduction to the Polar Plane Friday, May 8 ⬜ Attend office hours ⬜ Catch up or review the week’s work Statement of Academic Honesty I affirm that the work completed from the packet is mine and that I completed it independently. _______________________________________Student Signature
I affirm that, to the best of my knowledge, my child completed this work independently _______________________________________ Parent Signature
Monday, May 4
1. Story time! If technologically feasible, email me with a story! Also, please email me if you have questions about trigonometry. BUT. Please make your questions specific. At least give me a page number where you got confused. Better yet, tell me exactly what words you read that confused you. For today, we’re going to do a few more problems from “Relationships between Trig Functions” before moving on to radians:
2. Complete Problems 11-14 and 16 on pp. 122-123. Tuesday, May 5
1. Read “Radian Measure” on pp. 124-133. Wednesday, May 6
1. Complete Problems 1-14 on pp. 133-135. Thursday, May 7
1. Read “Introduction to the Polar Plane” on pp. 135-143.
Uni
t fou
r – 1
22
2.)
Let u
s co
mpl
ete
Exam
ple
1c. I
f sin𝛼
=𝑎
, wha
t are
the
six
Trig
ratio
s?
3.)
Now
let t
an𝛼
=𝑑
. (A
) W
hat a
re th
e six
Trig
ratio
s gi
ven
this?
(B
) Co
mpa
re th
is an
swer
to th
e pr
evio
us.
4.)
In E
xam
ple
1c, w
e le
t th
e le
ngth
of t
he h
ypot
enus
e of
the
tria
ngle
to
equa
l 1. I
s th
at O
K? W
hy d
on’t
we
let
the
hypo
tenu
se e
qual
𝑐,
to a
llow
for
any
and
all
poss
ibili
ties?
Now
sup
pose
sin𝛼
=𝑥 𝑐, a
nd th
e le
ngth
s of
you
r tria
ngle
are
𝑎,𝑏
, and
𝑐,
with
the
hypo
tenu
se e
qual
ing 𝑐.
(A
) W
hat a
re th
e six
Trig
ratio
s?
(B)
Com
pare
this
with
the
prev
ious
two
resu
lts.
5.)
Giv
en th
e Tr
ig fu
nctio
ns a
nd a
ngle
mea
sure
, writ
e th
e eq
uiva
lent
cof
unct
ion.
(A
) si
n3
0°
(B)
cos
10
° (C
) co
t7
° (D
) se
c64
° (E
) co
s3
1°
(F)
tan
14
° (G
) cs
c4
7°
(H)
sin2
5°
(I)
tan( 𝛽
+𝛾)
(J)
sin𝛽
6.
) W
rite
out a
ll of
the
cofu
nctio
n id
entit
ies,
incl
udin
g th
e on
es w
e di
scov
ered
in th
e re
adin
g. H
int:
The
re a
re s
ix o
f the
m.
7.)
Now
writ
e ou
t all
of th
e re
cipr
ocal
iden
titie
s. H
int:
The
re a
re s
ix o
f the
m.
8.)
Eval
uate
the
follo
win
g.
(A)
sin
23
0°
(B)
cos2
30
° (C
) ta
n2
30
° (D
) co
s24
5°
(E)
tan
24
5°
(F)
sin
26
0°
(G)
cos2
60
° (H
) ta
n2
60
° (I)
si
n( 3
0°2)
(J)
cos2𝛼
9.
) W
rite
out a
tabl
e of
val
ues f
or s
in2𝛼
,co
s2𝛼
, and
tan
2𝛼
, sta
rtin
g at
𝛼=
0, g
oing
up
by 5
° eac
h ro
w, a
nd e
ndin
g at
𝛼=
90
°. Yo
u w
ill n
eed
a ca
lcul
ator
for t
his
Exer
cise
. 10
.) U
se y
our r
esul
ts fr
om th
e pr
evio
us E
xerc
ise to
ans
wer
the
follo
win
g qu
estio
ns.
(A)
Wha
t is
the
max
imum
val
ue o
f sin
2𝛼
,co
s2𝛼
, and
tan
2𝛼
? (B
) W
hat i
s th
e m
inim
um v
alue
of s
in2𝛼
,co
s2𝛼
, and
tan
2𝛼
? (C
) Ar
e th
ere
any
simila
ritie
s or
diff
eren
ces
betw
een
sin
2𝛼
and
sin𝛼
? Co
mpa
re
your
resu
lts fr
om th
e pr
evio
us s
ectio
n.
11.)
One
of
the
mos
t im
port
ant
rela
tions
hips
in
Trig
onom
etry
is
the
Pyth
agor
ean
Iden
tity
we
disc
usse
d in
the
read
ing.
Writ
e th
is id
entit
y do
wn
now
. 12
.) Ev
alua
te th
e fo
llow
ing.
(A
) si
n2
60
°+
cos2
60
° (B
) co
s23
0°
+si
n2
30
° (C
) si
n2(𝛼
2)
+co
s2(𝛼
2)
(D)
sin
2( 3𝛼
+𝜋)
+co
s2( 3𝛼
+𝜋)
13.)
It is
ofte
n he
lpfu
l to
rew
rite
sin
2𝛼
or c
os2𝛼
. Use
the
Pyth
agor
ean
Iden
tity
to re
writ
e si
n2𝛼
and
co
s2𝛼
. 14
.) Si
mpl
ify th
e fo
llow
ing.
§3
Rel
atio
nshi
ps b
etw
een
the
Trig
func
tions
– 1
23
(A)
tan𝛼∙c
sc𝛼
(B
) ( s
in𝛼
+co
s𝛼)2
15
.) Ar
e th
ere
any
othe
r Pyt
hago
rean
Iden
titie
s? T
o fin
d th
is o
ut, u
se a
cal
cula
tor a
nd
try
the
follo
win
g fo
r diff
eren
t val
ues
of 𝛼
. (A
) se
c2𝛼
+cs
c2𝛼
(B
) ta
n2𝛼
+co
t2𝛼
(C
) Th
ere
are
two
othe
r Py
thag
orea
n Id
entit
ies.
Firs
t, us
ing
the
prev
ious
tw
o,
gues
s w
hat t
hey
mig
ht b
e. T
hen,
if y
ou c
an’t
figur
e it
out,
look
them
up
and
writ
e th
em d
own
now
. We’
ll di
scov
er h
ow to
arr
ive
at th
ese
resu
lts w
hen
we
have
som
e be
tter
tool
s. 16
.) An
swer
Tru
e or
Fal
se.
(A)
sin𝛼
=co
s(𝛼−
90
°)
(B)
sin
2𝛼
+co
s2𝛽
=1
iff 𝛼
+𝛽
=9
0°
(C)
sin
2𝛼
=si
n𝛼∙𝛼
(D
) si
n2𝛼
is s
omet
imes
neg
ativ
e.vi
vi A
ssum
e 𝛼∈ℝ
.
Uni
t fou
r – 1
24
Uni
t fiv
e Ra
dian
s an
d th
e U
nit C
ircle
“Deg
rees
are
fin
e fo
r ev
eryd
ay m
easu
rem
ents
. Bu
t T
rigo
nom
etry
mar
ks a
tu
rnin
g p
oin
t in
mat
h, w
hen
th
e st
ud
ent
lift
s h
is g
aze
from
th
e ev
eryd
ay t
owar
ds
larg
er, m
ore
dis
tan
t id
eas.
You
beg
in e
xplo
rin
g ba
sic
rela
tion
ship
s, d
eep
sym
met
ries
, th
e ki
nd
s of
pat
tern
s th
at m
ake
the
un
iver
se t
ick.
An
d t
o n
avig
ate
that
ter
rain
, you
nee
d a
not
ion
of a
ngl
es
that
’s m
ore
nat
ura
l, m
ore
fun
dam
enta
l, th
an s
lici
ng
up
th
e ci
rcle
into
an
arb
itra
ryn
um
ber
of p
iece
s. T
he
nu
mbe
r π
, str
ange
th
ough
it m
ay s
eem
, lie
s at
th
e h
eart
of
mat
hem
atic
s. T
he
nu
mbe
r 36
0 d
oesn
’t. C
lin
gin
g to
th
at B
abyl
onia
n a
rtif
act
wil
l on
ly
dis
trac
t yo
u a
nd
obs
cure
th
e el
egan
t tr
uth
s yo
u’r
e se
arch
ing
for.
”
Ben
Orl
in
§1
Rad
ian
mea
sure
– 1
25
Up
to t
his
poin
t, w
e’ve
mea
sure
d al
l of
our
ang
les
usin
g de
gree
s. In
thi
s un
it, w
e’ll
ende
avor
to fi
nd a
diff
eren
t and
per
haps
bet
ter m
etho
d of
mea
surin
g an
gles
. The
n w
e’ll
use
that
to g
raph
poi
nts
in a
new
type
of p
lane
. Fin
ally
, afte
r thi
s, w
e in
trod
uce
perh
aps
the
mos
t im
port
ant t
hing
in T
rigon
omet
ry:
The
Uni
t Circ
le.
§1
Radi
an m
easu
re
Deg
rees
wer
e in
vent
ed m
illen
nia
ago,
per
haps
by
the
anci
ent
peop
les
livin
g in
mod
ern
day
Iraq.
Kno
win
g th
e or
igin
s of
thi
s un
it co
uld
shed
som
e lig
ht o
n its
use
fuln
ess,
and
whe
ther
ther
e isn
’t a
mor
e us
eful
uni
t to
use.
Ther
e ar
e va
rious
the
orie
s as
to
why
deg
rees
wer
e us
ed a
nd w
hy t
hey
are
the
way
tha
t ar
e. A
lmos
t cer
tain
ly, h
owev
er, i
t has
to d
o w
ith a
circ
le. A
s w
ith a
nyth
ing,
it’s
ofte
n us
eful
to
cons
ider
por
tions
or
fract
ions
of t
he w
hole
.i The
anc
ient
s ch
ose
to c
hop
the
circ
le u
p in
to 3
60
equ
al p
ortio
ns, c
allin
g th
e an
gle
crea
ted
by e
ach
port
ion
a de
gree
, as
(par
tially
) sh
own
in F
igur
e 47
.
Why
36
0?
Perh
aps
beca
use
it is
a n
ice
num
ber
with
man
y fa
ctor
s. So
cut
ting
a ci
rcle
in
half
give
s yo
u a
nice
num
ber o
f 18
0°,
in th
irds
12
0°,
four
ths
90
°, an
d so
on.
Thi
s m
eans
i T
his
is w
hy, f
or e
xam
ple,
we
have
yar
ds. C
ould
you
imag
ine
mea
surin
g th
ings
if th
e sm
alle
st u
nit w
e co
uld
get w
as m
iles?
And
eve
n th
at is
n’t e
noug
h, w
hich
is w
hy c
ontin
ue to
sub
divi
de th
e un
its s
mal
ler a
nd
smal
ler.
Figu
re 4
7
Each
of t
he in
divi
dual
spo
kes
abov
e m
easu
res
a sin
gle
degr
ee. I
f we
wer
e to
con
tinue
cre
atin
g th
ese
spok
es, t
here
wou
ld
be 3
60
of t
hem
.
Uni
t fiv
e –
126
that
com
mon
ly u
sed
ratio
s ar
e le
ft w
ith a
who
le n
umbe
r. Th
is w
ould
n’t b
e th
e ca
se if
the
num
ber,
say,
10
, was
use
d. T
hen
only
a h
alf-
circ
le a
nd a
fifth
-of-
a-ci
rcle
wou
ld h
ave
who
le
num
bers
. Ano
ther
sup
posit
ion
is th
at t
here
are
app
roxi
mat
ely
36
0 d
ays
in a
yea
r. An
d sin
ce, e
ach
year
, sea
sons
rep
eat
them
selv
es, a
circ
le m
akes
a n
ice
repr
esen
tatio
n of
a
cale
ndar
.
Wha
teve
r the
reas
on, h
owev
er, w
e w
ant t
o se
e if
ther
e is
a b
ette
r way
of m
easu
ring
angl
es.
Of
cour
se, “
bett
er”
is r
elat
ive,
and
diff
eren
t si
tuat
ions
mig
ht c
all f
or d
iffer
ent
units
. So
whe
n w
e sa
y “b
ette
r,” p
erha
ps w
hat
we
shou
ld s
ay is
mor
e ap
prop
riate
for
our
wor
k in
Tr
igon
omet
ry.
Cons
ider
the
circ
le s
how
n in
Fig
ure
48. W
hat i
s th
e le
ngth
of t
he a
rc fr
om 𝐴
to 𝐵
?
Ther
e ar
e a
few
way
s w
e co
uld
answ
er t
his
ques
tion.
One
is
to m
easu
re i
t th
e ol
d-fa
shio
ned
way
. Tha
t, ho
wev
er, l
eave
s ro
om fo
r err
or, a
nd w
ould
n’t h
elp
us to
mea
sure
an
arc
from
a d
iffer
ent c
ircle
. Ano
ther
way
we
coul
d do
it is
to fi
nd th
e ci
rcum
fere
nce
of th
e ci
rcle
, the
n m
ultip
ly b
y th
e fra
ctio
n of
the
outs
ide
of th
e ci
rcle
repr
esen
ted
by 𝐴
��.ii T
his
isn’t
the
wor
st t
hing
in t
he w
orld
, but
the
n… H
ow w
ill w
e m
easu
re t
hean
gle
whi
ch w
ill
allo
w u
s to
find
the
fract
ion
of th
e ou
tsid
e of
the
circ
le th
at 𝐴��
take
s up
? As
you
can
see
, w
e ha
ve a
bit
of a
n is
sue.
As w
e’ve
don
e a
few
tim
es in
this
cou
rse,
we
shou
ld g
o ba
ck to
wha
t we
know
for c
erta
in.
We
know
that
the
circ
umfe
renc
e of
a c
ircle
is
𝐶=2𝜋𝑟,
ii F
or e
xam
ple,
if th
e ar
c w
ere
half
of th
e ou
tsid
e of
the
circ
le, y
ou w
ould
mul
tiply
the
circ
umfe
renc
e of
the
circ
le b
y 1 2, r
ight
?
Figu
re 4
8
§1
Rad
ian
mea
sure
– 1
27
whe
re 𝑟
is th
e ra
dius
of a
circ
le a
nd 𝜋
is th
e m
athe
mat
ical
con
stan
t app
roxi
mat
ely
equa
l to
3.14
. We
also
kno
w th
at e
very
radi
us in
a c
ircle
is c
ongr
uent
. And
that
’s ab
out i
t. Bu
t th
is do
es s
how
us
that
if w
e’re
tryi
ng to
figu
re s
tuff
out a
bout
a c
ircle
, it i
s us
ually
a g
ood
idea
to in
volv
e a
radi
us. T
hat’s
wha
t we’
ll do
in F
igur
e 49
.
In k
eepi
ng w
ith o
ur tr
aditi
on, w
e’ve
use
d th
e La
tin le
tter
s 𝑟
and 𝑎
for t
he le
ngth
s of
the
radi
us a
nd a
rc re
spec
tivel
y.
Now
, let
’s se
e w
hat h
appe
ns w
hen
we
rela
te th
e ra
dius
to th
e ar
c le
ngth
. Let
’s as
sum
e fo
r a
mom
ent t
hat 𝑟
=𝑎
, i.e
., th
at th
e ra
dius
is th
e sa
me
leng
th a
s 𝐴��
. Thi
s w
ould
allo
w u
s to
cr
eate
that
ang
le s
een
in F
igur
e 50
, rig
ht?
This
angl
e, w
hich
we’
ll ca
ll 𝛼
, is
uniq
ue. I
n ot
her w
ords
, the
re is
one
and
onl
y on
e an
gle
for w
hich
the
radi
us is
the
sam
e le
ngth
as 𝐴��
. Fig
ure
51 s
how
s th
is to
be
true
.
Figu
re 4
9
Figu
re 5
0
Uni
t fiv
e –
128
This
is in
tere
stin
g fo
r a c
oupl
e of
reas
ons.
Firs
t of a
ll, n
otic
e th
at th
ere
is on
e an
d on
ly o
ne
angl
e th
at c
omes
out
as
a co
nseq
uenc
e of
the
com
paris
on t
o th
e ra
dius
and
arc
leng
th.
Seco
ndly
, and
per
haps
mor
e im
port
antly
, the
siz
e of
the
circ
le (
and,
by
exte
nsio
n), t
he
leng
ths
of th
e ra
dius
and
arc
, won
’t m
atte
r.iii T
hus
we
mak
e th
e fo
llow
ing
defin
ition
.
Radi
an m
easu
re
The
angl
e fo
rmed
by
the
ratio
of t
he a
rc le
ngth
𝑎 to
the
radi
us 𝑟
in a
circ
le. S
ymbo
lical
ly,
𝛼𝑟𝑎𝑑
=𝑎 𝑟
,
whe
re 𝑟
is th
e ra
dius
and
𝑎 is
the
leng
th o
f the
arc
.
Ex
ampl
e 1a
Wha
t doe
s an
ang
le m
easu
re o
f 1 ra
dian
look
like
?
This
is a
n ex
celle
nt q
uest
ion,
and
a g
reat
star
ting
poin
t to
see
the
intu
ition
beh
ind
radi
ans.
A ra
dian
is a
ratio
bet
wee
n th
e ar
c le
ngth
and
the
radi
us, o
r 𝑎 𝑟
, whe
re 𝑟
is th
e ra
dius
of a
iii W
e’ll
show
this
expl
icitl
y in
the
fort
hcom
ing
Exam
ples
.
Figu
re 5
1
Her
e,𝑏
=1
.5𝑟.
As
a co
nseq
uenc
e of
this
incr
ease
d ar
c siz
e, th
e an
gle
is la
rger
, and
ther
efor
e 𝛼≠𝛽
.
§1
Rad
ian
mea
sure
– 1
29
circ
le a
nd 𝑎
is th
e le
ngth
of t
he a
rc w
e’re
con
cern
ed w
ith. U
sing
our d
efin
ition
, the
n, w
e ha
ve
1=𝑎 𝑟
.
But t
his
will
onl
y be
true
if 𝑎
=𝑟.
The
refo
re, a
mea
sure
of 1
radi
an w
ill lo
ok li
ke F
igur
e 52
, w
here
the
radi
us a
nd 𝐴��
are
equ
al.
Not
ice
that
our
ang
le m
easu
re is
1?
You
mig
ht b
e w
onde
ring
wha
t the
uni
ts o
f thi
s an
gle
mea
sure
are
, but
the
re a
ren’
t an
y! Y
ou c
ould
say
1 r
adia
n, b
ut if
an
angl
e m
easu
re is
re
port
ed w
ith n
o un
it, it
is a
ssum
ed to
be
mea
sure
d in
radi
ans.
iv
Did
you
not
ice
that
our
circ
le h
ad a
radi
us a
nd a
rc le
ngth
of 3
? W
e w
ante
d an
ang
le o
f 1,
and
that
is o
nly
true
whe
n 𝑟
=𝑎
. So
we
coul
d ha
ve a
lso
chos
en 𝑟
=𝑎
=5
, or 𝑟
=𝑎
=1
00
, (a
nd s
oon
) if w
e w
ante
d to
. Do
you
see
why
?
Exam
ple
1b
Wha
t doe
s an
ang
le o
f 2 lo
ok li
ke?
This
is a
sim
ilar q
uest
ion,
so
we
agai
n go
bac
k to
the
defin
ition
. The
equ
atio
n
2=𝑎 𝑟
mus
t be
tru
e. T
here
is a
n in
finite
am
ount
of
poss
ibili
ties
for
both
𝑎 a
nd 𝑟
, suc
h as
𝑎=
6,𝑟
=3
(whi
ch w
e sh
ow in
Fig
ure
53).
iv T
his
is a
noth
er re
ason
to p
refe
r rad
ians
.
Figu
re 5
2
We’
ve c
hose
n a
radi
us a
nd a
rc le
ngth
of 3
, but
we
coul
d ha
ve e
asily
cho
sen
any
othe
r len
gth
so lo
ng a
s 𝑟
=𝑎
.
Uni
t fiv
e –
130
The
prev
ious
tw
o ex
ampl
es w
ere
ther
e to
hel
p yo
u ge
t a
gras
p on
rad
ians
, but
we
still
ha
ven’
t see
n its
bes
t fea
ture
. We
expl
ore
that
now
.
Exam
ple
2a
Wha
t is
the
leng
th o
f the
radi
us g
iven
Fig
ure
54?
Now
this
is in
tere
stin
g. W
e ar
e gi
ven
an a
ngle
and
an
arc
leng
th, a
nd a
re to
ld to
find
the
leng
th o
f the
rad
ius.
Usi
ng t
he d
efin
ition
of a
rad
ian,
we
can
wor
k ba
ckw
ard
and
easil
y ge
t the
ans
wer
. Sin
ce, a
ccor
ding
to o
ur d
efin
ition
, we
have
3=24 𝑟
,
we
sim
ply
solv
e th
e pr
evio
us e
quat
ion
for 𝑟
and
get
𝑟=
8.
Easy
! But
to
real
ly a
ppre
ciat
e ra
dian
s, co
nsid
er F
igur
e 55
, whe
re w
e ha
ve u
sed
degr
ees
inst
ead.
Figu
re 5
3
As y
ou c
an s
ee, t
he ra
tio is
wha
t’s im
port
ant.
The
fact
that
the
arc
leng
th is
twic
e th
e le
ngth
of t
he ra
dius
is w
hat t
ells
us
we
have
an
angl
e m
easu
re o
f 2. A
ppre
ciat
e, a
lso, h
ow th
e an
gle
is cl
early
diff
eren
t fro
m th
e pr
evio
us E
xam
ple.
Figu
re 5
4
§1
Rad
ian
mea
sure
– 1
31
Coul
d w
e fin
d 𝑟?
The
ans
wer
is
no –
deg
rees
are
a m
easu
rem
ent
foun
d co
mpl
etel
y in
depe
nden
t of t
he s
ize
of a
circ
le. T
hus
it of
fers
us
no h
elp
at a
ll. R
adia
ns, t
here
fore
, giv
e us
free
info
rmat
ion!
Exam
ple
2b
Wha
t is
the
leng
th o
f the
arc
sub
tend
edv b
y an
ang
le o
f 6 a
nd a
radi
us o
f 10
?
No
pict
ure
is pr
ovid
ed, a
nd it
wou
ld b
e he
lpfu
l for
you
dra
w o
ne, b
ut it
is n
ot n
eces
sary
. W
e sim
ply
use
the
defin
ition
:
6=
𝑎 10
.
Hen
ce
𝑎=
60
.
Ther
e ar
e a
few
mor
e im
port
ant q
uest
ions
whi
ch m
ust b
e as
ked
if w
e ar
e to
suc
ceed
with
ra
dian
s. Fo
r exa
mpl
e, h
ow m
any
radi
ans
are
in a
full
rota
tion?
We
know
ther
e ar
e 3
60
° in
a fu
ll ro
tatio
n, b
ut w
hat a
bout
radi
ans?
Let u
s ans
wer
this
que
stio
n w
ith a
spec
ific
circ
le, a
nd th
en g
ener
aliz
e af
terw
ards
. Con
sider
a
circ
le w
ith a
radi
us o
f 1.vi
Sinc
e th
e de
finiti
on o
f a ra
dian
tells
us
that
𝛼𝑟𝑎𝑑
=𝑎 𝑟
,
And
we
have
𝑟=
1, w
e ha
ve
v T
his
is a
fanc
y, p
erha
ps o
ld-f
ashi
oned
wor
d w
hich
mea
ns fo
rmed
or c
reat
ed b
y. S
o th
e ar
c is
cre
ated
by
the
angl
e.
vi W
e co
uld
have
cho
sen
any
valu
e fo
r the
radi
us, b
ut w
e ch
ose
1. A
ny th
ough
ts o
n w
hy w
e w
ould
cho
ose
this
num
ber a
nd n
ot, s
ay, 2
3?
Figu
re 5
5
Uni
t fiv
e –
132
𝛼𝑟𝑎𝑑
=𝑎
.
If w
e ar
e co
nsid
erin
g a
full
rota
tion,
how
ever
, we
are
not
look
ing
at a
n ar
c, b
ut t
he fu
ll ci
rcum
fere
nce
of th
e ci
rcle
. The
refo
re, 𝑎
=2𝜋
and
hen
ce
𝛼𝑟𝑎𝑑
=2𝜋
.
This
is a
n im
port
ant f
act,
and
we
list i
t bel
ow fo
r you
r con
veni
ence
.
Radi
ans
in v
ario
us ro
tatio
ns
Full
rota
tion:
2𝜋
Hal
f rot
atio
n: 𝜋
Q
uart
er ro
tatio
n: 𝜋 2
Th
us, t
here
are
2𝜋
radi
ans
in a
full
rota
tion.
This
allo
ws
us t
o an
swer
our
nex
t m
ost
impo
rtan
t qu
estio
n:
How
do
radi
ans
rela
te t
o de
gree
s? In
oth
er w
ords
, how
doe
s on
e co
nver
t fro
m o
ne to
the
othe
r?
To fi
nd t
his
answ
er, l
et u
s fin
d ou
t ho
w m
any
degr
ees
are
in o
ne r
adia
n. T
o do
thi
s, w
e ju
st n
eed
to c
onve
rt. W
e w
ill u
se d
imen
sion
al a
naly
sis
to h
elp
us d
o th
is, a
s w
e sh
ow
belo
w.
36
0°
1 r
ota
tio
n
=
18
0°
≈
57
.30
° 1
ro
tati
on
2𝜋
rad
ian
s 𝜋
rad
ian
s 1
rad
ian
Th
e ro
tatio
ns c
ance
l, an
d le
ave
us w
ith o
ur re
sult
of a
ppro
xim
atel
y 5
7.3
° for
eve
ry 1
radi
an.
This
is a
str
ange
num
ber,
and
we
will
rar
ely
use
it. In
stea
d, t
he f
ract
ion
180
𝜋 is
wha
t yo
u
shou
ld m
emor
ize
and
beco
me
com
fort
able
with
. Tha
t sai
d,it
is he
lpfu
l to
know
how
muc
h 1
radi
an is
in d
egre
es, s
ince
it w
ill h
elp
you
get a
pic
ture
of w
hat y
ou’re
wor
king
with
.
Exam
ple
3a
Conv
ert 6
radi
ans
into
deg
rees
.
Sinc
e th
ere
are 180
𝜋 d
egre
es f
or e
very
one
rad
ian,
we
sim
ply
mul
tiply
thi
s nu
mbe
r by
6.
Thus
6 ra
dian
s is
34
3.8
°.
Exam
ple
3b
Conv
ert 7
5° i
nto
radi
ans.
This
is th
e op
posit
e of
the
prev
ious
pro
blem
. Thu
s, w
e ne
ed a
new
con
vers
ion
fact
or. W
e ap
ply
the
sam
e pr
inci
ple
to o
btai
n a
conv
ersio
n fa
ctor
:
§1
Rad
ian
mea
sure
– 1
33
2𝜋
1
ro
tati
on
=
𝜋
1
ro
tati
on
3
60
° 1
80
Th
is te
lls u
s th
at o
ne d
egre
e is
𝜋
180 r
adia
ns. A
nd s
ince
we
wan
t to
fin
d ou
t ho
w m
any
radi
ans
75
° is,
we
just
mul
tiply
the
prev
ious
by
𝜋
180. W
e ge
t
75∙𝜋
18
0=
75𝜋
18
0=
5𝜋
12
.
Whe
n w
orki
ng w
ith ra
dian
s, w
e ne
ver w
ant a
n ap
prox
imat
ion.
Did
you
not
ice
how
sim
ilar
the
conv
ersi
on f
acto
rs w
ere?
You
sho
uld
have
the
m
mem
oriz
ed, a
s yo
u’ll
need
them
in th
is s
ectio
n an
d be
yond
.
§𝟏 E
xerc
ises
1.)
Det
erm
ine
the
mea
sure
of 𝛼
(in
radi
ans)
giv
en th
e fo
llow
ing
radi
i and
arc
leng
ths.
(A)
𝑟=
10
,𝛼=2
0
(B)
𝑟=
30
,𝛼=
10
(C
) 𝑟
=1
5,𝛼
=1
00
(D
) 𝑟
=3 4
,𝛼=
16
2.
) D
eter
min
e th
e le
ngth
of 𝐴��
giv
en t
he f
ollo
win
g ra
dii a
nd a
ngle
s. Th
en s
ketc
h a
circ
le w
ith th
e gi
ven
info
rmat
ion
and
the
leng
th o
f the
arc
. (A
) 𝑟
=5
,𝛼=2
(B
) 𝑟
=3
,𝛼=
3
(C)
𝑟=
5 3,𝛼
=3 4
(D) 𝑟
=1
0,𝛼
=1 2
3.)
Det
erm
ine
the
leng
th o
f the
radi
us g
iven
the
follo
win
g ar
c le
ngth
s an
d an
gles
. (A
) 𝐴��
=3
5,𝛼
=7
(B
) 𝐴��
=5
,𝛼=
5
(C)
𝐴��
=1
0,𝛼
=3
(D
) 𝐴��
=6
,𝛼=
1 2
4.)
Sket
ch th
e fo
llow
ing
angl
es.
(A)
1
(B) 2
(C
) 3
(D
) 4
(E)
5
(F)
6
(G)
0.5
(H
) 6
.28
5.
) Co
nver
t the
follo
win
g an
gle
mea
sure
from
ang
les
to ra
dian
s or
vic
e ve
rsa.
(A
) 2
00
° (B
) 1
00
° (C
) 5
0°
(D)
1.5
(E)
𝜋 12
(F)
72
0°
(G)
𝜋 5
(H) 𝜋
6.
) Th
e fo
llow
ing
Figu
re is
a c
ircle
with
var
ious
rota
tions
on
it. A
ssum
e th
at 𝐶
is lo
cate
d at
the
orig
in,
and
that
the
re a
re f
our
Qua
dran
ts,
as n
orm
ally
def
ined
on
a co
ordi
nate
pla
ne. E
ach
angl
e be
gins
with
𝐴𝐶
, the
n ro
tate
s co
unte
r-cl
ockw
ise u
p to
the
next
poi
nt.
Uni
t fiv
e –
134
(A)
Writ
e ea
ch o
f the
ang
le m
easu
res
in ra
dian
s.vi
i (B
) W
hich
is th
e la
rger
ang
le, 𝜋 4
or 𝜋 2
? (C
) As
sum
e th
at w
e co
ntin
ue th
is pa
tter
n, s
o th
at in
Qua
dran
t II t
he n
ext a
ngle
is
30
° m
ore
than
90
°, th
en 4
5°
mor
e th
an 9
0°,
and
so o
n (in
clud
ing
Qua
dran
ts II
I and
IV).
Writ
e ea
ch o
f the
se a
ngle
s in
deg
rees
. (D
) N
ow w
rite
each
of t
he a
ngle
s yo
u fo
und
in (C
) in
radi
ans.
7.)
One
of t
he p
robl
ems
that
stu
dent
s ha
ve w
ith r
adia
ns is
tha
t th
ey’re
ter
rible
with
fra
ctio
ns. A
ccor
ding
ly, l
et u
s pr
actic
e ou
r fra
ctio
n in
tuiti
on.
(A)
Whi
ch n
umbe
r is
larg
er, 1 4
or 1 2?
How
can
you
tel
l with
out
havi
ng t
o di
vide
th
e nu
mer
ator
and
den
omin
ator
? (B
) M
ake
an a
rgum
ent f
or w
hy 1 4
is le
ss th
at 1 2
. (H
int:
Try
usin
g m
oney
!) (C
) Li
kew
ise,
whi
ch is
larg
er:
𝜋 3 o
r 𝜋 6?
(D)
Whe
n co
mpa
ring
a w
hole
num
ber t
o a
fract
ion,
it’s
ofte
n us
eful
to c
onve
rt
the
who
le n
umbe
r in
to a
frac
tion.
For
exa
mpl
e, w
hich
is la
rger
, 2 o
r 5 3?
To
see,
let’s
con
vert
2 in
to a
frac
tion
that
has
the
sam
e de
nom
inat
or a
s 5 3. N
ow
answ
er th
e qu
estio
n: W
hich
is la
rger
, 2 o
r 5 3?
(E)
Whi
ch is
larg
er: 𝜋
or 5
𝜋 6?
(F)
Whi
ch is
larg
er: 15𝜋
4 o
r 2𝜋
?
8.)
How
can
you
tel
l tha
t 5𝜋 6 is
less
tha
n a
half-
rota
tion?
(H
int:
Try
usin
g co
mm
on
deno
min
ator
s in
you
r fra
ctio
ns.)
9.)
Whi
ch Q
uadr
ant i
s 11𝜋
6 in
? H
ow c
an y
ou q
uick
ly te
ll? (H
int:
Use
the
Figu
re fr
om 2
.) to
hel
p yo
u vi
sual
ize.
) 10
.) O
f cou
rse,
you
can
alw
ays
conv
ert t
he ra
dian
s in
to d
egre
es to
che
ck w
hich
one
is
larg
er. C
onve
rt 7𝜋 4 a
nd 5𝜋 3 in
to d
egre
es a
nd th
en d
eter
min
e w
hich
one
is la
rger
.
11.)
Now
take
7𝜋 4 a
nd 5𝜋 3 a
nd g
et c
omm
on d
enom
inat
ors.
Whi
ch o
ne is
larg
er?
vi
i And
just
to re
itera
te:
Your
ans
wer
s m
ust b
e in
exa
ct fo
rm.
§1
Rad
ian
mea
sure
– 1
35
12.)
A w
heel
has
a ra
dius
of 12
inch
es.vi
ii (A
) If
the
whe
el m
akes
a fu
ll ro
tatio
n, h
ow fa
r ha
s th
e w
heel
tra
vele
d fro
m it
s st
artin
g po
int?
(B
) If
the
whe
el m
akes
a h
alf r
otat
ion,
how
far h
as th
e w
heel
trav
eled
from
its
star
ting
poin
t?
(C)
Supp
ose
the
whe
el h
as ro
tate
d 1
0𝜋
. How
far h
as it
trav
eled
? (D
) If
the
whe
el h
as tr
avel
ed 3
6 in
ches
, how
muc
h ha
s it
rota
ted?
13
.) Su
ppos
e a
whe
el is
24
inch
es a
roun
d.
(A)
If th
e w
heel
has
rota
ted 5𝜋 6
, how
far h
as it
trav
eled
? (B
) If
the
whe
el h
as tr
avel
ed 1
0 fe
et, h
ow m
uch
has
it ro
tate
d?
14.)
Supp
ose
a w
heel
has
trav
eled
10
feet
. (A
) If
it ha
s ro
tate
d 13𝜋
6, w
hat i
s its
radi
us?
(B)
If it
has
rota
ted 3𝜋 4
, w
hat i
s its
circ
umfe
renc
e?
§2
Intr
oduc
tion
to th
e Po
lar P
lane
Afte
r a b
rief h
iatu
s, w
e no
w re
turn
to o
ur T
rig fu
nctio
ns. W
e w
ill u
se w
hat y
ou le
arne
d in
U
nit f
our e
xten
sivel
y in
this
sect
ion;
you
will
nee
d to
be
able
to c
alcu
late
Trig
ratio
s ve
ry
quic
kly.
i We
will
also
beg
in to
wor
k w
ith T
rig fu
nctio
ns w
here
our
ang
les
are
in ra
dian
s.
To h
elp
us p
ract
ice
and
mem
oriz
e th
ese
Trig
func
tions
, we
will
now
intr
oduc
e a
new
way
to
grap
h.
Befo
re w
e do
thi
s, a
few
wor
ds o
n w
hat
mak
es t
he c
oord
inat
e pl
ane
so e
ffect
ive.
The
re
are
num
erou
s w
ays
one
coul
d se
t up
a gr
aphi
ng s
yste
m, b
ut, i
deal
ly, w
e w
ould
like
it b
e sim
ple
to u
se a
nd e
ffect
ive.
The
coo
rdin
ate
plan
e is
grea
t be
caus
e it
requ
ires
just
tw
o co
mpo
nent
s (a
n 𝑥
- an
d 𝑦
-val
ue) t
o pl
ot a
ny p
oint
. So
it’s
effe
ctiv
e an
d ea
sy to
use
.
So if
we’
re g
oing
to
com
e up
with
a n
ew w
ay t
o pl
ot p
oint
s, it
shou
ld b
e ju
st a
s sim
ple
and
effe
ctiv
e. In
oth
er w
ords
, we
shou
ld c
ome
up w
ith a
sys
tem
tha
t on
ly r
equi
res
two
com
pone
nts
to p
lot
any
poin
t. An
ythi
ng m
ore
than
tha
t w
ill r
ende
r ou
r sy
stem
far
less
ef
fect
ive.
So le
t us
cons
ider
a s
yste
m th
at u
ses
circ
les
inst
ead
of re
ctan
gula
r grid
lines
, as
show
n in
Fi
gure
56.
vi
ii Ass
ume
that
the
whe
el is
a p
erfe
ct c
ircle
. i O
r, as
we
sugg
este
d, y
ou s
houl
d si
mpl
y m
emor
ize
them
.
Uni
t fiv
e –
136
Let u
s em
phas
ize
that
we
have
circ
les
here
, not
rect
angu
lar g
ridlin
es. S
o us
ing
an 𝑥
- an
d 𝑦
-val
ue w
ill n
ot s
uffic
e. W
e ne
ed t
wo
diffe
rent
com
pone
nts
entir
ely.
How
abo
ut w
e us
e an
ang
lem
easu
re a
nd a
radi
us?
That
shou
ld a
llow
us t
o pl
ot a
ny p
oint
on
this
plan
e us
ing
only
two
com
pone
nts.
Cons
ider
Fig
ure
57. W
e ha
ve tw
o co
mpo
nent
s, 2
and
30
°. Le
t’s s
tart
with
the 2
. For
mal
ly,
it’s a
radi
us –
that
is, i
t is t
he d
istan
ce fr
om th
e ce
nter
to th
e po
int.
Anot
her w
ay o
f loo
king
Figu
re 5
6
Figu
re 5
7
§2
Intr
oduc
tion
to th
e Po
lar P
lane
– 1
37
at it
is th
at o
ur p
oint
mus
t be
on th
e se
cond
circ
le.ii T
he s
econ
d co
mpo
nent
, 30
°, te
lls u
s w
here
upo
n th
at s
econ
d ci
rcle
we
mus
t put
our
poi
nt. S
o w
e pl
ace
our p
oint
2 a
way
from
th
e or
igin
,iii th
en ro
tate
the
poin
t 30
° up.
We
show
this
proc
ess
in F
igur
es 5
8 a
and
b.
This
proc
ess
wor
ks v
ery
wel
l, so
let u
s no
w d
efin
e ou
r new
way
of g
raph
ing
Gra
phin
g on
the
Pola
r Pla
ne
A po
int 𝐴( 𝑟
,𝛼)
is pl
otte
d by
cre
atin
g a
line
segm
ent o
f len
gth 𝑟,
and
then
rota
ting
that
se
gmen
t 𝛼.
Just
like
with
the
coor
dina
te p
lane
, we
shou
ld la
bel o
ur c
ircle
s. Yo
u m
ay h
ave
notic
ed
lines
ext
rudi
ng fr
om th
e po
le, o
r orig
in o
f the
Pol
ar P
lane
. The
se li
nes
form
ang
les
with
po
sitiv
e 𝑥
-axi
s, an
d lo
ok a
wfu
lly s
imila
r to
a pr
oble
m fr
om th
e pr
evio
us s
ectio
n. W
e’ll
labe
l the
m b
elow
in F
igur
e 59
.
ii T
here
’s a
disa
dvan
tage
to th
is pe
rspe
ctiv
e, s
ince
we
coul
d al
so h
ave
1.5
as
the
radi
us. I
n th
is c
ase,
we
need
to d
raw
a c
ircle
hal
fway
bet
wee
n th
e fir
st a
nd s
econ
d ci
rcle
. So
it’s
not t
oo m
uch
of a
str
etch
to u
s th
is lo
gic.
iii W
hich
we’
ll gi
ve a
diff
eren
t nam
e sh
ortly
.
Figu
re 5
8a a
nd b
We
star
ted
by m
akin
g a
segm
ent o
f len
gth 2
, the
n ro
tatin
g th
at s
egm
ent u
p 3
0°.
Not
e th
at th
e se
gmen
t onl
y he
lps
to
plac
e th
e po
int;
wha
t we
real
ly c
are
abou
t is
the
poin
t.
Uni
t fiv
e –
138
As w
ith th
e co
ordi
nate
pla
ne, w
e do
n’t have
to c
hoos
e th
ese
angl
es. H
owev
er, t
here
is
good
reas
on to
cho
ose
thes
e pa
rtic
ular
ang
les,
and
we’
ll re
veal
that
ans
wer
sho
rtly
.iv
Also
of i
mpo
rtan
t not
e: T
he a
ngle
mea
sure
s al
way
s st
art o
n w
hat w
e no
rmal
ly c
all t
he
posit
ive 𝑥
-axi
s an
d ro
tate
up
from
ther
e.v
Exam
ple
1a
Plot
the
poin
t 𝐴( 2
,45
°).
To d
o th
is, w
e sim
ply
go t
o ou
r se
cond
circ
le, t
hen
rota
te u
p 4
5°.
We
show
the
plo
tted
po
int i
n Fi
gure
60.
iv If
you
’ve
not a
lread
y fig
ured
it o
ut y
ours
elf.
v Thi
s is
by
conv
entio
n. S
omeo
ne s
omew
here
dec
ided
that
they
wer
e go
ing
to d
o it
that
way
, and
we’
ve a
ll fo
llow
ed s
uit.
Figu
re 5
9
§2
Intr
oduc
tion
to th
e Po
lar P
lane
– 1
39
Exam
ple
1b
Gra
ph th
e po
int 𝐵
( 3.5
,10
°).
Nei
ther
of t
he t
wo
com
pone
nts
in 𝐵
is o
n a
line,
but
, lik
e th
e co
ordi
nate
pla
ne, w
e ca
n ea
sily
appr
oxim
ate
thei
r loc
atio
ns. W
e sh
ow th
is in
Fig
ure
61.
Figu
re 6
0
Figu
re 6
1
It m
ight
be
help
ful t
o dr
aw a
line
for a
10
° ang
le. I
t mig
ht a
lso b
e he
lpfu
l to
draw
a c
ircle
hal
fway
bet
wee
n th
e th
ird a
nd
four
th c
ircle
.
Uni
t fiv
e –
140
Exam
ple
1c
Gra
ph th
e po
int 𝐶
(2.2
,𝜋 2).
In th
is ex
ampl
e, w
e ar
e us
ing
radi
ans
to m
easu
re o
ur a
ngle
s an
d no
t deg
rees
. Rec
allin
g th
at 9
0° a
nd 𝜋 2
are
equ
ival
ent,
we
can
easi
ly g
raph
Fig
ure
62.
Man
y of
you
r Exe
rcise
s w
ill u
se ra
dian
s, so
the
abov
e gr
aph
we’
ve p
rovi
ded
may
not
be
the
mos
t he
lpfu
l. Yo
u w
ill w
ant
to c
reat
e a
Pola
r Pl
ane
with
the
rad
ians
list
ed a
nd n
ot
degr
ees.
But d
on’t
wor
ry, t
his
will
be
one
of y
our E
xerc
ises.
We
have
not
yet
dis
cuss
ed n
egat
ive
angl
es. U
p to
this
poin
t, yo
u m
ight
thin
k th
at a
ngle
s, lik
e le
ngth
s, ca
n on
ly b
e po
sitiv
e. B
ut a
ngle
s (un
like
leng
ths)
hav
e a
dire
ctio
n. S
o fa
r we’
ve
alw
ays
rota
ted
“up”
, whi
ch h
as a
mou
nted
to
a co
unte
r-cl
ockw
ise r
otat
ion.
Not
hing
is
stop
ping
us
from
rota
ting
in th
e op
posit
e di
rect
ion,
i.e.
, clo
ckw
ise,
but
we
have
n’t h
ad a
go
od w
ay t
o la
bel t
his
othe
r th
an s
pelli
ng it
out
ent
irely
. Let
us
ther
efor
e ag
ree
that
a
nega
tive
angl
e m
easu
re te
lls u
s to
rota
te c
lock
wise
, whi
le a
pos
itive
ang
le m
easu
re te
lls
us to
rota
te c
ount
er-c
lock
wise
.
Exam
ple
2
Gra
ph th
e po
int 𝐷
( 3,−
60
°).
Figu
re 6
2
§2
Intr
oduc
tion
to th
e Po
lar P
lane
– 1
41
The
nega
tive
angl
e te
lls u
s to
rota
te 6
0°
cloc
kwis
e. S
ince
we
know
our
poi
nt m
ust b
e on
th
e th
ird c
ircle
(du
e to
the
rad
ius
bein
g 3
), w
e ju
st n
eed
to d
eter
min
e ho
w t
o ro
tate
cl
ockw
ise. U
sing
the
sam
e Po
lar
Plan
e as
bef
ore,
but
cou
ntin
g in
the
oppo
site
dire
ctio
n (a
nd a
pply
ing
appr
opria
te la
bels)
, we
com
e up
with
Fig
ure
63.
But t
here
’s so
met
hing
qui
te c
urio
us a
s to
the
abov
e: T
his
is a
poi
nt w
e co
uld
have
mad
e us
ing
our P
olar
Pla
ne fr
om b
efor
e. If
we
take
the
sam
e po
int b
ut s
witc
h th
e la
bels
bac
k,
we
get F
igur
e 64
.
Figu
re 6
3
Perh
aps
you
notic
ed th
at w
e di
d no
t com
plet
e la
belin
g th
is Po
lar P
lane
. As
you
mig
ht h
ave
gues
sed,
yes
, thi
s w
ill b
e on
e of
yo
ur E
xerc
ises.
Uni
t fiv
e –
142
Not
ice
that
our
poi
nt is
now
loca
ted
at (
3,3
00
°). S
o it
appe
ars a
s tho
ugh
30
0° i
s equ
ival
ent
to −
60
°. Th
is is
inte
rest
ing!
Cote
rmin
al a
ngle
s Tw
o di
ffere
nt a
ngle
s th
at e
nd u
p in
the
sam
e sp
ot a
re s
aid
to b
e co
term
inal
. So
−6
0° a
nd 3
00
° are
cot
erm
inal
, sin
ce th
ey e
nd u
p in
the
exac
t sam
e sp
ot. A
noth
er w
ay
of lo
okin
g at
thi
s is
tha
t th
e po
int
crea
ted
by (𝑟,−
60
°) a
nd (𝑟,
30
0°)
will
be
the
sam
e (w
here
𝑟∈ℝ
>0
).
Exam
ple
3
Plot
the
poin
t 𝐸( 2
,84
0°)
.
This
prob
lem
con
tain
s an
othe
r str
ange
ang
le.A
fter a
ll, th
ere
are
only
36
0°
in a
rota
tion.
Bu
t who
’s to
say
that
we
can
only
do
one
rota
tion?
To
acco
unt f
or m
ultip
le ro
tatio
ns, w
e ca
n ha
ve a
ngle
s th
at e
xcee
d 3
60
°.vi H
ow m
any
rota
tions
is 84
0°?
Wel
l, if
36
0°
is on
e ro
tatio
n, th
en tw
o ro
tatio
ns w
ould
be
vi L
ikew
ise,
we
can
also
hav
e an
gles
that
are
less
than
−3
60
°.
Figu
re 6
4
We’
ve le
ft th
e la
bel f
rom
the
prev
ious
Fig
ure
for y
ou to
hel
p yo
u co
mpa
re.
§2
Intr
oduc
tion
to th
e Po
lar P
lane
– 1
43
36
0°
+3
60
°=
72
0°,
right
? Li
kew
ise, w
e ca
n se
e th
at t
hree
rot
atio
ns w
ould
be
1,0
80
°. So
it s
eem
s as
tho
ugh
we
have
two
full
rota
tions
and
then
som
e le
ft ov
er. T
o ac
coun
t for
this,
we’
ll ju
st s
ubtr
act
two
full
rota
tions
from
wha
t we
have
, 84
0°:
84
0°−
72
0°
12
0°.
So w
e ha
ve t
wo
full
rota
tions
and
the
n 12
0°
mor
e. T
his
help
s tr
emen
dous
ly w
hen
we
grap
h 84
0°,
since
now
all
we
have
to d
o is
iden
tify
the
12
0° a
ngle
on
our P
olar
Pla
ne. W
e sh
ow th
is in
Fig
ure
65.
This
also
tells
us
that
84
0° a
nd 12
0° a
re c
oter
min
al.
And
this,
in tu
rn, t
ells
us h
ow m
any
cote
rmin
al a
ngle
s ea
ch a
ngle
has
. Can
you
figu
re it
ou
t?
We’
ll fu
rthe
r exp
lore
thes
e id
eas
in th
e Ex
erci
ses,
as w
ell a
s pr
epar
e fo
r our
mor
e fo
rmal
in
trod
uctio
n to
the
Pola
r Pla
ne in
Uni
t sev
en. T
he b
revi
ty o
f thi
s se
ctio
n w
ill g
ive
you
time
topr
actic
e th
e ba
sics
and,
per
haps
mor
e im
port
antly
, pav
e th
e w
ay fo
r suc
cess
in th
e la
st
sect
ion
of th
is U
nit,
whi
ch is
per
haps
the
mos
t im
port
ant.
Figu
re 6
5