ch.IS
TrigonometricFunctions
-
t.ci) text
-
-
arctcncarctancarctcnx))
I
"
+ 4)
1
=
I
'
ltarctarix.lt/2cii)fcxI--arcsinCarctanCarccosx
ltarctoicarctorx)
))
1
'
f- (x)
C- 1)
I
=
.
"
ltarccosx
1- arctanzcarccosx)
2
1- ✗
ciiisfct) arctanctanx arctcnx)
-
-
-
I
1
'
f- (f)
=
lttoix arctoix
-
Civ)
) G)
-
=
arcsin
I
CHIT 'T
'
ltx
'
'
f- a)
_
=
I
1-
f- { /
-
(Iti )
1++2
=
1+112
C- ×)
I
.
"
2
(1++2)
✗
-
/
=
2
It ✗
/
'
-
,
, +×
,
arctan'C✗)
=
arctanx 1- tax
/ '× /
,,
But
/ seek
-
¥
-
{ zx
4- Citi /
.
-4
Sint
(;) Iim
2.
✗ + ✗3/6
-
Ciii) /im
3
✗→ o
✗
¥Ytct) llmgcx)
-
-
COSX
It ✗72
-
-
11m
=
,
×
✗→ o
Sin ✗ 1- ✗
-
11m
=
z×
✗ so
-
✗ +I
=
2
✗→ o
g
o
-
-
✗-20
'
d- (t)
cost
-
_
-
It
¥
Civ) Iim
=
=
Iim
°
-
×
24k
11m
✗→ o
cosxtl
12×2
I
cos ✗
=
24
24
0
g'G)
f- (f)
"
=
G) Iim
Sin✗ + ✗
-
arctorx
✗ + ✗313
=/im
×
-
g. G) =6✗
-
,
✗ so
_
0
=
=
0
g. (t)
"
✗→ o
f' (f)
"
=
-
g' (f)
limit
cos ✗ + I
=
=
=
11M
✗→ o
Bt L' Hcipital's Rule
limtct
× >◦
.
54-1=0
11M
✗ so
-
sink
✗
-
=
g' (f)
=
"'
sinx
24
limit"k×,
✗→ o
">
g
✗ + ✗To
4
✗ so
(f)
-
=
/
'
-
✗
COSX
✗→ o
sing)
-
=
lim
✗so
I
0
(t)
Bt L' Hcipital's Rule
,
limtct))
✗ so
-
gC✗
=
0
f
,;m
Sinxtxcosx
✗→ o
sing
=
ZCOSX
-
✗sing
SMX
0
-
✗
Xsinx
-
=
-
=
Cii) Iim
/im
)
,
(1++2)
0
°
=
"'
g. (f)
✗→ o
"
Gx
evil Iim
'C✗)
3×2
+ Z✗
=
✗→ ◦
"
/ + ×
-
(11-+2)
✗→ 0
6
"
Iim
'
-
-241-+25+2×-2 (ITI ) 2x
z×
'
limit"Ct)
1+1×2
✗→ 0
"
'"'
-
hm
=
✗→ o
g-
4×3
-
✗→ o
Sin ✗ + ✗
=
✗ so
Iim
=
✗→ o
-
,
Sint
Iim
It ✗42
=
✗→ o
g' (f) 3×2
f'4)
COSX
SINX
cosxtcosx
-
xsinx
+?
3.
fcx)
/ sins
=
✗≠o
✗
1.
✗ 0
=
(a) f' (o)
5inch)
tch) %)
-
h
-
'
g- (a) lim
=
=
h
h→o
=/im
Iim
l
,;m
0
≥
h
-
cosch) l
o
-
=
h >o
h
h→o
5inch) h
-
=
=
,,.m
=
0
2h
'
◦
hso
-5inch)
0
=
2
h so
-
(b) f"(0)
✗cost
'
For ✗ to , d- (f)
-
Sins
=
2
✗
hcosch) -5inch)
3- Ch) fico)
'
hcosch)-5inch)
-
f"Co)= lim
=
h
h so
-
=
h so
=
-
hsinch)
-
°
=
3h21
h so
-
-
=
◦
,;m
h
cosch)
=
-
0
§
-
3
h so
o
=
,
-
h
cosch)
lim
h so
-
Cosch)
Iim
ti
lim
4. (a) fct)=sinC2X)
sina.tl just measaneofnthatmochieshnct) by compressing the graph towards Zeo
f' (f)
or
=
ZCOSCZX)
-•÷•%F•••n
(b) Ect)
sink)
_
^
^
✗
act)=zxcosc×"
i.it#.:..g...E..T*.j
"
Thefofherouttnmothemcre compressed
FC
hit )= sin chit)
letgch)
MVT
-
-
=
-
-
>
.
o
hit
IX. ✗ C- Chitti) ^
I
it
(111-1) #
-
hit
ie , the distancebetween successive roots off
→
=
, ×,
11+1
h
-
gets smaller and smaller
.
=
z✓J
1
<
z✓Ñ
.
(c) Ect)
sin✗ + sinzx
-
_
local min
-
/
f (f)
COS✗ 1- ZCOS (ZX)
=
0
=
1-
it
A:1.÷:
looks like tour critical points
-
cos G) +2 (COSET)
sinzct))
-
Cosct)t2(cos (t)
cos ZX
It cost)
>
-
✓
1)
.io?m.Ecoscx)t4cos2Ct
Cos (X) + 2 (2054-1)
-
-
+◦ +
' 0cal
max
,
) -2=0
UCOSZX + cos ✗
D= I
cos ✗
-
2--0
4. 4. C-2)
-
+ to
local Max
✗2
33
=
•
=
1±✓zJ
•
.
✗i
84
=
&
=
^
.
•
59
×,
•
✗a
Sink) 1- Smut)
•
÷÷¥¥↑÷•n
-
sink)
=*"" "
'
I
'
f- (f)
copy,
*
+
¥
-
guy,
I
-
=
-
+
+
•
*
0
¥ IT
•
sect
=
+
*
31T
+
-
f-
*
I
=
tori ✗
25inch cos (f)
=
=
cos"Ct)
cos (t)
F
"
•
it
0
2-
'
d- co)
=
+
+
-
*
¥
-
*
¥ 31T 21T
•
*
I
0
f(HIT)
=
sinks)
21T
"
-
→
'
Zcosct) C- sink)
I
0
2
=
-
=
tachlt) hit
-
=
-
hit
adhitisaloati an inflection point
.
Sinzx
"
cos Ct)
=
0
→
sinG)
=
o
-
>
✗
=
0 or ✗
=
it
(e) tct) sink) ✗
-
_
-
a
:#
a
BO
tha
cosa
-
-
-
i
-
¥ it
•
IT
0
-
"
=
+
-
+
§•
31T
21T
e-
_
I
,
•
0
I
1T
,
+
-
*
¥ 21T
2
'
t
"
+
-
*
2
1- CHIT)
,
sink)
-
I
-
-
_
¥
I
I
j (t)
-
,
IT
-2
:
f' (11-215)--0
I
-
Een
""×
:{
,
✗
=
'
0
✗cost
-
Sins
✗to
=
,
×
I
f- (o)
"
f- (1)
=
-
3
cost
"
-
✗sinx
-
cost
=
sinx
=
-
4
'
=
0
✗cost
✗1=0
×
✗
f' 4)
>
0
=
F' (f)
÷
gq÷÷;¥¥%;÷÷%•
✗ to
×
From problem 3. oehnocr
f- co)
;"÷••¥
1.1
t
'
_
-
I
" "" '
-
→
Sins
=
,
0
→
✗ tan ✗
=
×
From part (d) chereoehadgcxt.tn/-xoesaothatthereoascnerootinerery(
"
2
(1^+2) 'T
i
z
)
(g) FCN
f' (f)
✗sin✗
sin✗ + ✗cos✗
=
d- C-×)
d- (1)
=
=
"
=
-
✗ sinC-×)
cos✗ + cos ✗
Do
0
=
→
✗
=
-
sin×
=
cosy
=
-
✗ C- Jinx)
✗sin✗
=
-
ton✗ or tax
✗sins Ict)
-
_
=
-
×
(even)
2Cosx-Xsin!xfCH-tanCti-x@og00.f
-
=
!
'
f- Ct)
=
+
cog ¢,
*
+
±
+
,
*
+
o
-
2
+
no
→
+
it s¥ zit
f-
critical points
'
2
-
f (t)
I >0
"
Zcosct) C- sin×)
25inch cos (f)
=
=
=
"
cos"Ct)
cos (t)
Sin2x
"
cos (f)
F
"
-
I
•
I
0
-
+
•
-
*
*
¥ IT IT 21T
2
'
+
-
*
z
g- co) 2
=
f-Chit)
=
forchit) + HIT hit
=
and hit is always an inflection point
f- CHIT)
'
=
2
d- has a root in each [
h
=
.
Zntl for n E I
.
hit
z
'
(4+215)
z
,
S
§
FCO)=
:
graphoffinpowcoordinatesishtperbolicspiralcr.cl
)
CHO) ,O)=( 910,0 )
-
-
O
r
0
→
ran
Is
as
1+12
Zalit
211Th 213
IT
a
/IT
111Th 113
3142
Za/31T
213159219
,,
a ,,,
,,
51172
2451T
2111Th 2115
,
,, , , , ,
,
a
_^
!aÉ
••
Za
31T
✗CO)=rcoSO=
ACOSO
0
)=rSinO=
asino
NO
11M / (G)
=
=
-
O so
0
11M CCOSO
=
a
0 so
-
-
11m ✗ Co) :O
D- so
6.
letgct) coscxtf)
-
_
g' (f)
=
g. (f)
=
-
"
-
sincttf)
coscxtt)
g"tg=O
gco) cost
-
_
gko)=
-
Sint
nobtthtlwehare
g.
=
C- sin
-115in + cost cos
✗
smcethisistwefcral-oehorev-xv-tcosc.tt
-1)
-
_
cost cost
-
smxsint
,,,
,,
(a) Sin2x Sin CXTX)
7.
-
cos 2.✗
COSH sink
sin3.✗
sin C2✗t✗)
=
>
>
cos ✗ I + cos ✗
=
_
=
25in✗ cost
=
-
-
=
205×-1
sin2x cost + sin ✗ cos 2x
=
or
Zsinxcosxtsinxccosx sink)
-
25Mt cos ✗ 1- Strict) (2052×-1)
>
=
=3 sin✗ cosh
45in✗ cos ✗
>
=
cos 3✗
2COSI
=
-
sink
sins
-
COSCZXTX)
=
-
COSZX Cos ✗
=
cos ✗
-
Sin 2x sins
or
-
sink) cost
>
2 sink cos ✗
-
(co)'t
=
cos ✗
-
Zsinxcosx
>
>
-
35in ✗ cost
Cos ✗
=
-
305×+3 COSI
>
4 cos ✗
=
(b)
sin
=
2)
IT/
cos [
cos
cos
=
2
(2.1+14)
¥
-
'
2
sin I
≥
-
205¥
=
3 cost
=
cos
=
-
✓I
*
4
a)
✗ 1- ✗
=
+
z
¥
1- tot
I
0
=
✓I
cos
→
=
2
Note chat inform.ae have implicitly used to choose the positive square root
1
IT
-
-
!
2
1-✗ DX
'
¥
→
IT
1-✗Zdx
=
1
0
'
] 1-✗2d✗= / 1-✗ DX
'
is ever out so
I ×
But
-
'
Also , Cos (f)
z
-
=
0
I
¥
T.F.at
¥
'
i
sin
>
Accost)
-
=
"
✗
Aco)
=
=
z
y
→
✗
+
=
(1+12)
,
cos is positive
.
sin (2.11-14)
:
we alsoknow
=
=
that cos ✗
-
1
.
=
-
I
>
-
Accost)
Similarly cos co)
,
=
I
=
A-C-1)
25in
=
z
(11-12)--0
ie cos CIT)
¥
"
cos ✗ -0 →
ie cos
sink) 20
in
cosctl > 0 for ✗ E (0,1+12)
i
-
-
=
¥
=
{
[011T)
.
Hence , cos cx) co for ✗ C- (Nz.IT ) and
Hereis the graph of Act)
••↳
=
→
✗ IT
=
# cos
=
✓2- sin 74
→
'T
Sin / 4
Ñ
=
2
1
tan (1+14)=1
sin (11-14)
/
=
COS (11-14)
sin (11-16)
{
=
cos (1+12)
>
cos (3-1+16)=4cos (11-16)
=
cos (1+16)
0 <
Noo ,
>
→
→
cos
¥
<
(IT / 6)
Sin CITI G)
=
¥
3.Cos (1+16)=0
(4050+161-3)=0
cos (11-16)>0
→
3
→
g
=
-
I -314
cos (1+16)
✓J
=
2
I
=
z
Asincxt B) canbeorittenasinxtbcosx
8. (a)
Now assume oeoe given a.b.Then,thereare And B
S.t.
letfct) Asincxt B) Then
asinxtbcosx-AS.int/ 1- B)
-
_
.
'
3- (f)
=
f. (f)
=
,
ACOSCXTB)
"
-
=
D= tan
_
Hence
→
Sin B
COSB
ASMCXTB)
→
b
a
let B S.t.
'
tab
=
&
,
(Bk)
f"tf=0
,
Define A- as
fco)
Asin(B)
=
b
a
A-
=
=
f- (O) ACOSCB)
'
costs
SMB
-
_
Then
a
Bfth 4 ,
-
ACOSB
b- ASMB
-
.
d- (f) =)' (o) Sinxtfco) cost
=
=
-
and a' +b'
A-cos (B) sincx) + ASMCBJCOSCX)
Then
asinxtbcosx
'
=
A
→
Asincxt B)
=
=
A
/
A
=
atb
=
Asin✗ cos ☐ + AsinBcosx
a
A
sin ✗ +
asmxtbcosx
Ita other pick D=
-
_
¥
.
b
A
cosx
)
if a -40
.
(c) fcx)
3 sin✗ + cos×
=
Let a- V5 , b--1
.
-
Bt part b) A
B tan (V51 ] )
"
3+1
=
,
=
2
=
,
aooecanorile
f-(t)
Asin (✗ + B)
=
for all ✗
Hoo Oooe find t.cn (V5/3)
"
tax
%
=
✓3- cos ✗
3cosh
=
=
=
✗?
sin✗
=
cost
1- cos't
3
9- 9 COSH
12052✗ 9
=
-
cos ✗
cos ✗
=
=
}
{
By >b, ✗ 1+16
=
.
Hence , 13=1+16 ow tch
=
25in (✗ 1-11-18)
9. (a)
tancxtt )
tax + tort
=
,
_
,
taytay
it ✗it , [✗+ f) are not hit + É
Proof
tcncxtf)
sincxtt)
=
sin✗ cost + Sint cost
sin✗cost + sintcosx
Cos✗ cost
=
=
coscxtt)
cos✗ cost
-
sinxsint
cost cost
-
sink Sint
cos ✗ cost
sin ✗
cost
Sint
+
tax + tart
cost
=
=
1-
sinxsint
1- torxtat
cost cost
(b) arctanxtcrctat
=
arctcn
( 1- + )
✗+
tt
Proof
From
tancx.tt )
tort 1- tart
,
'
=
,
a)
1- tax tort ,
,
let × , not , be the orator's of ✗ not
.
Recall that arctoshas domain IR and image (-72,1+12)
T.F.ard-axadarctatareinc-ITK.tt#xadycreinlR
The restrictions we hadfor a) werethat tax,
.
.tn/z.aUtaCXitt.lbedetined.iex.-/..x.s-t.noth1T+E
Therefore, ardent , ardent adcrctax today ≠ Kitt
.
Theory violation possible is arctaxtorctat
tancarctcnx + ardent )
✗+ t
=
1- ✗y
Take the actor of each side
arctorxtarctat
=
arctos
(
✗++
1- Xt
)
=
±
¥
¥ so this is theory restriction
,
.
.
10
.
arcsinttarcsinp-aro.in( ✗
I
-12+1
I
-
✗
2)
✗
if C- [ -1,1]
Proof
sincarcsinatarcsinpl-tcoscarcsinsltfcoscarcsinxlwhatiscoscarc.int
-÷•£•¥••
)?
'
[sinx]
-
-
sinx
cos ✗ for ✗ EIR
if ✗ C- C- 1.1) thence corset
•
'
'
[sinx]
^
-
-
cos ✗
ifarcsinksinx)≠o
=
arcsin' (SMX)
chichis true in C-lil)
arain
,
,
.
•
-
-
¥
hence
'
[ sin caroms)]
coscarcsinp)
=
also cos Carolina)
I
=
sincarcsinttarcsins)
'
I
=
=
'
arcsin Cs)
-12
2
-
✗
1-12+1 1- ✗
2
✗
=
Note that
Noooeucnttotahethearcsinofeachside
arcsin is the inverse of sincx) , ✗ C- [-1+12,1+12]
xp c- [-1,1] adarcsmx.aro.int C-
[
-
ÉÉ ]
'
Thus arcsinxtaro.int C- [ IT, IT] oeneedto
,
-
.
.
CASE I
Therefore, for -172 ≤ arcsinttarcsin ≤ 1+12 oehare
coraoerchcthappersfcrallsvch possible values of
1- ✗ 2)
arcsinxtarcsinp-arcs.in( ✗ 1-12+1
arcsinxtaro.int
CASE 2
Écarcsinatarcsins ≤ * then it is
It
arcsmcsincarcsinatarcsins) )
However , since sincx)
sin CIT
IT
-
-
arcsinx
arcsinx
-
-
-
-
longer true that
sincarcsinatarcsins)
SINCE ×)
-
arcsinp)
arcsinp
arcsinttarcsinf
=
no
=
✗
1-12+1 1- ✗
2
arcsinct
1-12+1 1- ✗2)
arcsin (✗
1-12+1 1- ✗2)
IT
=
=
-
CASE 3
It
-
* ≤ arcsinttarcsinc
sin C- IT
-
IT
-
-
arcsinx
arcsinx
-
-
=
+12 then since sin C-IT
arcsinp)
arcsinp
arcsinttarcsinf
_
_
=
IT
-
-
=
✗
×)
1-12+1 1- ✗
2
arcsin (✗
1-12+1 1- ✗2)
arcsin (✗
1-12+1 1- ✗2)
-
strict)
.
.
11 min any numbers then
.
sincmxsincnx)
=
sincmxcoscnx)
=
coscmx> coscnx)
=
{ lcoscm >× coscmtn) × ]
(it
{ lsincmtn>× Sinan )×]
Cii)
{ lcoscmtn>✗ coscm ) × ]
Ciii)
-
n
-
n
-
-
n
+
-
Proof
cos 1cm n)✗]
cos Cmx nx)
=
-
-
=
=
coscmx) cos C- nx)
=
Cos Cmxtnx)
coscmx) coscnx)
{[
{[
cos Cm n)✗
=
-
Sincmx) Sincnx)
coscmtn) × ]
-
-
-
]
n) ×
=
]
=
=
{ Zsincmx) Sincnx)
-
=
SMCMX) sincnx)
{ zcoscmx) coscnx) coscmxlcoscnx)
.
=
(i)
Ciii)
sin Cmx nx)
-
sincmx) cos c- nx) + sin C- nxscoscmx)
Slncmx) COSCNX)
sin [ Cmtn) ×]
=
-
cos Cm n> ✗ + coscmtn) ×
sin [ Cm
=
sincmx) sinC-nx)
coscmx) coscnx) * Sincmx) Sincnx)
cos [ Cmtn) × ]
=
-
=
-
Sincnx) co) (Mx)
sin Cmxtnx)
Sincmx) COSCNX) + sincnxlcosc.mx)
{ [ sincmtn>✗ Sinan ) × ] { Zsincnxkoscmx)
-
n
-
=
=
Sincnx) cos cmx)
Cii )
12
Min C- N
.
IT
sinmxsinnxdx
{
=
0
m≠n
IT
m=h
Proof
M n
=
IT
IT
,
/ siricmx)O✗= z / [Colo
IT
-
-
-
cos Zmx ]d×
{ [21T
.
=
IT
sin
m×)
"
]
-
2M
-
=
it
{(
-
o
-
o
)
=
IT
it
m≠n
it
it
sinmxsinnxdx
I / [ cos @✗ nx)
=
{[
=
o
-
o
]
=
-
-
-
) ]g×
cos (m×+n×
it
=
{/
sin cm×-n×)
sincmxtnx)
,
M
-
Mtn
n
)
"
-
IT
0
IT
coscmx) coscnxldx
{
=
0
m≠n
IT
m=h
Proof
m=n
IT
IT
/ coscmx)coscn×)d×
-
{[
=
COSCZMX)
+ 21T
( 2m
-
I
-
zm
=
IT
it
2M
1
IT +
=
/ coszcmx)d✗
=
IT
-
{
[ cosczmx) + coso ]d×
]
IT
)
IT
=
m≠n
IT
)coscn×)d× { % (coscmtn>
/ coscmx
-
IT
=
)×)g×
✗ + coscm n
-
*
{ [01-0]=0
=
IT
sincmx) coscnx)d✗=o
Proof
1T
IT
sincmxlcoscnx)d✗=
{ / [ sincmtn>✗
-
=
=
{/
0
,
cos [cmtn> ×]
-
Mtn
cos[Cm n> ×]
-
+
Sinan n)×]o×
-
-
IT
since cos (xx) is even
M
.
-
n
]
'T
-
it
=
{[
sinlcmtn> ×]
-
+
Mtn
sin[Crn n)×]
m n
-
]
'T
-
IT
13
(a)
.
tint on [ IT IT]
-
,
.
.
IT
min of
/ ltcx)
-
acoscnx> Tox
-
IT
*
I
occurs when a
=
4) cos cnxiox
*
Proof
IT
Leth(a) =/ ltcx)
-
acoscnx> Tox
IT
-
IT
/ [54-5 2kt) acoscnx) tacosKnx) ] ox
=
-
-
-
IT
IT
-
IT
IT
/ fcxidx
=
-
a.
IT
2¥) 4)
-
'
=
-
/Jct) coscnx) DX + Za / cosKnx) DX
2
-
it
a
IT
=
0
IN
IT, see
*
→
/ cosKnx) DX
-
IT
IT
h (a)
coscnx) DX + a
'
¥41
coscnx) d✗
prob 12
.
=
IT
Note thath is a quadratic poyn.in a.thus his cart e.w
.
.
local extrema occur at critical points of which there is just one
.
,
11m h(a)
11m hca)
=
a→ co
=
as
c. s - 0
-
IT
/Ect) coscnx) DX
Hence ,
-
a
*
is
=
,
globalmin
.
Note also that
IT
2 / cosKnx) DX > 0
"
h (a)
=
-
IT
BY
IT
min of / [1- Cx)
-
asincnx)]{✗ occurs at a
-
1
=
*
'T
4) sincnxiox
IT
proof
1T
Leth(a)
IT
/ [fcx)
=
-
-
-
a
-
IT
2
=
-
+
'
a
IT
IT
-
-
IT
IT
4) sincnx) DX
→
a
=
IT
h (a)
"
=
IT
) G) sincnx) 0×+2a / sincnx) DX
2¥
21T > 0
/
sincnx) + a- siñcnx) ox
/ sincnx> dx , a quadratic in a
-
IT
h' (a)
a
IT
/tcxisincnxsdx
-
-
IT
IT
/ tcxidx
-
/ (54-5-21-4)
=
-
IT
=
asincnx)]'d✗
IT
globalmin
.
(b) Define
1
an
=
IT
XJCOSCNX) DX
*
-
bn
n :O ,1,2 ,
.
.
.
IT
IT
1
-
=
*
(f) sincnx)d✗
Ci , di any numbers
n
1.2.3 ,
-
-
.
.
.
.
Then ,
IT
¥ / Fcx
)
-
N
/{
+
[ Ccncoscnx> + dnsincnx))) 'd✗
)
n =L
1
linear comb of coscnx) ad sincnx)
.
cxi
=
-
ztcx,
IT
-
IT
n -1,
-
,
.
.
.
.
N
/z
[ ( Cncoscnx) + dnsincnx))) + / {
+
[ ( Cncoscnx> + dnsincnx)))
+
n -1
n= '
-
N
IT
IT
)d✗
N
N
"
=/ cfcxiidx-co.fr/-)dx-/S12Cntct)coscnx)tzdntcx)sincnx)]
-
IT
-
h=1
it
/ (
+
-
IT
Co(
cncoscnx> + dnsincnx))
V0
[
oscnxiox
,
-
=/ (Knick
-
-
itcoao
-
2
# ( Cnantdnbn) +
+
z
h= '
IT
-
+
cncoscnx> + dnsincnx))
O
it
Co2 IT
N
IT
))
N
N
+
Sincnx)dX=0
so ,
É
)
,
icoscnxi + / di sincnx)
n ='
-
IT
-
IT
notethatevertothertermiso-thekrmncndmfcoscnxlsincmxldx.co
-
=/ cfcxiidx-ITCoao-2.FI ( Cnantdnbn) +
-
+
z
h= I
IT
some terms to
complete squares
=/ 0-4-150✗
-
+ ITC
ai
¥0
N
hit
her
ao
Co
.
z
-4-150×+1+2 (Cn
a
+
Hai
-
.
2-
-
an
I
number
Co2
2
)
-
n =\
( z
I
•°
+ *
thisportoecachasebtpidrmgcnavdn.no
-
+
Sbi
her
)
ITS Can't bi )
-
1
notethatan.bnoenumbers.wecnpkhcnaddn.itwepidrcn.sn
Thus the entire
-
z
I
and bn
N
N
Tai
-
Scare + bi ))
2
h= ,
)2+iTÉcdn but
nil
1
z +
-
Tai
2
N
N
-
N
IT
I
/
( scare + bit
+ *
t.it/&Cn-2SCnan+Ean)+it(Sdn=.i-2Sdnbn
n =L
IT
[ (Chittoor )
h =I
add and subtract
IT
,
.
.
.
,N
N
1T¥
_
2
tSCan + bit
ha
I
1
this is a number
dntherthemidotepotiaaboe chichis ≥o) is minimized at 0
etpr.is minimized
.
.
N
Co2 IT
N
IT
by problem 12
IT
.
14 (a) Sin ✗ + Sint
.
sincatb)
sin(a) cos(b) + sin(b) cos (a)
=
sin ca b)
sinca) cos C-b) + sinC-b) cos (a)
=
-
Sin(a) cosCb)
=
sin Catb) 1- sin ca b)
=
-
Giver at numbers ✗
✗
=
atb
-
>
a
=
+
=
sin (b) cos (a)
-
zsinca) cos (b)
only we coraloats find
✗ b
f-
a b
→
-
ad bat
-
✗
-
-
a
,
✗
-
2b
→
→
b.
t
=
2
a
=
✗
-
b
2✗
✗ +t
-
=
✗+ y
=
2
2
Thus ,
sin a) + sing)
=
zsin (
F) cos / ¥ )
✗
Also,
sincatb)
-
Simca
-
b)
=
25in(b) cos (a)
SO ,
sincx)
-
sinG)
(b) cos (atb)
Cos ca
-
b)
✗
a- ) cos ( {
cos ca
✗
+
-
cos (a) cosC-b)
=
-
t
)
Sin(a) SinCb)
Sin(a) sinC-b)
cos (a) cos Cb) 1- Sin(a) sinCb)
Cos catb) t cos ca
-
zsin (
Cos (a) cosCb)
=
=
coscatb)
=
-
-
b)
b)
=
=
2cos (a) cosCb)
-25in (a) sin Cb)
T F.
.
cos cut cos 4)
cos 4)
-
cos 4)
=
=
2 cos
-
( ¥ / cos (¥ )
zsin (
✗
{ ) sin / ¥ )
+
✗
✗
=
.
atb and -1
-
-
a
b.
-
15 (a) COSCZX)
COSH
=
.
cos (Zx)
sing
→
(b)
Cl Cos ✗1=205×-1
>
-
-
=
2
coil
=
cost
=
/ + cos (Zx)
>
cos ✗
→
sink
-
-
sink
I
=
-
Strix
sink
-
=
I -25in>✗
1- cos (Zx)
=
2
{
cos
' + •'✗
=
É
◦≤✗ ≤
2
Proof
COSCX)
=
cos [2.
E)
_
cos
→
(✗ /2)
' + COX
COSCX /2)
→
=
2
=
ltcosx
±
2
I
^
1
the positive sa root occurs
.
when COSCX/2) ≥ ochich
>
means
IT
≤
{ ≤ 2-
→
-
+ ≤ ✗ ≤ it
if He ✗ 221T then
^
≤
¥ ≤ IT
ad cos CX /2) ≤ 0 sooooooo need to
>
take the neg Sa Not
Same for
-
.
-
.
2t≤ ✗ ≤
-
since oeoeaslummg
IT
.
0 ≤
cetane the poi Sa Not
.
b
(C)
a)
sinzxdx
!
I
-
cosczx)
z
sinczx)
1
b- a
=
=
b
2
2
b
[ co5✗o×
(d) Ect)
_
DX
=
[
sinczb)
b-a
2
2
b-a
+
=
z
sincza)
4
a
' + cos "✗)
-
_
=
-
z
sin Gb)
-
sincza)
4
sink
-
'
d- (t) Zsinxcosx
-
-
a
=
sin CZX)
+
+
-
•
¥ go ¥
*
IT
a
sink
i.E://.TT?.F.ITinaxs--t
f- (t) ZCOSZX zsinzx
"
'
sink)
-
-
-
-
=
ZCOSCZX)
-
*
*
+
Zit
.
✗
12 ≤ 72 then
16
.no/-e:t--arctanx-sx--tany=sint
Sint
1-
Cost
sincarctax)=✗
1-
sincarctorxi ICI
-
-
=
=
sing
sin -1
-
arcsin'( sing )
Sinay
'
-
sincarctax) )
sincarctcn ✗141 + ✗2)
2
✗
=
'
×
sincarctax)?
ltxz
2
✗
sincarctanl)=
1++2
I
coscarctax)
17
✗ =tan(
.
arctax
:
1- siricarctax)
=
=
It ✗2
Y)
{
'
sincarctanx)
×
=
sinculz)
=
,+×
,
=
±
1- <↳ °
2×2
=
→
I
I
sincx)
sina.li/z)-.zsinCXlz)cosCX1z)
-
,
COSU
,+ ×
2
I
-
cosu-ltxcosctl-coscz.tn/z)--Cos2CX1z)-Sin2CX12)sinu=1-cos2u
-
_
-
>
"
=
I
Zsinzcx / z)
-
(' 1- ✗2)
=
-
'
4×2
(1- ✗2)
=
≥
→
(ITN )
1- cost
sinzcx / 2)
=
* 2x:-#
2
sink/2)
=
±
1- cost
-
V1 -2×2++1 )
2
another way
0
arct.cn/--z--U=2arctanX--Sinv--sinC2crctcnx)--2sinCcrctcnx)coscarcl-at)
2
'
'
×
2x
=
=
It ✗
2
It ✗
2
1++2
z×
=
1++2
11-+2
18
.
(a) sin(✗+ IT/2)
sin (✗ +1172)
sin X
=
* note
cos ×
=
-
cos ✗
O t
(b) From part a) , sincxttk)
=
I
.
we'd like to take arcsin of both sides
"
-11-12 ≤ ✗ 1-11-12 ≤
If
✗+ IT /2
-
:
-
[ I , I] →
-
:
IR
-
>
[-11-12,1+12]
[-11-12,1+12]
.
-
-
IR > [ I , I]
arcsin o cos
.
12 then ✗ C- [
arcsinccosx)
=
for all ×
,
:
arcsin
coscx)
=
cos ×
cos
IT 0] and
,
* ≤✗ ≤ 0
IF ✗ 1-11-12 C- [1+12,3*12] then ✗ C- [0,1T] and
sin (✗ + IT/2)
and since GT
-
-
_
Cos (f)
=
2) C- [-1+12,1+12] wehave
IT/
X
-
A- ✗-1+12
sin at ✗ IT/ 2)
=
1+12 ✗ arcsinccosx)
=
=
-
At this point, wehave
arcsinccosx)
{
=
✗+1+12
IT ✗
-
-
-
IT/2
it ≤ ✗ ≤ 0
0 ≤ ✗ ≤ IT
'
'
It ✗ Zithtx for ✗ C- [ IT, 0] then
-
=
✗ + IT /2=21+11+1+12 1- ✗
'
E [
_
21-211-11 , IT/21-211-11]
IT/
'
and since ✗ 1-11-12 E [-1+12,1+12]
sin (✗ 1-11-12)
=
arcsin CCOSX)
'
sin
=
✗
(211-111-+12++1)
=
'
sin (✗ 1- IT / 2)
=
SMCX 211-11+11-12)
-
211-111-1172
-
If ✗ 211-11 1- ✗ Jar ✗ C- [0,1T] then
'
=
✗ + IT /2=21+11+1+12 1- ✗
'
'
✗ 1-11-12 C-
sin [IT
hence ,
,
[1+12,3*12]
Sin CX 1- IT/ 2)
=
E [ 2*11+1+12 Zitht 31+12]
+12
-
-
=
sin (21+11++1+142)
✗ +2 ith
✗ 1- 21171
arcsinccosx)
=
-
't /
2)
=
=
sin Cx't 'T / 2)
sin ( H2
-
Sin (✗ -2*11+11-12)
=
)
✗ + 2*11
,
where 72
arcanecos ×)
=
21+11+1+12
✗ C-
[-11-+2*11,2*4]
IT/ 2- ✗ +2*11
✗ C-
[21+11,11-+211-11]
X
-
-
✗ -1-2*11--472
✗ C- [-1+12,72]
'
-
Now let's find arccoscsinx)
Note that cos CX
2)
IT/
-
coscx) cos C- IT /
=
IT/
-
2 C-
[0,1T]
✗
_
arccoscs.in ×)
'
×
=
1+12
=
→
¥ ≤✗≤ {
'
✗
_
-
✗
-
_
,
,
'
cos CX
=
arcoscsinct))
=
✗ ≤ IT
o ≤
sin✗
=
+1/21-21+11 ✗ ttz C- [ 0 IT]
-
_
+12
n
'
_
cos CX IT / 2)
✗ IT/ 2
[-1+12,1+12]
✗ C-
COS (IT/2- ×)
=
-
IT/ 2
+12
cos C-×) then
=
cos CX IT / 2)
_
-
→
-
since cos G)
✗
✗
=
E [ IT, 0 ]
IT/2
✗ E [IT/2,31T/2)
→
arccoscsinx)
-
sink)
=
✗
25° Strict) sinC-IT / 2)
-
✗
_
1-24-11
IT/ 2
cos CX IT/2)
=
,
cos CX
)
'
* 12 1-24-11
=
cos [✗
-
2)
IT /
=
cos CX
-
✗ 1-11-12
-
C- [ it o]
,
-
)
IT/ 21-2*11
=
cos
'
-
>
✗ E
'
=
cos [✗ +31T /2)
cos [21T
-
=
CXT-ᵗ
+
Zith)
cos (✗
✗ +211-11
Hence
=
-
211-11 +31T/2)
/2)
3 'T
-
=
cos (72
-
✗ 1-24-11
)
72
-
✗ 1-24-11
,
arccos (SMX)
=
2)
{
✗
_
*
172
-
12-24-11
11Th ≤ ✗ ≤ IT 1-2*11
✗+211-11
2 Ith
-
'
=
cos (✗ +
3¥ Zith)
[ IT ,2 it]
[0,1T]
arccoscs in ×)
IT /
~
-
→
-
[-1+12,72]
[IT, 21T]
=
Zith
1+12-2*11
'
=
-
'T
'
IT ≤ ✗ 12Th
+
Proof that coscx)
cosC-×)
=
recall the def of cos and sin
case 1
O ≤
:
-
✗ ≤ IT
Accost)
→
-
IT ≤ ✗ ≤ o
it ≤ ✗ 1- 21T -221T
→
cos ✗
cos CX)
→
)
Cos CZITC 1) t ✗ 1- 21T
-
=
Cos [21T ✗
-
Case2
:
IT ≤
cos (f)
→
-
21T)
=
Also , cos C- ×)
=
→
:
✗
-
cos C- ×)
=
-
IT →
0 ≤ ✗+ zit ≤ IT
sin✗
=
sin✗
=
'
cosC- ✗ )
'
=
In all three cases , cos CX)
Proof that strict)
case I :O ≤
✗
=
-
=
-
✗ ≤ IT
→
,
-
=
IT !
-
Bt def of sin ,
.
✗
=
sina.IT ×)
* ≤ ✗ ≤ 21T
cos [✗+21T)
-
-
'
=
Sin×
'
cos c- ×)
=
=
cos Cx)
=
.
a- ≤ ✗ ≤ 0
Sin (21T
-
✗ ≤ 21T
→
sinC- ×)
=
it 1- ✗ 1- 21T ≤ 21T
→
-
✗
-21T -2 ✗ ≤
-
-
-
21T )
IT
→
=
-
Sin C- ×)
0 ≤ ✗+21T ≤ IT
Sin CZITTX)
(✗+ Zit) -121T C- 1) , hence
sin Cx)
case 3
=
:
sin C- ×)
SMCXTZIT)
-
✗
=
=
Zitht ✗'
Sink's
Sin C-×)
=
'
✗
,
Sin C-✗ )
-
In all three calls , sink)
Proof that since ×)
-
sinCIT ×)
-
=
=
=
C- [ 0,21T]
'
'
=
=
=
-
=
-
sin C- ✗ -211-11)
sinC- ×)
Sin(f)
-
Sink)
SinC-×)
sinCÑc%C ) sinC-NcosCIT)
-
=
Strict)
-
t
+
'
=
'
cos C- ✗ -211-11)
=
✗ Zithtx
✗ C- [0,21T]
ad hence
sin CX) sin (✗ 1- 21T)
:
o≤ ✗ ≤ it
sinC-×)
CXTZIT) +21TC-1)
case 2
1- COSH
21+11 1- ×' , ✗' C- [0,21T]
=
,
Cos Kitt ×)
cos CX )
=
'
=
✗ C- [0,21T]
21T ≤ ✗ I
-
* ≤ ✗ ≤ 21T
✗ 21TKt ✗
'
cos ×
=
sin✗
case3
-
'
cos (21T C- 1) + ✗ 1- 21T )
=
COSCZIT ×)
cos C-×)
✗ ≤ 21T
-
=
o≤ ✗ ≤ it
cos (✗1- 21T)
=
cos ✗
=
¥
=
cos at ×)
-
=
=
=
cos Cit ) cos C-×)
-
-
cos C- ×)
cos G)
-
Sin
CÉ )° SinC- ×)
I
19 (a)
.
↓ Iz dt
,+
I
arctarict>
=
1++2
1+1×2
Since
is integrable , by FTCZ
I
↓ Iz dt
arctanci)
=
,+
-
arctcnco)
IT
'T
=
-
g
O
=
g
I. Ot
(b)
+
I
arctoict)
is diff
=
,
,+×
.
everywhere
.
Therefore
✗
FCX)
=
/ arctoictot
0
F'a) arctacx)
=
and
Cs
n
farctoictwt-limfarctoictidt-h.vn/arctacn)-arctaas]
n no
n > is
-
-
hence
o
1+1+2
dt
IT
=
z
Proof that Iim tax
Elim
-
_
co
✗→
sin✗
=
✗→
E-
Cost
Iim
✗→
E-
1- coil
=
COSX
Proof that limarctorcx)
✗→ co
let @ so
Then ,
✗→
E-
1- cosh
=
°
cogzx
I
.
IT
-
z
=
,;m
EL
IT
z
IT
Let ✗ > tact/ 2- E) Then z sarctancx) > F- @
.
Thatis , V-E>07M> oV- ✗ ✗> M → Iarctacx)
→
limarctorct1=11-12
✗ -70
-
¥12 @
"
=
'T
-
2
4
20
11m ✗sin C / ×)
'
.
=
✗→ as
Iim
sine' / ×)
=
/
✗→ as
g
✗
Note that
Iim
✗
-
cos Cx)
> as
_
-
l
'
×
=
✗
Iimcokx) chick doesn'texist
✗s o
1
→
can't v6 L'Hospital
2
Iim
sink)
✗ so
-
:/im
✗→ o
Iimsincl / ×)
✗ → as
M> 0
✗>M
V-E> 0 let
→
'
lim / ✗
=
1M
s
-
=
£ 1m
<
E→ M
=
£
Then × > µ .
_
.
I
0
✗→ as
sin is continuous
everywhere Hence
Iim Sincil×)
✗→ co
.
=
Sin (limit×)
✗→ as
=
sin 0=0
<E
=
✗
go
Cosa/
=
1
1
21 Cal Define
Recall that previously , sink had ✗ in degrees aclsinrx
.
had ✗in radios
sinocxi-sin.FI/Becav62iTraU--36OYoehad10=
cosocx)
( %o× )
cos
_
sink
'
Siri tells us the rate
to degrees
.
(b) lim
✗→ o
-
sin
)
'
/ }Éo✗) oithxindesrees
rad and
.
=
180
180
'T
.
=
=
✗
11m
✗→ o
)
-
Ñn°(×,
=
180
cos
Sin"×)
'
*COSTA
IT
.
) 180
(
tim ✗ Sint / ×
✗→ 0
Siri
360
pr.o.c.ofraurel.to degrees
Sino 'CX)= cos (
cos°Cx)=
-
21T
Sirin respect
change
-
-
.
(¥1.1T
180
180
1
1
=
Iim ✗ sina.CN
✗→ ◦+
IT
=
=
Iim
Smc'I
§
=
✗
✗not
'
Alton
cosc'% )
=/im
Iim
✗ → of
*
180
,
¥0 )
sin (
IT
IT
=
'
180
too
=
.
=
IT
-
*×
180
180
'
≥
22
For ✗ E [ I, I] , we defined ACN
^
-
.
✗ 1- ×
=
!
+
z
•
I
t dt
'
-
i
>
the graph OF chichis
•É•y
ACX)
-
we defined cos a) for ✗ C- [0,1T]
Theimage of A- is [0,11-12]
.
Act)
=
by Accost)
=
{
,
and
sin✗
1- COSH
=
.
{ C- [0,172]
✗ C- [0,1T] →
5. t
l
.
{ C- [0,11-12] there is at C- C- 1,1]
Hence,forever,
{ Furthermore + is unique since A.
,
is
decreasing Choice are are)
-
Therefore f- cos ✗ is defined adcniqreto every ✗ C- Cont]
_
.
.
Now , sin ✗ + cost 1 , so each point CCOSX , SMX) is on the unit circle
>
-
-
Now , for IT < ✗ ≤ 21T be here cos CX)
Note that 21T
-
=
Cos CZIT ×) and Sino)
-
=
-
.
Sin CRIT
-
×)
=
-
l
'
-
C- [OiIT] Therefore , For ✗ C- [M2'T] oeharecoscx) C- [ I , I] ,
✗
-
.
ad Sm CX) E [ 1,0]
-
suppose CX,f) is a point onthe unit circle ie ✗2++2=1
,
Then y
-
=
>
2
I
1- ✗
±
I
2
-
'
-
,
C-
✗
X ≥ o
[0 , I]
→
'
-
>
✗ ≤ I →
f E [
-
.
✗ E [ I , I]
-
I , I]
For any given ✗ there is a 2 C- [QIT] s t cos (2) ✗ ad a w E [IT 21T] 5.t ✗ Cos (w)
,
=
-
.
Thus ✗ cos (2)
=
=
Iff ≥ other -1
Iff ≤ other +
Thus
COS CW)
I
=
=
cos (21T ×)
-
-
Coste)
l
-
I
=
cost)
=
>
-
-
cos (U )
Sin (2)
=
=
-
sin (2)
l
-
=
Sin (w)
costar)
) , sink))
every point exit) isoform CCOSCO
-
=
Sinew)
=
.
-
23 (a) IT is Max
.
.
possible length of an interval
such an interval is
[2111T
-
on
chieh sin is one one
-
.
Form
¥ zhitt ¥ ]
2115T¥ 241+1)
[
or
,
,
it
¥]
-
Proof
^
+211-11
Any interval containing
or
>
¥+21T
has aninterior point it not
one ore
-
.
Proof Assume A-
( { 1-2*11-0
=
Then foray 0,20 oe hare
*
sin(172 + ×)
1-211-111-0)
,
+2*11-+0 , C- A ord Sin (72+21+11+4)
-
[2111T
>
sinCHTZ + ×)
(1%+211-11-81)
Sin
.
=
SMC"T2 ✗ )
-
cos C- ×)
=
¥+21This prated analogously
The result for
The largest
×)
-
cos ✗
=
-
sin (N2
-
.
possible intervals not containing ¥+2141 or ¥+21Thas intend points are d- term
-
¥ 2111T ¥ ]
ad sin is one dear them
-
(b) gcx) sin×
or
+
,
✗ C-
-
_
,
2115T¥ 241+11
[
it
-
¥]
.
C- 721-21111*12 +2111T)
÷
a
•
:÷→T:\
.
i÷
•
There is creatingto each h
.
Relative to arcsin , g- differs by 211-11
'
'
Therefore , (g- ' l
:
'
arcsin
.
-
-
¥
I
24 tcx)
seccx)
=
.
'
0 ≤ ✗ ≤ it
=
co, ×
/ cos ✗ not defined at ✗
+
=
th , h C- I.
a
•÷• ÷•gÉ
• COS
-
3*12
-
IT
/
I
cos
f-(X)
the image of 5 is [lit A) U C- 0 ,
-
I]
ad this is the domain off !
_
,
^
IT
•••
* 12
"
_
-
2s
I sinx sin-11<1 ✗ y 1 For all ✗≠ t
-
-
.
let ✗≠ -1
MVT
→
:-,
:
-
-
sin ✗
7C , c C- CX , -1 ) ^
>
I ≤
sin ✗
✗
.
-
-
sin ✗
✗
-
-
sin -1
Sint
✗
'
-
Sin (C)
t
-
lsinx sin-11 ≤ IX
.
-
t
cos (C)
=
equality any it c- Th h E I
≤ I
-
Sint
=
t
≤ I
-
-
,
-11
let ✗ Ct be any numbers , and let c c- [✗, f) 5. t
Sint
-
Sin✗
=
.
sinkc)
=
cos cc)
t ✗
-
Choose zs.t.tk
Sint
-
Sin ✗
=
¢ (✗ 2) Then
,
Sint
=
-
^
.
sinztsinz Sin ×
-
-
"
÷÷÷÷E÷E:i••
G- 2) cos CC,) + (2- ✗ ) cos (Cz )
where c, C- Gif) and Cz C- CX,2) , ad therefore Cz ≠ hit
→
Sint
-
Sin ×
≤ 1 Ct
=
≤
=
=
-
2) cos (G) It t.CZ ×) cos CC e)
-
1-1-21 / Cos (G)It 12 ✗IICOICG)I
-
1-1-21 t 12
t
-
✗
It ✗ I
-
-
✗I
/ cos ccz) I < 1 and
→
I
Icoscc, ) I ≤ I
Alternative Proofs
Fromproblem 14, sin Cx)
zsin (
✗
-
sinG)
a- ) cos ( Yt )
+
Assume sing ≤ sink)
=
=
1 Sint
=
-
Sin✗ I
G- ×)
-
2¥
Sint
-
✗
t
sin
2
-
2
cos
✗+ t
.
sin
≤ 2
2
.
sins
=
/ cost dt
=
t
t
t
✗
-1
Jot Jot + / costdt
_
✗
✗
=
Ct ×)
-
-
✗
ICI costlot
-
-1
+
'
+
✗
t
=
a- ) cos ( Yt )
2 sin (
G- ×)
=
-
z
/ sink1- I2) dt
✗
lethal sink✗ 12) ≥ 0 ad not identically zoo in at interval
=
:
.
y
Then / had× > 0 Hence
15inch2) Ot so ad thus I Smt sin✗I
recall sink×)
cos ✗
Def
Cos (4)
.
strict/ 2)
thus
-
.
,
>
=
>
-
sin ✗
=
I -2 sink
-
S is r
-
1- cos (x)
=
z
[✗it ES ^ ✗≠ -1
discrete
let road SCIR r discrete
→
1×-1-1 ≥ r
-
:
IR
-
s
IR
f > 0 for all ✗ ftp.\ g
'
-
sin ✗
r > O , S CIR
:
f
s
.
→
d- strictly increasing
]
'
=
-
< f-✗
'° ' "
2
=
It ✗ I
-
b
26
t.i.mg/FcxIsinCdx)dx
.
(a) Iim
✗→ •
%inX✗d✗
* FCX)
'
=
°
c
Proof
g- (f)
✗
→ as
-
cos <✗×)
1- G)
'
→
=
×
=
=
5inch×)
means the
graph is compressed extreme
sinCAN
tis cant everywhere
.
.
d
Isin child✗
FTCZ 1
-
a
'
-2 ≤
-
µ
ccosild
"
-
=
-
( cos (Ad) coscxc))
-
×
cosh) ≤
l
-
a
-2
d
→
¥11s inixox
=
0
(squeeze theorem)
(b) recall
fns defined on [a.b) is called a stepfunction if thereis partition P {to, .tn} of [crib] such thats isconstantan each
-
_
.
cti-i.fi)
.
Notethat thevalues of tatti arearbitrary
b
lets bestepfncn [a.b] Then /im / SCH sin did✗
=
0
.
A-s o
☐
Proof
Sis defined based on a partition P {to,
-
_
b
Iim / SCH sin child✗
✗→ as
a
=
Iim
✗→ O
i§
. .
.
.tn} of [a.b)
ti
si
/ sin Cdx)d✗
ti
-
,
=
0
.
.
.
(c) recall results
proved in 13-26
n
cast integrable on [a.b]
→
>07
stepfns fol It
3-
step ]nsz ≥ folds
V-E
,
f
b
b
≤
-
Is
,
b.b-
< E
b
b
(b) V-E>0,7s, ≤ fad Sz ≥f
-
1-15-15 ce
→
,
E
Sincil) ≥o
or
.
. .
s
f- integrable
→
,
.
.
e
b
5.
Assume )- integrable on [a.b] Given d.ie/-Px-- { to
have that _V✗ , ✗ C- Iti , ti]
,
_
.
a
.tn ,} be a partition of [a.b) s.t.tw at i=1,
V2 , ✗ C- [ti.at;]
→ sin CAX) < 0
.
lets , cross be step 5ns for Past 5. < taut>É, ie 5. < tcsz Then font [ti-i.fi] either
.
1) sin Cdt) ≥o
→
2) sin CAN ≤ o
→
.
s.CN Sincil) < tctssincix) < Sza) 5inch)
Szcx) Sincil)
<
tctssincxxscs.CN 5inch)
In either case iloeintesraleadtahethe limit
ti
ti
◦
=
Iims /
.
_
ti
◦
=
ti
sinxxdxclimb-cxssindxdxcl.ms/sinAxdx--oX-icsti-i
A-' o ti
,
A-' o
.
.
A-' • tie,
ti i
-
A-' o
ti
Iim b- Cx)sind✗d✗=0
✗→ O ti
i
-
b
na
ti
b-G) sincxxldx-SN-cttsincitldx.co
i= '
a
ti ,
.
b
ni
ti
limb-ctlsinchxldx-limIN-cttsincdtldx.io
✗ → as
.
Iims / sinxxdxclimb-cxssindxdxcl.ms/sinAxdx--o
A-i s
→
ti
ti
ti
a
✗→ i s
i=i
1- i. ,
b
ti,
. .
.
,
na we
•o
27
C
'
(d) Sin
^
.
•
A
sinks) Iim
Sinclth)
-
sink)
-
-
h
h→o
B•=ci,o,
•
◦
=
,,m
sink)coKh) +5inch)cosc×)
-
ˢM✗
→
2
{
<
<
Strix
zcosx
=/im
h 'o
Proof
-
f-
sink) (1- cosch))
h
>
AreaCOCB)
tant
=
=
=
2
z
{
,
2
2
tax
a
(b) cosy <
2
Slncx)
<1
✗
Proof
sinx
From a) , Sinx < ✗ <
COSX > 0 , ✗ 30
sink
cos ✗ <
cost
.
< ,
✗
Iimcosx
I
=
✗-20
11m
→
"n✗
-
1-"'✗
Iim
(c)
§
=
✗
✗→ o
Alton
I
=
✗
✗ so
SMX
:
Iim
✗ so
-
.
a-COSXJCITCOSX)
11M
✗ Cltcosx)
✗ so
-
=
Iim
f-so
1- cos'✗
✗Cltcosx)
sinx
=
Iim
✗
=
-
] 0
sinx
'
✗
1.0--0
sing /im
n,
=
SMX
-
sin ✗
AreaCOAB)=
It cost
sincx)
h
h so
(a) 0 < ✗ < Mz
-
=
,
0
Cosct)
1- cosh
h
Cosa,
+
°
+ coscxslim
h so
-
Sinh
h
]
slnch
h
28 fcx)
LCX)
✗ E C- 1,1 ]
'
I
=
.
-
X
length off on [ ✗ it]
=
I
(a)
LCX)
=
'
/
dt
1- fz
✗
Proof
•^
•
d- (f)
)
t
✗
-
'
-
t
'
↓
=
,
, ×
-
f (x)
,
"
×
l
-
'
-
×
-
=
✗ C- ✗)
-
=
I
'
-
✗
2
+✗
-
i
< °
=
"2
2
1- ✗
( 1- ✗2)
"2
( 1- ✗2)
2
1- ✗
tophalfofcn.it circle
In Problem 13
-
Zsoestcrtedoiththelensthlctip) of apdtgcnalarveinscribedinthegraphoff.no defined the length
Orton cab] as the wpoflct.PT
.
b
Oeshooeu that this number is
! 1+0-112 ittheintesracis intactintegrable
.
2
✗
f'Cti=
1- ✗
2
Itf'C×i=
I
1- ✗
2
I
"
1- (d- (t))
'
therefore,
/
✗
=
2
1- ✗
'
l
-
t
'
dtisinfactthelasthoffcncx.IT , it
recall that arcsiricx)
I
=
,
recall
,
,
_
×
,
_
×
✗ E C- 1. 1)
i•arcsin
•
'
,
^
thegraphofarcsin
arcsin is unbounded near 1. Hence
I
-
-
¥
/arcsinktlotis improper ad is defined
✗
e
Iim / arcsirictldt
Est
-
✗
=
Iim / arcsince)
@ → 1-
-
arcsincx)
]
=
I
-
arcsincx)
is integrable an [✗it]
.
I
-
(b) L' (f)
✗ C-
=
, ×
,
C- 1,1)
-
LCX)
] it
:
"
I
1- +
=
.
✗
! it
dt
-
.
↑ it
dt
.
isccnt.cn C- 1,1) hereby FTCI
,
,
1
-
L' CX)
✗ C- C- lil)
-
_
, ×
,
-
(c) IT =L C- 1)
LCCOICX)) :X
SMCX)
✗ C- [0,1T]
'
→
cos (f)
=
Strict)
=
-
Strict)
✗ C- (0,1T)
cosy)
1- COIZX
=
Proof
I
LCX)
:
'
/
l
✗
L' CX)
-
-
✗ C- [-1,1]
dt
t
'
1
✗ C- C- lil)
-
_
, ×
,
-
L' co
-
>
Lcieoeasingine ore
-
lent > Want
→
1-
'
isatin
-
.
Note that t.tl) IT LCI) :O , not decreasing
-
-
,
The domain of 1c- ( Lccosx))
'
is [0,1T]
,
[ (X)
'
'
=
the image is [ -1,1]
.
.
✗ C- [0,1T]
COSCX) =L (X)
_
'
1
I
'
(E) (f)
=
=
=
µ ,, ,,,,
,
_
,
-
5inch
✗ C- [0,1T]
1- cost
-
sinlcx)
zcoscx) cos'C✗)
-
coset) C- sinx)
=
=
2
>
1- cos ×
=
sincx)
Cos G)
✗ C- [0,1T]
✗
29 (a) ✗ (X)
.
=/ 11-1-21 lot
-
(
→
✗ odd ,
,µ §¥
increasing
liman exist
✗ s IN
-
Iim ✗(f)
✗
-
=
-
liminal
✗ >→
> as
-
* aime ,
* wine,a,,
→
✗ →as
Proof
f- diff
✗
✗G)
! (11-1-21 dt
_
=
1+1
'
is differentiable and even
,
'
FTCI → ✗ 4-1
.
11min41
(1++2)-1 > ofcrallx
=
IF ✗(f) is odd
✗
get) =) CH fl ×)
-
Ot
.
-
-
'
g. G)
then
:
-
'
=
.tk/-)+-f' C- ×)
'
Itf isodd.ie if )- G)
'
I
fatty
Letta beadiffer.tn and
increasing
.
×. .
✗ → Cs
'
✗
→
.
Iim / ( 11-1-21 dtexistsbecavll
=
-
Proof
.
=
-
f' C- ×) , then
✗
!
01-+1<1+1-4 Ot
'
_
exists
'
g. Ct)=o
'
"
→
g. is constant
cat , hence integrable
,
onion]
~
.
§ f. Of
,
/f-
"
=
-
t
I
-
-
.
since oc
'
'
c
1++2
since gco)=o then get)=O
+l
-
→
s
dx
-
-
l
Hence , JCX) =) C- ×) , iefis even
°
+2
-
fatty
then
-101 exists
11m ✗ CH
=
✗
✗→ i s
.
f- diff
Iim / ( 11-1-21 Ot exists because
-
s
-
b
Iim
-
o
todd
→
o
/ (11-1-25'ot
Iim
=
✗
✗→ - 0
f'even
'
o
=
.
-
✗
_
.
-
( Fatf ) Ot fatty Ot ]
-
'
✗
✗→ i s
-
+
-
c-
catch [-1,0] hence int
-
✗ s
-
-
.
Ét→ot= 11m f- f- Ii /
'
Iim
✗
co
✗→
'
0C
<
11-1-2
-
Proof
'
'
=
Iim
✗ so
get)=tCt)tfC-✗)
(1+1)=1
-
as
'
g. (f)
◦
I
C- 0,0)
an
µ
(11-1-27
→
'
_
Ot exists
.
=J'cx) f- C-×)
'
-
d- even of 'Ct)= ] C-×)
'
-
'
-
✗
-
)=ftcHot=
Since ✗ isodothorxcx
✗
Iim
✗→ i s
-
Finally
✗
,
°
=
-
✗ C- ×)
and
→
✗
=
-
lim / fctldt
✗→
_
✗ is increasing choice , are
on the image of ✗ chichis
Haidt
o
✗
!ktWt= -11m / f- Chot
→ as
-
×
is
-
C- 172,4127
°
=
glct) -0
-
I
2
one) ada'≠o so
g constant
→
✗
-
'
is defined
901=0--5--0
-
sfct)
→
?⃝
=
.
tooo
-
3- C- ×)
'
(b) (✗ 1) (X)
-
It [✗
_
=
≥
'
(x)]
Proof
(✗
-
'
I' (f)
1
I
It [ ✗
-
=
=
=
gyg.gg,,
,
'
CHI
'
It [✗ (x))
"
recap
✗
✗G)
! (11-1-25'dt
=
"
atty
'
✗ CX)
(1++2)
-
=
'
✗
so
cretinous ' is defined or (-1+12,1+12) and
-
(d- I' CN
-
"
It [✗ Ct) )
_
'
'
'
(c) ✗ C- C- +12.1421
tact)
Sinai
>
define
hereoecresimpynamingafnoeawecdthnwexistscnc-MZ.IT/2l
✗ G)
"
_
tax
adoeoealeaothertntromthepreviwsae
:
lttoix
limsincx)
E-
Cil
✗ (x)
n
I
-
✗→
Proof
✗ (x)
"
sincx)
tancx)
tact)
=
=
=
'
It [✗ Ctl]
"
tan'Cx)
/ + tank)
11m tacxl-cscnotswehootoshoothilkrmo.ly here)
✗→
E-
11M
1+1-94×1=8
✗ → it,
Usmgllltopital , tim sina.tl
tax
Iim
=
=
•
Itta ✗
>
✗→ %
-
✗→ E-
'
It [✗ (t)]
"
as
=
Iim
a.µ, ,✗
.
✗→
E-
.
,q,
'
Iim
✗→
E-
"
[ ' + [✗ "t] )
"
=
'
d- (f) (✗ e) (t)
'
-
'
( It [✗
"
=/im
'
"
(t)] )
=
"
✗
(f) (( + [✗ (f)5)
_
" "m
'
'
( It [ d- (x)] )
'
✗
-
'
'
It [✗' G) ]
"
(t)
'
≤
lim
✗ '
-
E-
Sincx)
Ie , limsincx)
✗ → E-
ad
→
=
Iim
✗→
÷
.
I
sing)
.
It /imsincx) exists thence hare Iimsincx) =L , lim
✗→
E-
✗→
2--1 Therefore 12=1--1--1--1 Howerd since
.
tact) > o
,
→
sin (f) so
.
.
,
oeoe
E-
✗→
E-
l
'
=
l '
Sincx)
considering ✗sonar limit then act)> o
-
'
✗ G) > 0
ii) limsincx)
✗→
_
=
-
I
It
Proof
limsincx)
Analogous to a) ,
→
11m sinct)
✗→
-
IT
=
-
✗→
-
It
tax
Iim
=
✗→
_
-
'
as
=
=
co
'
72--1
/ + fay
lim
× .
+
.
.
I
/ zt
'CX)={
Strict)
Ciii) Sin
fancy,
I
✗ 1=0 and ✗ C- C- 172,72)
✗:O
Proof
sincx)
tax
=
lttoix
lttoix-tanxtax.CI/-toix)
Clttanx)
sin'C✗)
lttarix
=
lttonzx
≥
Clttoix)
=
-
toixclttazx)
I
=
Citta't)"
✗ :O
→
Strict) =\
✗ to
→
tact) -1-0
"
Civ) sin (f)
=
→
-5inch
Strict)
lttoix
=
tax
tact) .CI/-toix)
"Ct)
sin
-
|+taz×
lttoix
lttoix
✗ C- (-11-12,412)
=
Smx
tax
-
tax
=
=
lttoix
-
sins
tax
sink
30
Suppose thereis Into not always zero WI to + to -0
"
-
.
(a)
.
fit Cfo 5 is constant and either t.co) -1-0 or f. co) -1-0
'
'
,
.
Proof
to't
"
.
'
to to
+
o
-
-
Hito +2 to to o
'
"
-
-
[ Cto I' + ti ]
'
→
(ti )'t ti
'
o
-
-
is constant
since toG) is not always zero then thereis some ✗ 01 to Ct)
'
'
Then Cto (A)
constant -1-0 because i > o ao
'
+ c
Assume to(o)
=
=
'
Assume to co)
Then
chico))
0 Then
toco)
0
=
.
-
?
.
-
=
h>o
h> o
→
c≠ 0
Gotts5 ≥ o
fico) ≠ 0
→
=
t.co) -1-0
.
.
.
.
(b) there is5ns S.t.
"
S t S
=
Sco)
0
=
S (o)
'
=
0
1
Proof
Oe are stillunder the initial
lets
assumptionthat thereis Into not always zero WI to + to -0
"
-
"
atotbto
'
-
-
"
ato + b. to
'
'
Then s
s
=
'
=
ato
-
b-to
"
since to
=
-
-
at
+
bti
"
"'
"
S +S
to
"
.
.
=
ati't b to + atotbto
'
'
"
sco)
=
'
s co)
=
>
at (o) +b fico)
=
◦
a
t.co
cos
to cost b. to (o)
"
'
tied
-
=
0
=
a
0
>
0
,
because C- to + to)
Cto + to) + b (to't to )
"
"
'
"'
=
I
/ 1;) =/ ! /
to co)
-
determinant
=
-
to Coi
Thus thereis a solution
.
to Coi
(to Coit to Coi )
'
'
-
=
_
=
constant -1-0 by
portal
.
to
+
'
to
=
0
(c) Define
sin
Cos
Recap
"
we assume 1-
s
-
-
S
=
+5--0 has a solution other tha the)no
-
.
'
Oe name this assumed to exist 501'n to
-
-
Then , cos cannotbe
positive for all ✗so
.
.
Then, this 501'n satisfies another D.E. aswell
Proof
:
to:-( to)
'
constant
proved
In 1111-1-6 we
'
ad at 0 , to auto cannotbothbe 0
f- twice differentiable
✗ 20
text > 0
→
×, ×> on g- (×,
"
→
f- decreasing
◦
=
f-'co) -0
Recall whats it
1
.
/ (f) tbfocx)
SCX)
=
a
Sco)
=
sin0=0
'
-
_
atotbfo
"
.
S +5--0 ,
justlike to
.
.
Noo assume SCO) -0,5403=1
-
◦
.
These should single outaJn lespe.at/aadb
COS 0=1
=
'
-
chat property does thisKhare?
-
s co)
Noo define a 5ns
.
,
"
Whereto is f.t.to + to -0 , adisnotakrats zero
-
.
we showed itisakrats
.
possibleto solve the linear equation that
result
.
Cos ✗
=
skx)
axial 1- by :(t)
-
-
At this point we've constructed this Fnsst
Assume _V✗ , ✗so
→
cos ✗ > 0
.
.
S 't ) -0
'
-
Then Skt) > 0
Scot
.
since sco) other _V✗ , ✗So → sct) so
_
"
'
→
s is
S"C✗)= -50-7
→
-
5401--1
-
Butit is alsotrue that s +5=0
0
-
Note that
decreasing
S
"
-
-
'
S
s
"'
=
-
s
bt assumption so atleast twicediff
,
butthen it must be thricediff
.
.
At this pointwehave
chat does all this men?
cos
'
-
-
s is decreasing
'
s is continuous , S co) -1 , ad tx, ✗so f. t.sk/-o
✗ ≥ o → cos CX) so
Costco)
S (o)
"
=
=
-
s co)
=
Sinco)
-
=
'
-
0
thus the set {× :X> on skis ≤ o } has ainfimum
cos is twice differentiable
Let's call such a number ✗
Bf
"
1111-1-6 , FX , ✗> on cos (f)
But cos"Ct)
so
'"
=
S
(1)
=
7- ✗ , ✗> or cos (f)
-54×1
=
0
Thus , 7×0 ✗ > on cos ✗ ≤ 0
,
◦
.
.
1-
=
0
we define it as Zxo
=
.
-
cos (x)
.
o .
.
.
(d) sin (1+12)=1
Proof
"
S
=
-
s
SCX)
s Ct)
=
s "C✗)
SCH
-
s
s (t)
Strict)
-
=
cos (t)
'
'
'" '
-
cos G)
=
"
S (X)
=
Sin (t)
=
'
"
SCL)
→
=
(t)
-
s (t)
=
s Cx)
-
"
=
-
=
SCX)
by definition 5kHz)
Sin (f)
=
cos #2)
_
,
'
=
'
h (X)
,
>
hence , sinkHz ) + cos CITI 2)
scax)
=
'
S Ca✗1. a
=
=
S (o)
sin CITK) l or sin CITK)
-
-
=
-
I?
-
SICH so
→
=
1
Therefore,
S
sinco) -0 ad on (0,1+12) ,
=
Sin(11-12)--1
IF the conditions in the lemma are satisfied by
(e) COS IT, Sin IT, Cos 21T, sin21T
.
"
g. tg
] + f- 0
-
=
"
-
=
-
=
suppose f- defined everywhere and that
Hero ,
f + f- O
"
d-co)
'
d- (o)
-
g-
-
O , ie
0
=
O
'
f-bs + as
0
=
.
=
(1)
-
=
a
-
g-co) b'(o) b ta 0
"
g
-
gco) ] co) b. 0
Recall
lemma
a
0
=
'
we know
:
.
"
sink11721--1
:
cos cax)
=
5 1-5=0
I
SCO)
chichis it
-
°
>
→
s CX)
"
=
Wehave
so set CS 12=1
'
1
Sin(f)
=
Let hcx)
0
.
01-1
=
=
5+4't? constant
As shocnin part (a) ,
Skolt (Sho))
SCH
=
bsin tacos
0
This allows us toprove
then F- o
-
sincttf)
Theorem 4
"
d- defined everywhere
cos att)
f + f- O
=
=
Sin✗ cost 1- sin-1 Cos ✗
cos ✗ cost
-
sin✗ Sint
"
-
d-co)
-
=
d- (o)
'
-
_
a
s
f- b sin tacos
-
Thus
b
COS IT cos (1+121-1+12)
=
-
-
Proof
sin
let gcx)
=
tct) bs
-
"
g.
4)
=
f' Ct)
-
Coif
'
-
Sin iT=
b s't as
f- (f) tbs + as'
"
=
-
0
as
cos 21T
'
=
'
-
ther
g. (1)
2- Sin"T2 Cos"Tz
=
sin21T 25in
:
cos
=
'
2- sinCITK)
cos CTU)
0
I
=
-
1
(F) cos and sin are periodic v1 period 21T
Proof
Recall
:
a
5nF is periodic 01 period T if fcxtt) =)(f) forall ✗
sin (✗ t 21T
-
)
cos (✗1- 21T)
>
=
=
Sin✗
=
cos G) color IT)
=
0
Sin(t) cos (21T) + 5in (21T) cos CX)
cos ✗
-
ZITI
Sin Cx) Sin [
.
31
Recall 11-24
.
Graph of RationalFunction
P
Rational Fns
let
p[✗I
"
=
an✗
Pia polynomial fns , anotalways zero
q
,
t
+ a. ✗ tao
. .
-
and qcx)
bm✗7-
=
phas at most n roots , qhas atmost m roots
••
roots of p , itnot roots of a, oe roots of
•
Pla is not defined at roots of9
.
-
-
t b ✗ tbo
,
.
.
p
a
.
.
oe Contador out common roots in denom adnvmer
.
.
•
Theresulting expressionis not defined atthese commaroots ,but for other points that aren't a commonroot nor a root ofa
we can cancel the commonfactors
.
"
an✗
From S 32 oehncw that Iim
+
"
-
^
+a_✗ + a ◦
=
-
✗→ •
bm✗7-
-
.
-
✗
Iim
,
n
)
box
m
-
-
-
a ◦✗
-
-
(bm + bm
-
.
,
✗
'
-
+
.
.
.
+
)
(a) A rationalfn has afinite number of roots unless it is zero everywhere)
Sin has infinite roots
exists if] m ≥ n
-
m
✗→ as ✗
t b ✗ tbo
"
Can + an , ✗ +
.
.
sin (21+11)--0 , h E E
:
ao sin is not 0 everywhere
.
(b) sin isn't defined implicitly by an algebraic eq
.
ie , there do not exist rationalFns Fo,
"
(sin ✗)
+
" '
Fn Cx) (sin✗)
. .
.
.tn
-
,
+
. . -
+
.
.
5.t
.
,
To G)
=
For all ✗
0
Proof
✗
=
ITH
→
"
(sin ✗)
sin Cth)
=
0 →
To(ith)
'
Fn Cx) (sin✗In +
-
+
. . -
+
.
.
Sin✗
[ sin ✗ + In Cx) (sin✗5" '
" '
_
+
,
✗ + Fn , (x) (sin×)
" '
=
. -
-
+
""
_
Butthen ,by continuity , Vx , Sin
Therefore, I , (f)
→
'
_
✗ ≠ ith → sin
0
0 since
1- (ith)
=
_
to
=
F.(t) sin Cx)
_
=
=
+
. -
-
Csince to is rational, either itis o at a finite numberof
O
=
o
T.CN/--o
+
F.Ct)
=
0
✗ + In , Cx) (sin✗5- +
'
_
0
.
. -
-
+
F.Cx)
points or it is zero)
.
let's prove by induction that the set
A {n
=
n :O
} a fn 0 in (sin ✗)
n EN t {0
:
"
=
To -0 as weproved previously
'
Fn Cx) (sin✗In +
-
+
.
.
-
→
Assume Ji
=
for i 0,1 ,
0
=
"
(Sin ×)
Then
sincx)
✗ ≠ ith
It
.
.
Fn (x) (sin✗)
.
_
,
/ (sin✗In
-
(sin ✗In
-
→
.
" '
+
""
=
. .
'""'
.
'""'
+
.
. .
+
""
Fry (t) (sin ×)
,
Enact) (sin✗In
"
+
Fn (f) (sin×)
_
,
" """
-
+
-
+
""''
+
Enacts ]
Enacts
=
0
Butthis expression is continuous except at possiblyafinite set of points
But if
. . -
+
To G)
=
0
For all ✗
}
∅ , , ∅ , satisfy
32
.
Qi't g lo
,
=
,
0
'
④i + g, 0/2--0
92 > 9 i
"
(a) ∅ ,
de olio
-
Cgz g.) 0,02--0
-
-
,
Proof
¢, ∅, + 9, 01,02
"
∅z ∅ + 92¢20
"
,
-
Oli ∅, + 9, ∅ Oz
"
'
∅ ∅,
=
-
.
,
"
0 ∅,
(2)
0
=
,
(1) (2)
a)
0
=
-
,
did,
-
-
g. 0,0/2--0
∅ & (ga g,)
-
,
-
o
b
(b)
∅, CX) so
∅, (x) > o
a) [ ¢ ∅
"
"
tara" ✗ in Ca b)
→
,
,
-
∅, ¢ ] so
,
proof
✗ e ca ,b)
→
0.0, Cgz g.) so →
-
did
-
did, 0.02 Cgz g.) > o
=
-
,
b
i.F.
a) [ click olio ]
-
,
so
'
since [ click -0 . Oli ]
=
di di + didi
b
a) [ ¢ ∅
"
Then, Fitz
,
_
,
-
0:10 ]
,
Assume ∅, (a)
=
=
[ loicblozcb)
,
,
=
d. ∅
-
∅ Cb)
∅ Cb)
"
=
,
[ loicblozcb) 6.(b) di Cb))
=
=
(C) Cannot have ∅, (a)
didi -0,19
"
-
0
.
=
-
loicasozca) )
-
-
-
did
,
[olicasozca) 0,(a) di Ca) ]
-
[6.(b) di Cb)
-
∅ Catolica) ) so
,
0
Then
loicblozcb) QicatchCa) > 0 from partb)
-
,
Noo ,
.
Oli (a) ≥ o ao 0!Cb) ≤ 0 otheroiceoe'dhare negativevalves
of ∅, an Ca,b)
,
.
But then loicblozcb)
-
loicasozca) < O
±
Hence Q , (a) to or Q , (b) 1=0
.
C it ∅, is continuous then
02cal ad ∅z0b) ≥ 0)
(d) ∅ (a)
0, (b)
=
,
is also
0
=
it ¢ , > 0 , Oz < 0
b)
an Ca,
b)
ii)
0,20 0 > o
an Ca,
iii)
¢ <0,0220
an Ca,
,
,
b)
,
In it wehave 0 , > 0 , 012<0
✗ E ca ,b)
"
,
-
b)
an Ca,
Chloe Cgz g.) < o → did, did,
-
→
-
b
a) [ d. ∅
impossible if
did ] < 0
,
=
[ olicblozcb)
-
loicasozca) )
Assume ∅,(a) ¢,Cbl -0 Then ∅:(b) Odb)
'
In iil ,
[ olicblozcb)
-
-
∅ Catolica) ) < 0
,
-
.
,
.
b)
an Ca,
Ioicasozca) )
now assume ∅ ,(a)
Then
[6.(b) dicb)
-
0,20 0 > 0
,
-
,
.
But ∅ , Cb) < 0 ,
Chloe Cgz g.) < o
Olicatch(a) co
¥ IT
Qica) > 0 so olicblozcb) Qicatch(a) ≥o -1
-
=
-
=
∅ Cb) -0
-
[6.(b) dicb)
-
∅ Catolica) ) so bt pot b) calculations
,
-
=
.
olicblozcb) Qicatch(a) < 0
-
.
assumptions cchich also imply Q!(a) < 0 Q!(b) so)
But consider arinitid
,
:
olicblozcb) Qi(a) ∅ (a) ≥o -1
-
+
≥o
,
-
.
≥o
Simitoy in iii.
part b) plus ∅ ,(a)
=
d.(b)
=
0 meds
olicblozcb) diCaio Ca) > o
-
,
.
but since ¢, <0,0220 and di Casco ao Qicb) > 0 , then olicblozcb)
+
≤o
-
loicasozca) ≤ 0
-
≤o
.
-1
.
.
Interpretation
we
proved
∅ , , ∅ , satisfy
"
+9,0/1=0
014+9,0/2--0
92 > 91
Then
∅,C✗) > 0
¢,(×, > ◦
tara" ✗ incaib)
→
Cannot have ∅, (a)
=
∅ (b) =o
,
Therefore the contrapositive Sats
0.ca) -0,41=0
-
→
QQ) ≤ ofcrsomexincaib) or Qzct)≤ ofcrsomexincaib)
Ifwe assume Q, > ocnc-ibltheraadboecacecutiveroots.no/-herewltsctsthatOzmusl- hareazeoin
G.b)
.
sin [ Cht
33 (a)
.
;-) ×]
SinIch
-
-
f-1×3=2 sincxlzlcoschx)
This identity
Proof
Recall from Problem 14
sink)
-
sing)=2sin(
✗
a- / cos / ¥ )
+
Hoke
sin [ Cht
=
25in
{
(b)
;-) ×]
-
sinlch f) ×]
-
(E) coschx)
sin /Cnt
+ cosxtcoszxt
-
-
-
tcosnx
;-) ×]
=
25in
(É)
leads to thisformulator Écoscix> +
in
chkhoecalatercpotdluceto compute
Proof
b
sin [ Cht ;-) ×]
I
z
-
sin[Ch f- 1×1=2
-
1- COS✗ + cos 2.✗ +
=
.
.
-
+
×,
zsin(
,,
'
25in(✗ 12)
1- C.
.
.
zsincxyz,
I
1
z
+
/ sin [C' + E) ×]
/ sin [Czt ;-) ×]
/ sin [↳ + E) ×]
25in( 12)
;-) ×]
=
25in
-
sink'
since
-
-
f) ×]]
f-1×3]
(E)
-
sin[Cn
-
f) ×]]
/ sin [Cnt ;-) ×] sink' f) ×]]
-
✗
sin /Cnt
-
)
I
+
sincllzlcoschxocosbf Computinglower sums
tcosnx
I
{
+
=
{
-
.
G)
sinxtsinzxt
. .
_
Stn
+ sinnx
=
'
(Y ×/ sin /{ ×)
sin
Proof
consider the expression cos [ Cht
/ E)
'z)×] cos / Ch E) ×]
-
-
From problem 14
cos Catb)
cos ca
-
Cos (a) coscb)
=
b)
=
=
cos (a) cosC-b)
-
-
-
since) sinC-b)
cos (a) cos Cb) 1- Simca) sincb)
coscatbltcosca b)
coscatb)
since) Sincb)
-
cos ca
-
b)
Zcos (a) coscb)
=
=
-25in (a) sin Cb)
Usingthelatter result wehae
cos [ Cht
→
sinchx)
'z)×] cos / Ch E) ×]
-
-
cos
-
h
-
É)
I
-
=
×
zg,n(
11--1
zsinchx) sin(E)
-
'z)×]
-
25in(
[ 5inchf)
_
cos [ Cht
/ (h f) ×]
=
=
, ,,
25in(
/
cos (✗ 12)
-
É)
cos (3×12)
+ cos
(3×12)
-
cos
(51-12) t
.
.
.
+
cos ( Cn
-
-
'
cos [Cnt / 2) ×]
-
Cosa/2)
=
=
zsincxlz)
-25in
'
/×
/ sin ( y )
7
coscx)
-
cos 4)
=
-
zsin (
✗
F) sin / ¥ )
✗
cos
E) ×)
( Cnt E) ×)
(d) let P bepartition of lab] into equal subintervals
Let t.CH cos ✗ , fact) sink
Suppose be [01*12] Then
-
-
_
-
Loi , P)
bn É
=
.
1
Since
Loi ,p)
=
.
.
(i. bn )
=
I [{
§ cos (i. bn ) )
+
"'
sin [Cnt
1- COS✗ + cos 2✗ +
z
.
cos
i= ,
.
.
.
-
t cosh✗
25in
by
ntt ) nb ]
sin[ (
;-) ×]
=
(E)
-
zbn
then
b
,
zsinzbn)
an
Wehner that lowersums increase as we increasepoints d-at givenpartition
[
11m sin Cnt
h so
f) bn ]
=
sinb
-
Slncblzn)
11M
=
b/2n
n so
✗ so
-
-
Hora 11m With)
h
-
11m
=
Smx
=
×
I
sinb
sb
b
proving Isin
=
I
-
cosCb) is analogous
.
.