3.5: Trigonometric Functions Reference Evans 6.1
Consider a right-angled triangle with angle θ and side lengths x, y and h asshown:
θx
yh
The trigonometric functions sine, cosine and tangent of θ are defined as:
sin θ =opposite
hypotenuse=
y
h, cos θ =
adjacent
hypotenuse=
x
h
tan θ =opposite
adjacent=
y
x=
sin θ
cos θ
71
Reciprocal Trigonometric Functions
The reciprocal trigonometric functions secant, cosecant and cotangent are de-fined as:
sec θ =1
cos θ
cosec θ =1
sin θ
cot θ =1
tan θor
cos θ
sin θ
72
Example
θ12
513
For the above right-angled triangle, find:
1. sin θ
2. cos θ
3. tan θ
4. sec θ
5. cosec θ
6. cot θ
You can now attempt Sheet 3 Q19
73
The Unit Circle
We can form right-angled triangles in a unit circle (circle of radius 1).
If θ is the anticlockwise angle between the positive x-axis and the ray−−→OP then for
all θ ∈ R:
sin θ =y
1= y , cos θ =
x
1= x and tan θ =
y
x
1
x
y
x
y
θ
1
1O
P(x,y)
1
x
y
x
y
θ
1
1O
P(x,y)
1
x
y
x
y
θ
1
1
negative θ(clockwise)
O
P(x,y)
74
Signs of Trigonometric Functions
The signs of the trigonometric functions for different values of θ can be determinedby noting the signs of x and y in the unit circle:
1
x
y
x
y1
1
y
x
1st quadrant x positive y positive
4th quadrant x positive y negative
2nd quadrant x negative y positive
3rd quadrant x negative y negative
75
‘CAST’ Diagram
Since in the unit circle sin θ = y, cos θ = x and tan θ = sin θ
cos θ = y
x, we obtain the
following ‘CAST’ diagram as a shorthand for remembering signs of trigonometricfunctions in different parts of the circle:
x
y1
1
ALL sin, cos, tan positive
only COS positive
only SIN positive
only TAN positive
C
AS
T
76
Degrees and Radians
In the unit circle, if we measure the length of the arc from A to B in an anticlock-wise direction, we have another way of measuring angles.
A
B
1
1
.
.θ
This length is called the radian measure of the angle θ and we can signify it bythe symbol θc rather than θo, but usually we don’t show units for radians.
77
Radians on the Unit Circle
Since the circumference of the unit circle is 2πr = 2π · 1 = 2π, a completeangle around the whole circle is 2π radians. Hence if we divide the circle into fourquarters or quadrants we have radian measures as shown:
0, 2ππ
π2
3π2
78
Converting Between Degrees and Radians
To convert from degrees to radians multiply by π180.
To convert from radians to degrees multiply by 180π
.
Examples
Convert 360◦,1◦,90◦,135◦ to radians.
Convert π,1, π3,7π6 radians to degrees.
79
Two Important Triangles
The following triangles help us work out the exact values of sin, cos and tan ofcertain special angles.
π
π π
π3 3
6 6
1 1
2 23
2
1
1
π
π4
4
80
Exact Values using Triangles
Using the two important triangles on the previous slide and your knowledge of theunit circle complete the table with exact values:
θ sin θ cos θ tan θ
0π
6
π
4
π
3
π
2
π
3π2
***Important to know these ratios, either the triangles or the table
81
Examples: Find the exact values of the following trig ratios:
1. cos�5π6
�
2. sin�4π3
�Ask yourself:
• quadrant?
• angle from x-axis?
• ± CAST?
• value of ratio?
3. tan�3π4
�
4. sin�11π6
�
5. tan�−π
4
�
6. cos�−2π3
�
82
Finding Angles
Here we need to work backwards considering both the value of the ratio and itssign, as this determines the quadrants of the resultant angles.Find all values of θ between 0 and 2π satisfying:
1. sin(θ) = 12
2. tan(θ) = 1
3. cos(θ) = − 1√2
83
Solving Trigonometric Equations
Solve the following trigonometric equations for x ∈ [0,2π).
1. cosx+ 12 = 0
2. 2 sin2 x+ sinx− 1 = 0
84
3. cosx+ sinx = 0
4. 2 tan2 x− 6 = 0
You can now attempt Sheet 3 Q20-21
85
Sine and Cosine Graphs Reference Evans 6.2
Plotting the values of f(x) = sinx and g(x) = cosx for x ∈ R gives thefollowing graphs
�2 Π �Π Π 2 Π 3 Π 4 Πx
�1
1f �x�
�2 Π �Π Π 2 Π 3 Π 4 Πx
�1
1g�x�
86
Properties of Sine and Cosine Graphs
• The sine and cosine functions repeat themselves after an interval (or period) of2π units. That is:
. . . sin(x− 2π) = sinx = sin(x+2π) = sin(x+4π) . . ., and
. . . cos(x− 4π) = cos(x− 2π) = cosx = cos(x+2π) . . ., for all x ∈ R
These functions are therefore said to be periodic or cyclic, with period 2π.
• The maximum and minimum values of sinx and cosx are 1 and −1 respec-tively. The graphs of f(x) = sinx and g(x) = cosx are therefore said to havean amplitude of 1.
87
Graph of Tangent Function Reference Evans 6.6
The function h(x) = tanx = sinx
cosx is defined for x ∈ R such that cosx �= 0.
Plotting h(x) = tanx over its domain x ∈ R\{(2k + 1)π2 | k ∈ Z} gives thefollowing graph
88
Properties of Tan Graph
• The tangent function h(x) = tanx is undefined at x = {(2k +1)π2 | k ∈ Z}(this is where cosx = 0).
The graph has vertical asymptotes at these x-values, which are usually indicatedby dotted or dashed vertical lines.
• The tangent function repeats itself after an interval of π units. That is:
tan θ = tan(θ + π), for all θ ∈ R\{(2k +1)π
2| k ∈ Z}
This function therefore has period π.
89
Dilations of Trig Graphs Reference Evans 6.2, 6.3 & 6.6
Sketch the following graphs - the standard sin or cos graph is shown - state theperiod and amplitude of the transformed graph in each case.Note: a dilation in the x direction (e.g. y = cos bx) changes the period
�to 2π
b
�
and one in the y direction (e.g. y = a sinx) changes the amplitude (to a)
1. y = sin2x 2. y = 3cosx3. y = −5
2 sinx 4. y = cos�x
2
�
90
Translations of Trigonometric Graphs
Sketch the following graphs on the axes below –adding or subtracting a value tothe trig function will translate the graph up or down, while adding or subtracting avalue to the x term will translate the graph to the left or right respectively.
1. y = tan(x− π
4)
91
2. y = sin(x) + 3
3. y = cos(x+ π)
92
Multiple TransformationsSketch the following sequence of graphs on the same set of axes.y1 = sinx y2 = −2 sinx y3 = −2 sin 3x y4 = −2 sin
�3x− π
2
�
You can now attempt Sheet 3 Q22-23
93
DE
PA
RT
ME
NT
OF
MA
TH
EM
AT
ICS
AN
DS
TA
TIS
TIC
S
MAST10012In
troduction
toM
ath
ematics
Semester1,2011
REVIS
ION
-TRIG
ONOM
ETRY
A:
Findingtrig
ratiosin
theUnit
Circle
1.Id
enti
fyth
equad
rant
that
the
angl
eis
in:
•Q
1has
angl
esfr
om0→
π 2
•Q
2has
angl
esfr
omπ 2→
π
•Q
3has
angl
esfr
omπ→
3π 2
•Q
4has
angl
esfr
om3π 2→
2π
Q1
Q2
Q3
Q4
Thi
sis
just
the
firs
tre
volu
tion
ofth
eu
nit
circ
leW
eca
nof
cou
rse
fin
dbi
gger
angl
esby
mov
ing
arou
nd
the
circ
lem
ore
than
once
orn
egat
ive
angl
esby
goin
gin
the
oppo
site
dire
ctio
n
2.D
ecid
eif
the
rati
oyo
unee
dto
find
(usu
ally
sin,
cos
orta
n)
isp
osit
ive
orneg
ativ
ein
the
quad
rant
you
found
inst
ep1
(mos
tst
ude
nts
use
CA
ST
tore
mem
ber
the
sign
s)
3.U
seth
esp
ecia
ltr
iangl
esto
find
the
rati
ore
quir
ed
2
1
3
3
6
2
1
4
4 1
(rem
embe
rS
OH
-CA
H-T
OA
)
e.g.
sinπ 6
=1 2,
cosπ 6
=√3 2,
tanπ 6
=1 √3
4.F
oran
gles
that
give
poi
nts
onth
ex
ory
axes
we
use
the
bas
icdefi
nit
ions:
•co
sθis
thexco
ord
inate
•si
nθis
theyco
ord
inate
•ta
nθ
=si
nθ
cosθ
Ther
efor
ew
eca
nfind
valu
eslike
:
cosπ
=−
1or
sin( −π 2
) =−
1or
tan
3π 2
=(s
in3π 2)
(cos
3π 2)
=−
1 0=
undefi
ned
Itm
ayhel
pto
look
atth
eunit
circ
leab
ove
tose
eth
ese
poi
nts
(and
rem
emb
erth
at−π 2
and
3π 2
are
just
diff
eren
tnam
esfo
rth
esa
me
poi
nt
onth
eci
rcle
.
1
PRACTIC
EEXERCIS
EA
1.F
ind
the
valu
esof
the
follow
ing
trig
rati
os-
the
step
sm
enti
oned
onth
epre
vio
us
pag
ehav
eb
een
spel
tou
tfo
rth
efirs
tfe
wques
tion
s(t
hen
you
are
onyo
ur
own
todo
the
rest
by
follow
ing
the
sam
epro
cess
):
(a)
sin
5π 4
i.5π 4
isin
Q..
....
(as
5π 4
=π
+π 4)
ii.
InQ
....
..si
nis
....
..(+
or–)
iii.
Fro
mtr
iangl
e2
we
know
that
sinπ 4
=..
....
=⇒
sin
5π 4
=
(b)
cos11π 6
i.11π 6
isin
Q..
....
(as
11π 6
=2π−
π 6)
ii.
InQ
....
..co
sis
....
..(+
or–)
iii.
Fro
mtr
iangl
e1
we
know
that
cosπ 6
=..
....
=⇒
cos11π 6
=
(c)
tan
8π 3
i.8π 3
isin
Q..
....
(as
8π 3
=3π−
π 3
=π−
π 3)
ii.
InQ
....
..ta
nis
....
..(+
or–)
iii.
Fro
mtr
iangl
e1
we
know
that
tanπ 3
=..
....
=⇒
tan
8π 3
=
(d)
cos
5π
i.5π
ison
the
....
..ax
is(x
ory)
ii.
Ithas
coor
din
ates
(...,...
)
iii.
As
cos
isth
ex
coor
din
ate
we
know
:
=⇒
cos
5π=
2
2.N
owtr
yth
esa
me
pro
cess
wit
hth
ese
ques
tion
s:
(a)
sin(5π 3)
(b)
cos(
5π 4)
(c)
tan(7π 6)
(d)
sin(2π
)
(e)
cos(−
7π 2)
(f)
sin(−
7π 4)
(g)
tan(−
17π 6)
Ther
ear
eot
her
trig
rati
osw
eca
nuse
but
they
are
bas
edon
the
stan
dar
don
es.
So
tofind
cose
c,se
cor
cot
we
calc
ula
tesi
n,
cos
orta
nre
spec
tive
lyan
dth
enju
st”t
urn
them
upsi
de
dow
n”
bec
ause
ofth
edefi
nit
ions:
cose
cθ
=1
sinθ
secθ
=1
cosθ
cotθ
=1
tanθ
=co
sθ
sinθ
Exam
ple
s:W
efo
und
onpag
e1
inse
ctio
n3
that
:
sinπ 6
=1 2
=⇒
cose
cπ 6
=2 1
=2
and
cosπ 6
=√3 2
=⇒
secπ 6
=2 √3
(or
2√3
3)
tanπ 6
=1 √3
=⇒
cotπ 6
=√
3
3.F
ind
the
follow
ing
trig
rati
os(y
oum
ayfind
your
answ
ers
toQ
2use
fulin
som
eca
ses)
:
(a)
cose
c(5π 3)
(b)
sec
(−7π 2)
(c)
cot(
7π 6)
(d)
sec
(3π 4)
(e)
cot(
6π)
3
B:
SolvingTrigequations
1.R
earr
ange
the
equat
ion
tom
ake
the
trig
rati
o(u
sual
lysi
n,
cos
orta
n)
the
sub
ject
e.g.
sinx
+1
=0
=⇒
sinx
=−
1√
2co
s(x
+5π 6)−
1=
0=⇒
cos(x
+5π 6)
=1 √2
2.F
ind
the
bas
ican
gle
that
sati
sfies
this
rati
o–
this
may
invo
lve
look
ing
atth
ean
gles
inth
etw
osp
ecia
ltr
ian
gles
orlo
okin
gat
the
coor
din
ates
ofpo
ints
onth
eu
nit
circ
lew
here
they
inte
rsec
tw
ith
the
two
axes
.
3.D
ecid
ew
hic
hquad
rants
the
answ
ers
must
be
in-
look
atth
esi
gnof
the
trig
rati
oe.
g.if
the
sin
rati
ohas
aneg
ativ
ean
swer
then
angl
esm
ust
be
inQ
3an
dQ
4if
the
cos
rati
ohas
ap
osit
ive
answ
erth
enan
gles
must
be
inQ
1an
dQ
4so
we
are
real
lylo
okin
gat
our
CA
ST
diag
ram
”bac
kwar
ds”
4.U
seknow
ledge
ofth
eunit
circ
leto
find
the
bas
ican
gle
inth
eri
ght
quad
rants
So
the
unit
circ
leon
the
righ
tm
ight
be
use
ful
her
e
Not
eth
atθ
isth
ebas
ican
gle
found
inst
ep2
5.C
hec
kth
edom
ain
ofth
eques
tion
haveyou
found
all
theso
lutions?
you
may
nee
dto
add
orsu
btra
ct2π
ifyo
un
eed
bigg
eran
gles
orn
egat
ive
angl
esfo
rth
edo
mai
ngi
ven
inth
equ
esti
on-
that
is,
angl
esou
tsid
eth
est
anda
rd0→
2πw
hich
ison
lyon
ere
volu
tion
arou
nd
the
un
itci
rcle
PRACTIC
EEXERCIS
EB
Sol
veth
efo
llow
ing
trig
equat
ions
over
the
give
ndom
ains
(as
inE
xer
cise
Ayo
uar
egu
ided
thro
ugh
the
firs
tfe
wques
tion
sth
enyo
ush
ould
use
the
sam
epro
cess
toco
mple
teth
ere
st):
1.2
cosx
=√
3x∈
(0,2π
)
cosx
=...
...
bas
ican
gleθ
=π 6
see
spec
ial
tria
ngl
e1
cos
is+
inQ
...(θ)
and
Q..
.(2π−θ)
=⇒
x=
π 6,
2π−
π 6
=⇒
x=
π 6,
... 6
(mentalcheck
:th
ese
valu
esar
ein
the
dom
ain
0→
2π)
Not
e:if
we
added
orsu
btr
acte
d2π
toei
ther
ofou
ran
swer
sw
ew
ould
get
valu
esou
tsid
eth
edom
ain,
sow
edon
’tnee
dto
do
anyth
ing
else
her
e
4
2.ta
n(x−
π 4)−√
3=
0x∈
(0,2π
)
tan(x−
π 4)
=...
bas
ican
gleθ
=...
see
spec
ial
tria
ngl
e1
tan
is..
.(+
/–)
inQ
...(θ)
and
Q3
(......)
=⇒
x−
π 4=......,
......
=⇒
x−
π 4=......,
......
=⇒
x=......
+π 4,
......
+π 4
add
π 4to
RH
Sof
equat
ion
tofindx
=⇒
x=......,
......
use
LC
Dto
add
frac
tion
s
(men
tal
chec
k:
thes
eva
lues
are
insi
deth
edo
mai
nof
0→
2π,
and
ifw
ead
ded
orsu
btra
cted
2πto
eith
erof
them
we
wou
ldge
tan
swer
sou
tsid
eth
edo
mai
n,
sow
edo
n’t
nee
dto
doan
ythi
ng
else
here
)
3.8
sin(x
+π 6)
+4
=0
x∈
(−2π,2π
)
sin(x
+π 6)
=...
bas
ican
gleθ
=...
see
spec
ial
tria
ngl
e1
sin
is–
inQ
3(π
+θ)
and
Q..
.(......
)=⇒
x+
π 6=......,
......
=⇒
x+
π 6=......,
......
=⇒
x=......−
π 6,
......−
π 6su
btr
actπ 6
toR
HS
ofeq
uat
ion
tofindx
=⇒
x=......,
......
use
LC
Dto
add
frac
tion
s
and
x=......−
2π,
......−
2πdom
ain
nee
ds
neg
ativ
ean
gles
too
Not
e:w
en
eed
all
fou
ran
swer
s–
the
two
neg
ativ
ean
dth
eor
igin
altw
opo
siti
veon
es
So
the
final
solu
tion
sar
e:
x=......,
......,
......,
......
4.√
2co
sx
+1
=0
x∈
(−2π,2π
)
5.si
n(x
+π 3)
=1
x∈
(0,4π
]
6.co
s(x−
3π 4)
=0
x∈
(0,2π
)
7.2
cos2x
+3
cosx
+1
=0
x∈
[0,2π
] 5
Trig
on
om
etry
Frid
ay,
22
Ap
ril 2
01
1
3:0
5 P
M
Rev
isio
n t
op
ics P
ag
e 1
Rev
isio
n t
op
ics P
ag
e 2
Rev
isio
n t
op
ics P
ag
e 3
Rev
isio
n t
op
ics P
ag
e 4
Rev
isio
n t
op
ics P
ag
e 5
Rev
isio
n t
op
ics P
ag
e 6
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