Sine
and
Cosine
Sine
and
Cosine
Sine
and
Cosine
Recall
that: Px
2IR —*where
IR where px(x,
x is projection
the projection
onto
the
y)x =
is isthe
onto
the
Recall
that
2R2:—+R
y)y)==
PxpX :R
px(x,
Recall
that
→
R
where
p
(x,
x
the
projection
onto
the
X
x-axis,
and py
that: R
: 1R
2R —÷
IR where
py(x, y)y is= the
is the projection
ollto the
py 2—+
2
where
py(x,
onto
the
x-axis,
and
that
y)y)== y isy theprojection
x-axis,
and
that
p
:
R
→
R
where
p
(x,
projection
onto
the
Y
Y
y-axis.
y-axis.
y-axis.
(x)(x
-----1
Px (x,)
—
Examples:
Examples:
Examples:
= 2
• p (2, 8) =2 2
=
)
8
,
2
•px(
X
= 5
)
5
,
3
pY(—
• p •(−3,
5) =5 5
•
(1 /1
√
,
2
PX
1
2 11 — 2
•pX(4)
• pX 2 , 23 =
=
2
=
)
5
,
3
•pY(—
Y
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
Definitioncosine
of cosine
Definition
Definitionof
of cosine
L
t 5,
1
LD
*
*
*
*
functioll is the function cos : R —÷ IR defined as
The cosine
The
Thecosine
cosinefunction
functionisisthe
thefunction
functioncos
cos: R
: R—+→IRRdefined
definedasas
0
cos(9) = Px wind(&)
cos(O)
cos(θ)=pxowind(O)
= pX ◦ wind(θ)
* sws.
àTK?I
IiS.
C
?LO7
8
204
223
204
*
*
*
*
= pX ◦ wind
π
4
= pX
√1 , √1
2
2
= pX ◦ wind
π
3
= pX
√ 3
1
,
2 2
00 0
oo
1Jj)
0
00
1Jj)1Jj)
Cl)
0
00
Cl)Cl)
00 0
π
3
o
C,)C,)C,)
π
4
•• •
• cos
.. .
224
‘1‘1‘1
n0n0 n0
00 0
• cos
cJcJ cJ
= pX
3 1
2 ,2
cI
cI
cI
00 0
= pX ◦ wind
π
6
π
6
• cos
√
C..,C..,C..,
t’)
t’)
t’)
Cl)
CD
CDCD
Cl)Cl)
Examples.
√
3
2
=
=
=
1
2
√1
2
• Px
2
*
--j —
*
*
*
*
*
*
*
*
*
*
*
*
Definition
sine
cosine
Definition
of of
c.I’Q 5,
sinefunction
functionisisthe
thefunction
functioncos
sin: :RR—+→RRdefined
definedasas
cosine
TheThe
sin(θ) = °pYwind(9)
◦ wind(θ)
cos(9)
=
Px
•‘LO1
0
‘I,
0
‘I,
= pY ◦ wind
√
0
II
= pY
3 1
2 ,2
= pY
√1 , √1
2
2
=
II
II
π
4
π
6
II
= pY ◦ wind
-d
Cl)
0
π
6
II
-d
0
II
Cl)
—0
• sin
8 8
—
204
Examples.
1
2
ci ci
-
-v-J-v-J
π
4
ii
-d
0
U)
.
0
C.)
• sin
=
ii
-d
0
U)
0
C.)
.
---
c-) c-)
225
√1
2
I
I
-ICN
-d
0
II
C!)
0
C.)
.
I
I
-ICN
-d
0
II
C!)
0
C.)
.
—
Recall that Px : JR
2 —+ JR where px(x, y) = x is the projection onto the
2 —+ R where py(x, y) = y is the projection onto the
x-axis, and that py : JR
y-axis.
√ 3
1
= pY (x,i)
,
=
2 2
.
I
0
π
3
-ICN
= pY ◦ wind
I
II
C!)
0
C.)
.
π
3
-d
• sin
√
3
2
—
Px
Examples:
) =
8
,
2
•px(
2
) =5
5
,
3
•pY(—
Cosine and sine are the coordinates of wind
v\_1
If θ ∈ R,1)
then
is the x-coordinate of
— cos(θ) is pX ◦ wind(θ). That is, cos(θ)
5n (e
(as(e),
the point wind(θ). Similarly, sin(θ) is the y-coordinate of the point wind(θ).
Taken* together,
* we* have*
*
*
*
*
*
*
*
*
*
wind(θ) = cos(θ), sin(θ)
•Px(’,
Throughout
the point on the unit circle obtained by beginDefinition
of mathematics,
cosine
ning at the point
0) and winding a length of θ is usually written as
is(1,
the function cos : JR
JR defined as
The cosine function
cos(θ), sin(θ) , and that’s the way we’ll usually write it from now on.
cos(6) = Px o wind(9)
C6
*
c
SV
I*M$.
I—
Ecas (e, Sin (e))
L7
c.os(e)
204
It will be important to keep
in mind that a point on the unit circle is a point
of the form cos(θ), sin(θ) , and that any point of the form cos(θ), sin(θ)
is a point on the unit circle.
The next page contains a list of some common values of θ that arise in
trigonometry, along with their values from cos and sin.
226
θ
−π
6
wind(θ)
√
3
1
2 , −2
√
3
2
1, 0
0
√
3 1
2 ,2
π
6
π
4
cos(θ)
1
√
√1 , √1
2
2
√1
2
√1
2
1
2
π
2
0, 1
0
3π
4
5π
6
π
7π
6
√
3
2
√1 , √1
2
2
−
√
3 1
2 ,2
−
− 1, 0
√
−
3
1
2 , −2
√
3
2
− 12
− √12
√
−
3
2
−1
227
0
1
2
√ 3
1
2, 2
− 12 ,
− 21
3
2
π
3
2π
3
sin(θ)
√
−
3
2
1
√
3
2
√1
2
1
2
0
− 21
Graphs of sine and cosine
Graphs of sine and cosine
C05(8’)
s n (e)
Zr
Identities for sine and cosine
An identity is an equation in one variable that is true for every possible
value of the variable. For example, x + x = 2x is an identity because it’s
Identities for sine and cosine
always true. It does’t matter whether x equals 1, or 5, or
it’s always true
An
identity
is
an
equation
in
one
variable
that
is
true for every possible
that x+x=2x.
value ofofthe
example, xof+identities
x = 2x is
because it’s
an assortment
foran
theidentity
functions
The remainder
thisvariable.
chapter isFor
3
always true. It does’t matter whether x equals 1, or 5, or − 5 ; it’s always true
sine and cosine.
that x + x = 2x.
Lemma The
(The Pythagorean
identity)
Foranany
number 6,of important identi(7). remainder
of this chapter
contains
assortment
ties for the functions
sine +and
cosine.
2
cos(O)
2
sin(0)
= 1
—,
Lemma (7).
(The
Pythagorean
Forthe
anyunit
number
Proof: Regardless
the point identity)
wind(9) is on
circle.θ,
of what
0 equals,
2
2
Recall that wind(0) = (cos(0),sin(0)),
so that
(cos(0),sin(0))
is a point on
cos(θ)
+ sin(θ)
=1
the unit circle.
2
Proof:
equation
is
x2 +y
= 1. sin(0))
Since (cos(θ),
sin(θ))
= 1.
Since
is a
The equation
for The
the unit
circleforis the
2 +unit
x
(cos(0),
2 circle
y
is a unit
pointcircle,
on theit unit
it istoathe
solution
of this
That is,
point on the
is a circle,
solution
equation
for equation.
the unit circle,
2+y
x
2 = 1. That is,
cos(θ)2 + sin(θ)2 = 1
2 + sin(0)
cos(0)
2 = 1
228
209
Lemma (8). For any number θ,
π
sin θ +
= cos(θ)
2
Proof:The marked point on the y-axis in the picture on the left is
sin θ + π2 . It’s the y-coordinate of the point obtained by winding around the
circle a distance of π2 and then winding another θ more. We can rotate the
picture on the left by a quarter turn clockwise, which would match sin θ + π2
with the x-coordinate of the point obtained by winding
around the circle a
π
distance of θ, the number cos(θ). Thus, sin θ + 2 = cos(θ).
S
CI
Lemma (9). For any number θ,
π
cos θ −
= sin(θ)
2
Proof: We’ll use Lemma 8 to prove this lemma. Notice that Lemma 8 tells
us
h
h
πi π
π i
sin θ −
+
= cos θ −
2
2
2
Simplifying, we have
π
sin(θ) = cos θ −
2
which is what we had wanted to show.
229
O+7r)
=
—cos(9)
and
sin(O+7r)
=
—sin(O)
umber ir is exactly half the length of the unit circle. There
wind(9 + 7r) is the point on the unit circle that is exactly
the unit circle from the point wind(&).
Lemma
(10).
any number θ,
(10).
Lemma
For For
any number ,
cos(θ +=π)—cos(9)
= − cos(θ)and and sin(O+7r)
sin(θ +=π)—sin(O)
= − sin(θ)
cos(O+7r)
e
Proof:
number
π isI exactly
the length
of unit
the unit
ThereProof:
The The
number
ir is exactly
half half
the length
of the
circle.circle.
There
point
cos(θ 7r)
+ π)
sin(θ
+ π)onisthe
the unit
pointcircle
on the
unit circlethat is
fore,fore,
the the
point
wind(9
is , the
point
that
+
is exactly
exactly
halfway
around
the
unit
circle
from
the
point
cos(θ)
,
sin(θ) .
halfway around the unit circle from the point wind(&).
2
10). For any number 9,
e
hat the vector
—wind(s)
of the
same length of I
= —cos(&)
andis a vector
sin(&-j-7r)
os(O+ir)
= —sin(6)
oints in the opposite direction of wind(U).
e number it is exactly half the length of the unit circle. There
nt wind(9 + it) is the point on the unit circle that is exactly
nd the unit circle from the point wind(6).
Lemma (10). For any number 9,
(cos(e+’r, Sri(9+7r))
1n
(e)
—cos(&)
and is sin(&-j-7r)
cos(O+ir)
—sin(6)
Also notice
that the=vector
—wind(s)
a vector of= the
same length of
wind(O),
that
points
the negative
in the
opposite direction
of
wind(U).
Also
notice
that
of
the
vector
cos(θ)
,
sin(θ)
, which
is the
Proof: The number it is exactly
half the length of the unit circle.
There
−cos(θ)
, − +sin(θ)
is the
vector
in thethat
opposite
direction
fore,vector
the point
it) is , the
wind(9
on that
point
the points
unit circle
is exactly
halfway
around, sin(θ)
the unit
of cos(θ)
. circle from the point wind(6).
pictures above that wind(6 + ir) and —wind(E) are
the two
2
(cos(e+’r, Sri(9+7r))
Therefore,
(e) 1n
C
C
os(&+),
sin(O+7r))
=wind(9+ir)
that the vector
—wind(9)
is a vector of the same length of
points in the opposite =
direction
of wind(9).
—wind(6)
—(
(
cos(&), sin(s))
We can see in= the two pictures above that wind(6 + ir) and —wind(E) are
—cos(&), —sin(9))
the same vector.= Therefore,
cos(’), -sn-i
vectors are equal, their
first coordinates
are equal
sin(O+7r))
=wind(9+ir)
Also notice(cos(&+),
that the vector
—wind(9)
is a vector of the same length of
—cos(&)
cos(9+ir)
wind(9),
in the opposite direction
that =
points
of wind(9).
= —wind(6)
We can see in the two pictures above that the vectors drawn are the same.
coordinatesThat
are equal.
cos(&), sin(s))
is,
= cos(θ + π) , sin(θ + π) =
−
cos(θ)
,
−
sin(θ)
sin(O+7r) = —sin(&)
—cos(&), —sin(9))
=
in the two pictures
above
that
wind(&
—wind(O)
it) andcos(’),
are
+
Because these vectors are equal, their first-sn-i
coordinates
are equal
.
Because these vectors are equal, their first coordinates are equal
or. Therefore,
211
cos(θ + π) = − cos(θ)
= —cos(&)
(cos(9 + it), sin(6 + it))
wind(6 cos(9+ir)
+ it)
and their second coordinates are equal.
and their second coordinates
= —wind(O) are equal.
sin(θ + π) = − sin(θ)
= —sin(&)
sin(O+7r)
cos(&),
sin(9))
=
We can see in the
two pictures above that wind(& + it) and —wind(O) are .
—cos(&), —sin(9)) 230
the same vector. Therefore,
=
211
se vectors are equal. their
first
coordinates
are eqilal
sin(6 + it))
(cos(9
wind(6 + it)
+ it),
(&))
(—
—(
(
(—
(&))
—(
(
cos(&+ir)
=
ond coordinates are equal.
—cos(9)
=
—wind(O)
cos(&), sin(9))
Lemma (11). For any number θ,
Lemma (11). For any number 0,
cos(−θ) = cos(θ)
cos(—9) = cos(9)
Proof: Whether we wind clockwise around the circle a
Proof: Whether
we wind
around the
counterclockwise
a length
of θ,clockwise
the x-coordinates
willcircle
be thea
counterclockwise
of 9, the i-coordinates will be the
to
say that cos(θ) a=length
cos(−θ).
to say that cos(0) cos(—0).
length of θ, orY1i
‘J
length
of 0, o
same. Which
is
same. Which is
e
Lemma (12). For any number θ,
Lemma (12). For any number
0, = − sin(θ)
sin(−θ)
= —sin(O)
Proof: The picture on the sin(—9)
left shows
sin(θ). The picture in the middle is
theProof:
first picture
flipped
over thearound
x-axis. the
Allcircle
of theay-coordinates
exchanged
If we wind
clockwise
length of 9, orare
counterclock
with
negatives,
so y-coordinates
the picture in will
the middle
shows
sin(θ).
The
picture
wise their
of 9, the
most likely
not− be
equal.
However,
a length
on
the right
shows
sin(−θ).
the rightmost
the same
from the
notice
picture
belowNotice
that that
the only
difference picture
betweenissin(9)
and
picture
as
the
one
in
the
middle,
and
thus,
−
sin(θ)
=
sin(−θ).
sin(—9) is that one is the negative of the other. That is, sin(O) = sin(—0).
—
e
e
231
212
Even and odd functions
An even function is a function f (x) that has the property f (−x) = f (x)
for every value of x. Examples of such functions include x2 , x4 , x6 , and by
Lemma 11, cos(x).
An odd function is a function g(x) that has the property g(−x) = −g(x)
for every value of x. Examples of such functions include x3 , x5 , x7 , and by
Lemma 12, sin(x).
Period of sine and cosine
The period of the winding function is 2π, meaning that
wind(θ) = wind(θ + 2π)
Therefore,
cos(θ) , sin(θ) =
cos(θ + 2π) , sin(θ + 2π)
Because these vectors are equal, their first coordinates are equal
cos(θ) = cos(θ + 2π)
and their second coordinates are equal
sin(θ) = sin(θ + 2π)
These last two identities show that sine and cosine are, just as the winding
function, periodic functions. Their period is 2π.
232
Exercises
For #1-14, identify the given value.
1.) cos
5⇡
4
8.) sin
5⇡
4
2.) cos
4⇡
3
9.) sin
4⇡
3
3.) cos
3⇡
2
10.) sin
3⇡
2
4.) cos
5⇡
3
11.) sin
5⇡
3
5.) cos
7⇡
4
12.) sin
7⇡
4
6.) cos
11⇡
6
13.) sin
11⇡
6
7.) cos 2⇡
14.) sin 2⇡
Suppose that ↵ is a real number, that 0  ↵  ⇡2 , and that cos(↵) = 23 .
Use Lemmas 7-12, and that the period of sine and cosine is 2⇡ to find the
following values.
15.) sin(↵)
20.) cos( ↵)
16.) sin(↵ + ⇡2 )
21.) sin( ↵)
⇡
2)
22.) cos(↵ + 2⇡)
18.) cos(↵ + ⇡)
23.) sin(↵ + 2⇡)
17.) cos(↵
19.) sin(↵ + ⇡)
233
Match the numbered piecewise defined functions with their lettered graphs
below.
24.) f (x) =
(
25.) g(x) =
(
26.) h(x) =
(
27.) p(x) =
(
cos(x)
sin(x)
if x 0; and
if x < 0.
sin(x)
cos(x)
if x 0; and
if x < 0.
cos(x)
sin(x)
if x > 0; and
if x  0.
sin(x)
cos(x)
if x > 0; and
if x  0.
A.)
B.)
C.)
D.)
234
For #28-39, identify the given value.
28.) log2 (8)
29.) log5 (125)
32.) log9 (3)
33.) log2
36.) log7 (49)
37.) loge (e7 )
30.) log3 (9)
1
4
34.) log6
1
6
p
38.) loge ( e)
31.) log13 (13)
35.) log10 (10, 000)
39.) loge
1
e
Find the solutions of the equations given in #40-42.
40.)
x 1
x
2x = 6
41.)
x 1
3x+2
235
=4
42.) x +
1
x
=4