Math 103
Lecture 6
Dr. Mustafa El-Agamy
Summary
Scaling Property
𝒚 = 𝟐 𝐬𝐢𝐧 𝒙
𝒚 = 𝐬𝐢𝐧 𝒙
𝒚 = 𝟎. 𝟓 𝐬𝐢𝐧 𝒙
The three functions have the same properties except the range.
𝒚 = 𝐬𝐢𝐧 𝒙
𝒚 = 𝐬𝐢𝐧(𝟐 𝒙)
Period= 𝝅
𝒚 = 𝐬𝐢𝐧 𝒙
𝒚 = 𝐬𝐢𝐧(𝟎. 𝟓 𝒙)
Period= 𝟒 𝝅
The three functions have the same properties except the period and the zeros.
Scaling Property
Function
𝒚 = 𝑨 𝐬𝐢𝐧(𝝎 𝒙)
𝒚 = 𝑨 𝐜𝐨𝐬(𝝎 𝒙)
𝒚 = 𝑨 𝐭𝐚𝐧(𝝎 𝒙)
Domain
ℝ
ℝ
ℝ−
𝑘𝜋
2𝝎
𝑘: odd integer
Range
Period
Zeros
−𝑨, 𝑨
2𝜋
𝝎
𝑛𝜋
𝝎
−𝑨, 𝑨
2𝜋
𝝎
𝑘𝜋
2𝝎
ℝ
𝜋
𝝎
𝑛𝜋
𝝎
𝑨>𝟎
𝑛: integer
𝑘: odd integer
𝑛: integer
Sketch the function 𝒚 = 𝟐 𝐬𝐢𝐧 𝟑𝒙. Then, state the domain, range, period
zeros and symmetry.
Domain: −∞, ∞ ≡ ℝ
Range: −2, 2
Period: 2𝜋/3
Zeros: 𝑛 𝜋/3
𝑛: integer
Symmetry: Odd function
Sketch the function 𝒚 = 𝟑 𝐬𝐢𝐧 𝒙 . Then, state the domain, range, period
zeros and symmetry.
Domain: −∞, ∞ ≡ ℝ
Range: 0, 3
Period: 𝜋
Zeros: 𝑛 𝜋
𝑛: integer
Symmetry: Even function
Find the range of the function 𝒚 = 𝟖 − 𝟑 𝐜𝐨𝐬 𝟒𝒙. Also, find the period
and the zeros.
(Multiply by -3)
⇒
∵ −1 ≤ cos 4𝑥 ≤ 1
(Add 8)
⇒
11 ≥ 8 − 3 cos 4𝑥 ≥ 5
Period:
2𝜋
4
=
𝜋
2
3 ≥ −3 cos 4𝑥 ≥ −3
⇒
11 ≥ 𝑦 ≥ 5
Zeros:
∴ 𝑦 ∈ 5 , 11
No 𝑥 − intercepts
Trigonometric and Hyperbolic Functions
Trigonometric Identities
cos(𝐴 − 𝐵) = cos 𝐴 . cos 𝐵 + sin 𝐴 . sin 𝐵
(1)
cos(𝐴 + 𝐵) = cos 𝐴 . cos 𝐵 − sin 𝐴 . sin 𝐵
(2)
sin(𝐴 − 𝐵) = sin 𝐴 . cos 𝐵 − cos 𝐴 . sin 𝐵
(3)
sin(𝐴 + 𝐵) = sin 𝐴 . cos 𝐵 + cos 𝐴 . sin 𝐵
(4)
If 𝐴 = 𝐵 = 𝑥 , we get
from (1)
cos 2 𝑥 + sin2 𝑥 = 1
÷ cos 2 𝑥
÷ sin2 𝑥
1 + tan2 𝑥 = sec 2 𝑥
cot 2 𝑥 + 1 = csc 2 𝑥
from (2)
cos(2𝑥) = cos 2 𝑥 − sin2 𝑥
cos(2𝑥 ) = 2 cos 2 𝑥 − 1
cos(2𝑥 ) = 1 − 2 sin2 𝑥
1
cos 2 𝑥 = (1 + cos 2𝑥)
2
1
sin2 𝑥 = (1 − cos 2𝑥)
2
from (4)
sin(2𝑥) = 2 sin 𝑥 . cos 𝑥
8
Trigonometric and Hyperbolic Functions
Prove that
2 cos 7𝑥 . cos 3𝑥 = cos 4𝑥 + cos 10𝑥
Solution
By adding equation (1) and (2), we get
cos(𝐴 − 𝐵) + cos(𝐴 + 𝐵) = 2 cos 𝐴 . cos 𝐵
Then put 𝐴 = 7𝑥 and 𝐵 = 3𝑥.
Prove that
tan(𝐴 + 𝐵) =
tan 𝐴 + tan 𝐵
1 − tan 𝐴 . tan 𝐵
Solution
tan(𝐴 + 𝐵) =
sin(𝐴 + 𝐵) sin 𝐴 . cos 𝐵 + cos 𝐴 . sin 𝐵
=
cos(𝐴 + 𝐵) cos 𝐴 . cos 𝐵 − sin 𝐴 . sin 𝐵
÷ cos 𝐴 . cos 𝐵
sin 𝐴 . cos 𝐵
cos 𝐴 . sin 𝐵
(cos 𝐴 . cos 𝐵) + (cos 𝐴 . cos 𝐵)
tan 𝐴 + tan 𝐵
=
=
sin 𝐴 . sin 𝐵
1 − tan 𝐴 . tan 𝐵
1−(
cos 𝐴 . cos 𝐵)
Note that:
Replace 𝐵 by (−𝐵)
For 𝐴 = 𝐵 = 𝑥
tan(2𝑥) =
2 tan 𝑥
1 − tan2 𝑥
tan(𝐴 − 𝐵) =
9
tan 𝐴 − tan 𝐵
1 + tan 𝐴 . tan 𝐵
Trigonometric and Hyperbolic Functions
Geometry
Angle between two lines (𝝋)
𝛼1 = 𝜑 + 𝛼2
𝜑 = 𝛼1 − 𝛼2
𝝋
tan 𝜑 = tan(𝛼1 − 𝛼2 )
𝜶𝟐
tan 𝛼1 − tan 𝛼2
tan 𝜑 =
1 + tan 𝛼1 . tan 𝛼2
∴ tan 𝜑 =
𝜶𝟏
𝑚1 − 𝑚2
1 + 𝑚1 . 𝑚2
where 𝒎 = 𝐭𝐚𝐧 𝜶 is the line slope.
Parallel lines (𝜑 = 0° or 180°):
0=
𝑚1 − 𝑚2
⇒ 𝑚1 − 𝑚2 = 0 ⇒
1 + 𝑚1 . 𝑚2
𝑚1 = 𝑚2
Perpendicular lines (𝜑 = 90°):
1
𝑚1 − 𝑚2
=
⇒ 1 + 𝑚1 . 𝑚2 = 0 ⇒
0 1 + 𝑚1 . 𝑚2
10
𝑚1 . 𝑚2 = −1
𝒙
Trigonometric and Hyperbolic Functions
Radius of Earth
One thousand two hundred years ago, the great scientist Abu Al-Rayhan
Al-Biruni invented an easy calculation method to measure the radius of Earth
by means of Astrolabe.
Astrolabe
is an instrument
used to measure
the angles of
elevation and
depression.
1 He measured the height of a mountain 𝒉 above sea level using trigonometry.
He measured an elevation angle 𝜽𝟏 , then moved toward the mountain
a distance 𝒅, then measured another elevation angle 𝜽𝟐 .
tan 𝜃1 =
ℎ
𝑑+ℓ
tan 𝜃2 =
ℎ
ℓ
𝓵
tan 𝜃1
ℎ
=
𝑑+(
ℎ
tan 𝜃2 )
tan 𝜃1 =
ℎ tan 𝜃2
𝑑 tan 𝜃2 + ℎ
11
ℎ=
𝑑 tan 𝜃1 tan 𝜃2
tan 𝜃2 − tan 𝜃1
Trigonometric and Hyperbolic Functions
2 Al-Biruni ascended on the mountain that knows its height 𝒉 and set the
pointer of Astrolabe so that it indicated the last point in the sea C. So that,
he could measure the angle of depression 𝜶.
The tangent AC
(Line of sight)
is perpendicular to the
radius of the circle
𝜶
𝒉
(Earth)
cos 𝛼 =
𝑅
𝑅+ℎ
𝑹
𝑹
𝑅 cos 𝛼 + ℎ cos 𝛼 = 𝑅
𝜶
𝑶
𝑅=
ℎ cos 𝛼
1 − cos 𝛼
Al-Biruni repeated the measurement several times and averaged the results to
increase the accuracy of the measurement. Al-Biruni was able to measure only
four quantities (three angles and a distance) to calculate the radius of Earth with
an accuracy 99%.

It is truly amazing how mathematics enables a man sitting on top of a mountain
to measure the size of the world without traveling from where he is. Did you
know the importance of mathematics now?!
12
Trigonometric and Hyperbolic Functions
Hyperbolic Functions
Certain even and odd combinations of the exponential functions 𝒆𝒙 and 𝒆−𝒙
arise so frequently in mathematics and its applications. In many ways they are
analogous to the trigonometric functions, and they have the same relationship
to the hyperbola that the trigonometric functions have to the circle. For this
reason, they are collectively called hyperbolic functions and individually
called hyperbolic sine, hyperbolic cosine, and so on.
Hyperbola
The graphs of hyperbolic sine and cosine
𝑒 𝑥 − 𝑒 −𝑥
𝑦 = sinh 𝑥 =
2
𝑒 𝑥 + 𝑒 −𝑥
𝑦 = cosh 𝑥 =
2
can be sketched using graphical addition as in the following figures.
13
Trigonometric and Hyperbolic Functions
Domain: 𝑥 ∈ ℝ
Domain: 𝑥 ∈ ℝ
Range: 𝑦 ∈ ℝ
Range: 𝑦 ∈ [1, ∞[
Odd function
Even function
sinh(−𝑥) = − sinh 𝑥
cosh(−𝑥) = cosh 𝑥
Zero: 𝑥 = { 0 }
sinh 𝑥 𝑒 𝑥 − 𝑒 −𝑥 𝑒 2𝑥 − 1 1 − 𝑒 −2𝑥
𝑦 = tanh 𝑥 =
=
=
=
cosh 𝑥 𝑒 𝑥 + 𝑒 −𝑥 𝑒 2𝑥 + 1 1 + 𝑒 −2𝑥
Domain: 𝑥 ∈ ℝ
Range: 𝑦 ∈ ]−1,1[
Zero: 𝑥 = { 0 }
Odd function
tanh(−𝑥) = − tanh 𝑥
lim tanh 𝑥 = 1
𝑥→∞
lim tanh 𝑥 = −1
𝑥 → −∞
Horiz. asymp.: 𝑦 = ±1
14
Trigonometric and Hyperbolic Functions
1
2
2 𝑒𝑥
𝑦 = csch 𝑥 =
=
=
sinh 𝑥 𝑒 𝑥 − 𝑒 −𝑥 𝑒 2𝑥 − 1
Domain: 𝑥 ∈ ℝ − { 0 }
Range: 𝑦 ∈ ℝ − { 0 }
Odd function
csch(−𝑥) = − csch 𝑥
lim csch 𝑥 = −∞
𝑥 → 0−
lim csch 𝑥 = ∞
𝑥 → 0+
lim csch 𝑥 = 0
𝑥 → −∞
lim csch 𝑥 = 0
𝑥→∞
Asymptotes: Vertical (𝑥 = 0) and Horizontal ( 𝑦 = 0)
1
2
2 𝑒𝑥
𝑦 = sech 𝑥 =
=
=
cosh 𝑥 𝑒 𝑥 + 𝑒 −𝑥 𝑒 2𝑥 + 1
Domain: 𝑥 ∈ ℝ
Range: 𝑦 ∈ ]0,1]
Even function
sech(−𝑥) = sech 𝑥
lim sech 𝑥 = 0
𝑥→∞
lim sech 𝑥 = 0
𝑥 → −∞
Horizontal asymp.: 𝑦 = 0
15
Trigonometric and Hyperbolic Functions
1
cosh 𝑥 𝑒 𝑥 + 𝑒 −𝑥 𝑒 2𝑥 + 1 1 + 𝑒 −2𝑥
𝑦 = coth 𝑥 =
=
=
=
=
tanh 𝑥 sinh 𝑥 𝑒 𝑥 − 𝑒 −𝑥 𝑒 2𝑥 − 1 1 − 𝑒 −2𝑥
Domain: 𝑥 ∈ ℝ − { 0 }
Range: 𝑦 ∈ ℝ − [−1,1]
Odd function
coth(−𝑥) = − coth 𝑥
lim coth 𝑥 = −∞
lim coth 𝑥 = ∞
𝑥 → 0−
𝑥 → 0+
lim coth 𝑥 = −1
lim coth 𝑥 = 1
𝑥 → −∞
𝑥→∞
Asymptotes: Vertical (𝑥 = 0) and Horizontal ( 𝑦 = ±1)
16
Trigonometric and Hyperbolic Functions
Prove that
cosh 𝑥 + sinh 𝑥 = 𝑒 𝑥
Solution
𝑒 𝑥 + 𝑒 −𝑥 𝑒 𝑥 − 𝑒 −𝑥 2 𝑒 𝑥
LHS = cosh 𝑥 + sinh 𝑥 =
+
=
= 𝑒 𝑥 = RHS
2
2
2
Note that:
cosh 𝑥 + sinh 𝑥 = 𝑒 𝑥
cosh 𝑥 − sinh 𝑥 = 𝑒 −𝑥
Prove that
(cosh 𝑥 + sinh 𝑥)5 = cosh(5𝑥) + sinh(5𝑥)
Solution
LHS = (cosh 𝑥 + sinh 𝑥)5 = (𝑒 𝑥 )5 = 𝑒 5𝑥 = cosh(5𝑥) + sinh(5𝑥) = RHS
Note that:
(cosh 𝑥 + sinh 𝑥)𝑛 = cosh(𝑛 𝑥) + sinh(𝑛 𝑥)
(cosh 𝑥 − sinh 𝑥)𝑛 = cosh(𝑛 𝑥) − sinh(𝑛 𝑥)
for any 𝑛 ∈ ℝ
17
Trigonometric and Hyperbolic Functions
Hyperbolic Identities
cosh(𝐴 − 𝐵) = cosh 𝐴 . cosh 𝐵 − sinh 𝐴 . sinh 𝐵
(5)
cosh(𝐴 + 𝐵) = cosh 𝐴 . cosh 𝐵 + sinh 𝐴 . sinh 𝐵
(6)
sinh(𝐴 − 𝐵) = sinh 𝐴 . cosh 𝐵 − cosh 𝐴 . sinh 𝐵
(7)
sinh(𝐴 + 𝐵) = sinh 𝐴 . cosh 𝐵 + cosh 𝐴 . sinh 𝐵
(8)
If 𝐴 = 𝐵 = 𝑥 , we get
from (5)
cosh2 𝑥 − sinh2 𝑥 = 1
÷ cosh2 𝑥
÷ sinh2 𝑥
1 − tanh2 𝑥 = sech2 𝑥
coth2 𝑥 − 1 = csch2 𝑥
from (6)
cosh(2𝑥) = cosh2 𝑥 + sinh2 𝑥
cosh(2𝑥) = 2 cosh2 𝑥 − 1
cosh(2𝑥) = 2 sinh2 𝑥 + 1
1
cosh2 𝑥 = (cosh 2𝑥 + 1)
2
1
sinh2 𝑥 = (cosh 2𝑥 − 1)
2
from (8)
sinh(2𝑥) = 2 sinh 𝑥 . cosh 𝑥
18
Trigonometric and Hyperbolic Functions
Note that:
(cosh 𝑥 + sinh 𝑥)𝑛 = cosh(𝑛 𝑥) + sinh(𝑛 𝑥)
For 𝑛 = 2 :
LHS = (cosh 𝑥 + sinh 𝑥)2 = cosh2 𝑥 + sinh2 𝑥 + 2 sinh 𝑥 . cosh 𝑥
= cosh(2𝑥) + sinh(2𝑥) = RHS
Prove that
sinh(2𝑥) = 2 sinh 𝑥 . cosh 𝑥
Solution
RHS = 2 sinh 𝑥 . cosh 𝑥
𝑒 𝑥 − 𝑒 −𝑥
𝑒 𝑥 + 𝑒 −𝑥
𝑒 2𝑥 − 𝑒 −2𝑥
= 2(
= sinh(2𝑥) = LHS
).(
)=
2
2
2
Prove that
cosh2 𝑥 − sinh2 𝑥 = 1
Solution
𝑒 𝑥 + 𝑒 −𝑥 2
𝑒 𝑥 − 𝑒 −𝑥 2
LHS = cosh 𝑥 − sinh 𝑥 = (
) −(
)
2
2
2
2
𝑒 2𝑥 + 𝑒 −2𝑥 + 2
𝑒 2𝑥 + 𝑒 −2𝑥 − 2
4
=(
)−(
) = = 1 = RHS
4
4
4
19
Trigonometric and Hyperbolic Functions
Another solution technique
LHS = cosh2 𝑥 − sinh2 𝑥
= (cosh 𝑥 + sinh 𝑥). (cosh 𝑥 − sinh 𝑥) = 𝑒 𝑥 . 𝑒 −𝑥 = 1 = RHS
Prove that
cosh(𝐴 + 𝐵) = cosh 𝐴 . cosh 𝐵 + sinh 𝐴 . sinh 𝐵
Solution
RHS = cosh 𝐴 . cosh 𝐵 + sinh 𝐴 . sinh 𝐵
𝑒 𝐴 + 𝑒 −𝐴
𝑒 𝐵 + 𝑒 −𝐵
𝑒 𝐴 − 𝑒 −𝐴
𝑒 𝐵 − 𝑒 −𝐵
=(
).(
)+(
).(
)
2
2
2
2
𝑒 𝐴+𝐵 + 𝑒 𝐴−𝐵 + 𝑒 −𝐴+𝐵 + 𝑒 −𝐴−𝐵
𝑒 𝐴+𝐵 − 𝑒 𝐴−𝐵 − 𝑒 −𝐴+𝐵 + 𝑒 −𝐴−𝐵
=(
)+(
)
4
4
2𝑒 𝐴+𝐵 + 2𝑒 −𝐴−𝐵 𝑒 (𝐴+𝐵) + 𝑒 −(𝐴+𝐵)
=
=
= cosh(𝐴 + 𝐵) = LHS
4
2
Prove that
1 + tanh 𝑥
𝑒𝑥 = √
1 − tanh 𝑥
Solution
sinh 𝑥
1 + tanh 𝑥 1 + (cosh 𝑥 ) cosh 𝑥 + sinh 𝑥
𝑒𝑥
=
=
= −𝑥 = 𝑒 2𝑥
sinh
𝑥
1 − tanh 𝑥 1 −
cosh 𝑥 − sinh 𝑥 𝑒
(
)
cosh 𝑥
1 + tanh 𝑥
RHS = √
= √𝑒 2𝑥 = 𝑒 𝑥 = LHS
1 − tanh 𝑥
20