Chapter two
Functions
2-1- Exponential and Logarithm functions :
Exponential functions : If a is a positive number and x is any number ,
we define the exponential function as :
y = ax with domain : -∞ < x < ∞
Range : y > 0
The properties of the exponential functions are :
1. If a > 0 ↔ ax > 0 .
2. ax . ay = ax + y .
3. ax / ay = ax - y .
4. ( ax )y = ax.y .
5. ( a . b )x = ax . bx .
x
6. a y a x ( a ) x .
y y
7. a-x = 1 / ax and ax = 1 / a-x .
8. ax = ay ↔ x = y .
9. a0 = 1 ,
a∞ = ∞ , a-∞ = 0 , where a > 1 .
a∞ = 0 , a-∞ = ∞ , where a < 1 .
The graph of the exponential function y = ax is :
Logarithm function : If a is any positive number other than 1 , then
the logarithm of x to the base a denoted by :
y = logax where x > 0
At a = e = 2.7182828… , we get the natural logarithm and denoted by :
y = ln x
Let x , y > 0 then the properties of logarithm functions are :
1. y = ax ↔ x = logay and y = ex ↔ x = ln y .
2. logex = ln x .
3. logax = ln x / ln a .
١
4. ln (x.y) = ln x + ln y .
5. ln ( x / y ) = ln x – ln y .
6. ln xn = n. ln x .
7. ln e = logaa = 1 and ln 1 = loga1 = 0 .
8. ax = ex. ln a .
9. eln x = x .
The graph of the function y = ln x is :
y2
1.5
1
0.5
0 X
e
-0.5 0 1 2 3 4 5
-1
-1.5
-2
Application of exponential and logarithm functions :
We take Newton's law of cooling :
T – TS = ( T0 – TS ) et k
where T is the temperature of the object at time t .
TS is the surrounding temperature .
T0 is the initial temperature of the object .
k is a constant .
EX-1- The temperature of an ingot of metal is 80 oC and the room
temperature is 20 oC . After twenty minutes, it was 70 oC .
a) What is the temperature will the metal be after 30 minutes?
b) What is the temperature will the metal be after two hours?
c) When will the metal be 30 oC?
Sol. :
ln 5 ln 6
T T S ( T0 T S )e tk 50 60 e 20 k k 0 .0091
20
a) T 20 60 e 30 ( 0.0091 ) 60 * 0.761 45 .6 o C T 65.6 o C
b) T TS 60 e 120 ( 0.0091 ) 60 * 0.335 20.1 o C T 40 .1 o C
c) 10 60 e 0.0091 t 0.0091 t ln 6 t 3.3 hrs .
٢
2-2- Trigonometric functions : When an angle of measure θ is placed in
standard position at the center of a circle of radius r , the trigonometric
functions of θ are defined by the equations :
y 1 x 1 y 1 Sin
Sin , Cos , tan
r csc r sec x Cot Cos
r
y
θ
o x
The following are some properties of these functions :
1) Sin 2 Cos 2 1
2 ) 1 tan 2 sec 2 and 1 Cot 2 csc 2
3 ) Sin( ) Sin .Cos Cos .Sin
4 ) Cos( ) Cos .Cos Sin .Sin
tan tan
5 ) tan( )
1 tan . tan
6 ) Sin 2 2 Sin .Cos and Cos 2 Cos 2 Sin 2
1 Cos 2 1 Cos 2
7 ) Cos 2 and Sin 2
2 2
8 ) Sin( ) Cos and Cos( ) Sin
2 2
9) Sin( ) Sin and Cos( ) Cos and tan( ) tan
1
10 ) Sin .Sin [ Cos( ) Cos( )]
2
1
Cos .Cos [ Cos( ) Cos( )]
2
1
Sin .Cos [ Sin( ) Sin( )]
2
٣
11 ) Sin Sin 2 Sin .Cos
2 2
Sin Sin 2Cos .Sin
2 2
12 ) Cos Cos 2Cos .Cos
2 2
Cos Cos 2 Sin .Sin
2 2
θ 0 Π/6 Π/4 Π/3 Π/2 Π
Sinθ 0 1/2 1/√2 √3/2 1 0
Cosθ 1 √3/2 1/√2 1/2 0 -1
tanθ 0 1/√2 1 √3 ∞ 0
Graphs of the trigonometric functions are :
1.5
0.5
-2Л -Л
0 Л 2Л
-0.5
-1
-1.5
y Sinx D x : x
R y : 1 y 1
٤
1.5
0.5
0
-2Л -Л Л 2Л
-0.5
-1
-1.5
y Cosx D x : x
R y : 1 y 1
-2π -π 0 π 2π
2n 1
y tan x D x : x
2
R y : y
٥
-2π -π 0 π 2π
y Cotx D x : x n
R y : y
-2π -π 0 π 2π
-1
2n 1
y Secx D x : x
2
R y : y 1 or y 1
٦
1
-2π -π 0 π 2π
-1
y csc x D x : x n
R y : y 1 or y 1
Where n 0 ,1 ,2 ,3 ,......
EX-2 - Solve the following equations , for values of θ from 0o to 360o
inclusive .
a) tan θ = 2 Sin θ b) 1 + Cos θ = 2 Sin2 θ
Sol.-
Sin
a ) tan 2 Sin 2 Sin
Cos
Sin ( 1 2Cos ) 0
either Sin 0 0 o ,180 o ,360 o
1
or Cos 60 o ,300 o
2
Therefore the required values of θ are 0o,60o,180o,300o,360o .
b ) 1 Cos 2.Sin 2 1 Cos 2( 1 Cos 2 )
( 2Cos 1 )( Cos 1 ) 0
1
either Cos 60 o ,300 o
2
or Cos 1 180 o
There the roots of the equation between 0o and 360o are 60o,180o
and 300o .
٧
EX-3- If tan θ = 7/24, find without using tables the values of Secθ and Sinθ.
Sol.-
y 7
tan r 7 2 24 2 25
x 24 ٧
r 25 y 7
Sec and Sin
x 24 r 25
٢٤
EX-4- Prove the following identities :
a ) Csc tan .Sec Csc .Sec 2
b ) Cos 4 Sin 4 Cos 2 Sin 2
Sec Csc tan Cot
c)
tan Cot Sec Csc
Sol.-
1 Sin 1
a) L .H .S . Csc tan .Sec .
Sin Cos Cos
Cos 2 Sin 2 1 1
. Csc .Sec 2 R .H .S .
Sin .Cos
2
Sin Cos 2
b ) L .H .S . Cos Sin ( Cos 2 Sin 2 ).( Cos 2 Sin 2 )
4 4
Cos 2 Sin 2 R .H .S .
1 1
Sec Csc Cos Sin 1
c) L .H .S .
tan Cot Sin Cos Sin Cos
Cos Sin
1
Sin Cos Sin .Cos tan Cot
2 2
. R .H .S .
Sin Cos 1 Sec Csc
Sin .Cos
1
EX-5- Simplify when x a .Csc .
x a
2 2
1 1 1 1
Sol.- tan .
x2 a2 a 2 Csc 2 a 2 a Cot 2 a
EX-6- Eliminate θ from the equations :
i) x = a Sinθ and y = b tanθ
ii) x = 2 Secθ and y = Cos2θ
Sol.-
٨
x a
i) x a .Sin Sin Csc
a x
y b
y b tan tan Cot
b y
2 2
a b
Since Csc 2 Cot 2 1 2 2 1
x y
x
x2 4
2
ii ) x 2 Sec Cos
x
2
y Cos 2 y Cos 2 Sin 2
4 x2 4
y 2 2
x2 y 8 x2
x x
EX-7- If tan2θ – 2 tan2β = 1 , show that 2 Cos2θ – Cos2β = 0 .
Sol. –
tan 2 2 tan 2 1 Sec 2 1 2( Sec 2 1 ) 1
1 2
Sec 2 2 Sec 2 0 0
Cos Cos 2
2
2Cos 2 Cos 2 0 Q .E .D .
EX-8- If a Sinθ = p – b Cosθ and b Sinθ = q + a Cosθ .Show that :
a2 +b2 = p2 +q2
Sol.-
p a .Sin b .Cos and q b .Sin a .Cos
p 2 q 2 ( aSin bCos ) 2 ( bSin aCos ) 2
a 2 ( Sin 2 Cos 2 ) b 2 ( Cos 2 Sin 2 ) a 2 b 2
EX-9- If Sin A = 4 / 5 and Cos B = 12 / 13 ,where A is obtuse and B is
acute . Find , without tables , the values of :
a) Sin ( A – B ) , b) tan ( A – B ) , c) tan ( A + B ) .
Sol. -
5 13
5
4 A
B
-3 12
٩
a ) Sin ( A B ) SinA .CosB CosA .SinB
4 12 3 5 63
. .
5 13 5 13 65
tan A tan B
b) tan( A B )
1 tan A .tan B
4 5
63
3 12
4 5 16
1 .
3 12
tan A tan B
c) tan( A B )
1 tan A .tan B
4 5
33
3 12
4 5 56
1 .
3 12
EX-10 – Prove the following identities:
a) Sin ( A B ) Sin ( A B ) 2.SinA .CosB
Sin ( A B )
b) tan A tan B
CosA .CosB
SecA .SecB .CscA .CscB
c) Sec ( A B )
CscA .CscB SecA .SecB
Sin 2 Cos 2 1
d) Cot
Sin 2 Cos 2 1
١٠
Sol.-
a ) L .H .S . Sin ( A B ) Sin ( A B )
SinA .CosB CosA .SinB SinA .CosB CosA .SinB
2 .SinA .CosB R .H .S .
Sin ( A B ) SinA .CosB CosA .SinB
b) R .H .S .
CosA .CosB CosA .CosB
tan A tan B L .H .S .
1 1 1 1
. . .
SecA .SecB .CscA .CscB
c) R .H .S CosA CosB SinA SinB
CscA .CscB SecA .SecB 1 1 1 1
. .
SinA SinB CosA CosB
1 1
CosA .CosB SinA .SinB Cos ( A B )
Sec ( A B ) L .H .S .
Sin 2 Cos 2 1 2 Sin .Cos ( Cos 2 Sin 2 ) 1
d) L .H .S .
Sin 2 Cos 2 1 2 Sin .Cos ( Cos 2 Sin 2 ) 1
2 Sin .Cos 2 Cos 2 Cos
Cot R .H .S .
2 Sin .Cos 2 Sin
2
Sin
EX-11 – Find , without using tables , the values of Sin 2θ and Cos 2θ, when:
a) Sinθ = 3 / 5 , b) Cos θ = 12/13 , c) Sin θ = -√3 / 2 .
Sol. –
a)
٥ 3 5
3
θ θ
٤ -4
3 4 24
Sin 2 2.Sin .Cos 2. .( )
5 5 25
4 3 7
Cos 2 Cos 2 Sin 2 ( ) 2 ( ) 2
5 5 25
١١
b)
١٣
5
θ
θ
١٢
-5
١٣
5 12 120
Sin 2 2.Sin .Cos 2( ).( )
13 13 169
12 5 2 119
Cos 2 Cos 2 Sin 2 ( ) 2 ( )
13 13 169
c)
-١ 1
θ θ
-√3
-√3
2 2
3 1 3
Sin 2 2 Sin .Cos 2( ).( )
2 2 2
1 3 2 1
Cos 2 Cos 2 Sin 2 ( ) 2 ( )
2 2 2
EX-12- Solve the following equations for values of θ from 0o to 360o
inclusive:
a) Cos 2θ + Cos θ + 1 = 0 , b) 4 tan θ . tan 2θ = 1
Sol.-
١٢
a ) Cos 2 Cos 1 0 2Cos 2 1 Cos 1 0
Cos( 2.Cos 1 ) 0
either Cos 0 90 o ,270 o
1
or Cos 120 o ,240 o
2
90 ,120 ,240 ,270 o
o o o
2 tan
b) 4. tan . tan 2 1 4. tan . 1
1 tan 2
9 tan 2 1
1
either tan 18.4 o ,198.4 o
3
1
or tan 161.6 o ,341.6 o
3
18.4 ,161.6 ,198.4 o ,341.6 o
o o
2-3- The inverse trigonometric functions : The inverse trigonometric
functions arise in problems that require finding angles from side
measurements in triangles :
y Sinx x Sin 1 y
-1 1
200
150
100
50
0
-50
-100
-150
-
-200
y Sin 1 x D x : 1 x 1
R y : 90 y 90
١٣
