Download CIRCULAR (TRIGONOMETRIC) FUNCTIONS RECIPROCAL

CIRCULAR (TRIGONOMETRIC) FUNCTIONS
This Area of Study Involves:
•
Definition and graphs of sin, cos and tan and the reciprocal functions cosecant, secant
and cotangent and simple transformations of these.
•
The identities: sec 2 ( x ) = 1 + tan 2 ( x ) and cosec 2 ( x ) = 1 + cot 2 ( x ) .
•
Compound and double angle formulas for sine, cosine and tangent:
sin ( A ± B ) , cos ( A ± B ) , tan ( A ± B ) ,sin ( 2 A ) , cos ( 2 A ) and tan ( 2A )
•
Graphs of the restricted circular functions of sine, cosine and tangent over principal
domains and their respective inverse functions sin −1 , cos −1 , tan −1 and graphs of these
inverse functions, including simple transformations. (students need to be familiar with
alternative notations for sin −1 , cos −1 , tan −1
RECIPROCAL TRIGONOMETRIC FUNCTIONS
f ( x ) and g ( x ) are a pair of reciprocal functions if f ( x ) =
1
1
and g ( x ) =
.
g( x)
f ( x)
The following expressions are examples of reciprocal trigonometric functions:
f ( x ) = sin( x ) and g ( x) =
1
= cosec( x)
sin( x)
f ( x ) = cos( x ) and g ( x ) =
1
= sec( x )
cos( x )
f ( x ) = tan( x ) and g ( x ) =
1
= cot( x )
tan( x )
EXACT VALUES
sin ( 0 ) = 0
cos ( 0 ) = 1
π  1
sin   =
6 2
3
π 
cos   =
6 2
2
π  1
sin   =
=
2
2
4
3
π 
sin   =
3 2
2
π  1
cos   =
=
2
2
4
π  1
cos   =
3 2
π 
sin   = 1
2
π 
cos   = 0
2
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tan ( 0 ) = 0
3
π  1
tan   =
=
3
3
6
π 
tan   = 1
4
π 
tan   = 3
3
π 
tan   is undefined
2
Unit 3 Specialist Mathematics – Reference Notes
Page 31
GRAPHS OF TRIGONOMETRIC FUNCTIONS
The graphs of y = cosec( x) , y = sec( x ) and y = cot( x ) can be derived by using the theory
of reciprocals. Simply plot the reciprocal of each value of y at select values of x .
Note that the X intercepts on the graphs of y = sin( x ) , y = cos( x ) and y = tan( x ) become
vertical asymptotes on the graphs of the corresponding reciprocal functions.
y = sin( x )
y
2
π 
 ,1
2 
1
 5π 
,1
 2 
•
x
0
−0.25π
 −π

, −1

 2

0
0.25π
0.5π
0.75π
π
1.25π
1.5π
1.75π
2π
2.25π
-1
•
 3π

 , −1 
 2

-2
y = cosec( x )
y
2
•  5π ,1
1
0
−0.25π
 −π

, −1 

 2

•
 2
π 
 ,1
2 
0
0.25π
0.5π

x
0.75π
π
1.25π
-1
1.5π
 3π

 , −1 
 2

1.75π
2π
2.25π
-2
x=0
x =π
x = 2π
Domain of y = cosec( x) is R \ {± mπ , m = 0, 1, 2, } .
Range of y = cosec( x) is R \ {−1 < y < 1} .
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y = cos( x )
y
2
( 0,1)
1
•
( 2π ,1)
x
0
−0.25π
0
0.25π
0.5π
0.75π
π
1.25π
1.5π
1.75π
2π
•
2.25π
-1
(π , −1)
-2
y = sec( x )
y
2
1
(2π ,1)
(0,1)
x
0
−0.25π
0
0.25π
0.5π
0.75π
π
1.25π
1.5π
1.75π
2π
2.25π
-1
-2
x=
−π
2
x=
π
2
x=
3π
2
x=
5π
2

 (1 ± 2m)π
, m = 0, 1, 2, 
2


Domain of y = sec( x ) is R \ 
Range of y = sec( x ) is R \ {−1 < y < 1}
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y = tan( x )
y
2
1
x
0
−0.25π
0
0.25π
0.5π
0.75π
π
1.25π
1.5π
1.75π
2π
2.25π
-1
-2
x=
−π
2
x=
π
x=
2
3π
2
x=
5π
2
y = cot( x )
y
2
1
 −π 
,0

 2 
•
x
0
−0.25π
0
0.25π
0.5π
0.75π
π
1.25π
1.5π
1.75π
2π
2.25π
•  5π , 0 

 2


-1
-2
x =π
x=0
x = 2π
Domain of y = cot( x ) is R \ {± mπ , m = 0, 1, 2, }
Range of y = cot( x ) is R .
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Unit 3 Specialist Mathematics – Reference Notes
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THE FUNDAMENTAL IDENTITIES
The principles and definitions in trigonometry are based on the UNIT circle, which is a circle
with centre ( 0, 0) and a radius of 1 unit. The equation of this circle is x 2 + y 2 = 1 .
If θ is an angle measured anti-clockwise from the positive direction of the X axis:
sin (θ )
represents the y coordinate of a point on the circle.
cos (θ ) represents the x coordinate of a point on the circle.
tan (θ )
represents the gradient of the radius line that passes through a point that lies on
the circle ( tan (θ ) =
sin (θ )
).
cos (θ )
Substituting x = cos (θ ) and y = sin (θ ) into x 2 + y 2 = 1 gives: sin 2 (θ ) + cos 2 (θ ) = 1
Dividing by cos 2 (θ ) produces:
tan 2 (θ ) + 1 = sec 2 (θ )
Dividing by sin 2 (θ ) produces:
1 + cot 2 (θ ) = cosec 2 (θ )
The following identities are provided on the VCAA formula sheet.
sin 2 ( x ) + cos 2 ( x ) = 1
tan 2 ( x ) + 1 = sec 2 ( x )
1 + cot 2 ( x ) = cosec 2 ( x )
These identities may be used to solve for the solution to a trigonometric expressions, when
information regarding another trigonometric function is known.
(See Question 25)
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We can also use triangles to help us solve these types of problems.
A knowledge of the following triads will assist in the construction of triangles:
3, 4, 5
5, 12, 13
8, 15, 17
7, 24, 25
(See Question 26)
THE ADDITION THEOREMS
(COMPOUND ANGLE FORMULAE)
sin( x + y ) = sin ( x ) cos ( y ) + cos ( x ) sin ( y )
sin( x − y ) = sin ( x ) cos ( y ) − cos ( x ) sin ( y )
cos( x + y ) = cos ( x ) cos ( y ) − sin ( x ) sin ( y )
cos( x − y ) = cos ( x ) cos ( y ) + sin ( x ) sin ( y )
tan( x + y ) =
tan ( x ) + tan ( y )
1 − tan ( x ) tan ( y )
tan( x − y ) =
tan ( x ) − tan ( y )
1 + tan ( x ) tan ( y )
These formulae
are provided on the
VCAA formula
sheet.
(See Question 27)
THE DOUBLE ANGLE FORMULAE
sin ( 2 x ) = 2sin ( x ) cos ( x )
cos ( 2 x ) = cos2 ( x ) − sin 2 ( x ) = 2 cos2 ( x ) − 1 = 1 − 2sin 2 ( x )
tan ( 2 x ) =
2 tan ( x )
1 − tan 2 ( x )
These formulae
are provided on the
VCAA formula
sheet.
(See Questions 28 – 31)
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INVERSE CIRCULAR FUNCTIONS
The basic trigonometric graphs of y = sin ( x ) , y = cos ( x ) and y = tan ( x ) repeat
themselves indefinitely. For example:
y = sin ( x )
y
1
x
0
2.5π
−2π
−1.5π
−π
−0.5π
0
0.5π
π
1.5π
2π
2.5
-1
In order for inverse trigonometric functions to exist, the domains of the trigonometric
functions must be restricted so that they are one-to-one functions.
GRAPHS OF INVERSE FUNCTIONS
The inverse sine function is obtained by restricting the domain of the sine function to
−
π
≤x≤
π
2
2
−1 ≤ y ≤ 1.
, a standard convention. A one-to-one function is obtained with a range of
y = sin( x )
y = sin −1 ( x )
 π π
The inverse function of the restricted sine function: f : − ,  → [−1, 1] , f ( x ) = sin( x ) is
 2 2
called the inverse sine function or the arcsine function. It is denoted by sin −1 ( x) , arsin( x ) ,
arcsin(x) or asin( x ) . It has a domain of − 1 ≤ x ≤ 1 and range of −
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π
2
≤y≤
π
2
.
Unit 3 Specialist Mathematics – Reference Notes
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The inverse cosine function is obtained by restricting the domain of the cosine function to
0 ≤ x ≤ π . A one-to-one function is obtained with a range of − 1 ≤ y ≤ 1 .
y = cos−1 ( x )
y = cos( x )
The inverse function of the restricted cosine function: f : [0, π ] → [ −1, 1] , f ( x) = cos( x) is
called the inverse cosine function or the arccosine function. It is denoted by: cos −1 ( x) ,
arcos( x ) or acos( x ) . It has a domain of − 1 ≤ x ≤ 1 and range of 0 ≤ y ≤ π .
The inverse tangent function is obtained by restricting the domain of the tangent function
to −
π
2
<x<
π
2
. A one-to-one function is obtained with a range of R .
y = tan( x )
x=
π
2
y = tan −1 ( x )
x=−
x=−
π
x=
2
π
2
π
2
The inverse function of the restricted tangent function:
 π π
f :  − ,  → (−∞, ∞) , f ( x) = tan( x)
 2 2
is called the inverse tangent function or the arctangent function. It is denoted by: tan −1 ( x) ,
artan( x ) , arctan(x) or atan( x ) . It has a domain of R and range of −
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π
2
< y<
π
2
.
Unit 3 Specialist Mathematics – Reference Notes
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SUMMARY
sin −1 ( x)
cos −1 ( x)
tan −1 ( x)
Domain
[−1, 1]
[−1, 1]
R
Range
− π π 
 2 , 2 
[0, π ]
−π π 
, 

 2 2
Note:
We can only find the inverse of a value that lies within the specified domain.
 3
−1  1 
 , cos 
 etc.
 2
 2 
For example: sin −1 
Expressions such as sin-1 (1.2) and cos-1 (2) cannot be evaluated, as the bracket value lies
outside the domain. (Such expressions can be evaluated over the complex number field, but
this is beyond the scope of the VCE course).
The answers obtained must be placed within the given range.
(See Question 32)
Note:
cos−1 ( cos ( x ) ) = x if x ∈ [0, π ]
 π π
sin −1 ( sin ( x ) ) = x if x ∈  − , 
 2 2
If x does not lie in the required range then you must find an equivalent angle that does fall in
the range.

 5π  


3π  

 3π  
3π
−1
−1
For example: cos −1  cos 
  = cos  cos  −
  = cos  cos    =
 4 
 4 
 4  4



Recall that:
cos( −θ ) = cos(θ )
sin( −θ ) = − sin(θ )
(See Question 33)
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MAXIMAL DOMAINS AND RANGES
Maximal Domains:
Place the angle contained in the bracket between the upper and lower limits of the
given domain and solve for x .
Maximal Ranges:
State the standard range for the function. Perform the same operations on the range values
that have been performed on the function.
(See Questions 34 and 35)
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TRANSFORMATIONS OF FUNCTIONS — SUMMARY
Translations
Vertical Translation: f ( x) → g ( x) = f ( x) + D .
Example 1
The graph of f (x) is shifted in the vertical direction by D units.
f(x) = x2
If D is +ve the shift is up. If D is −ve the shift is down.
g(x) = x2 + 2
= f(x) + 2.
The point (α , β ) on f (x) is shifted to the point (α , β + D) .
D=2
Horizontal Translation: f ( x) → g ( x) = f ( x − δ ) .
Example 2
The graph of f (x) is shifted in the horizontal direction by δ units.
f(x) = x2
If δ is positive the shift is to the right. If δ is negative the shift is to
the left.
The point (α , β ) on f (x) is shifted to the point (α + δ , β ) .
Dilations
Vertical Dilation/Dilation from x-axis/Dilation parallel to
y-axis: f ( x) → g ( x) = Af ( x) .
The graph of f (x) is compressed ( A < 1 ) or stretched ( A > 1 ) in
the vertical direction (change of scale along the y-axis by A).
δ=1
Example 3
f(x) = sin x
g(x) = 3sin x
= 3 f(x)
Dilation factor is A.
The point (α , β ) on f (x) is dilated to the point (α , Aβ ) .
A=3
Horizontal Dilation/Dilation from y-axis/Dilation parallel to
x-axis: f ( x) → g ( x) = f ( Bx) .
Example 4
The graph of f (x) is compressed ( B > 1 ) or stretched ( B < 1 ) in
1
the horizontal direction (change of scale along the x-axis by
).
B
1
Dilation factor is
.
B
α 
The point (α , β ) on f (x) is dilated to the point  , β  .
B 
Reflections
g(x) = (x − 1)2
= f(x − 1)
Dilation factor = 3
f(x) = cos x
g(x) = cos 4x = (4x)
B=4
Dilation factor = ¼.
Reflection in x-axis: f ( x) → g ( x) = − f ( x) .
Example 5
The graph of f (x) is reflected around the x-axis.
The point (α , β ) on f (x) is reflected to the point (α , − β ) .
f(x) = e x
Reflection in y-axis: f ( x) → g ( x) = f ( − x) .
Example 6
The graph of f ( x) is reflected around the y-axis.
The point (α , β ) on f ( x) is reflected to the point (−α , β ) .
f(x) = e x
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g(x) = − e x = − f(x)
g(x) = e − x = f(−x)
Unit 3 Specialist Mathematics – Reference Notes
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In general, a sequence of translations, dilations and reflections can be used to transform a
basic function f ( x) into a new function g ( x ) :
f ( x) ⎯Transforma
⎯ ⎯ ⎯tions
⎯→ g ( x) = Af ( Bx − C ) + D = Af ( B[ x − CB ]) + D = Af ( B[ x − δ ]) + D .
g ( x) = Af ( B[ x − δ ]) + D is the STANDARD TRANSFORMATIONS FORM.
The sequence of translations, dilations and reflections have been applied in the following
order:
Horizontal
f ( x) ⎯Dilation
⎯ ⎯⎯→ f ( Bx)
1.
2.
Horizontal dilation factor is
1
.
B
Reflection around y-axis, if required.
Horizontal
f ( Bx) ⎯Translatio
⎯ ⎯⎯n → f ( Bx − C ) = f ( B[ x − CB ]) = f ( B[ x − δ ])
3.
Horizontal translation is δ.
(Note the algebraic manipulation that is required here).
Vertical
4.
f ( B[ x − δ ]) ⎯Dilation
⎯⎯→ Af ( B[ x − δ ])
5.
Reflection around x-axis, if required.
6.
Af ( B[ x − δ ]) ⎯Translatio
⎯ ⎯⎯n → Af ( B[ x − δ ]) + D = g ( x)
Vertical dilation factor is A.
Vertical
Vertical translation is D.
For particular examples, other transformations orders may also be valid.
If the basic function f ( x) is the sine or cosine function so that
sin ( nx − C ) + B = A sin ( n[ x − C ]) + B = A sin ( n[ x − δ ]) + B , then:
g ( x) = A cos
cos
cos
n
A is the amplitude.
2π
is the period (measured in radians).
n
δ (the horizontal translation) is the phase.
Note:
If the order of transformations is not particularly stated then the usual order is DRT.
D: Dilation
R: Reflection
T: Translation
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There are three key stages in applying the transformations idea:
1.
Identify the basic function f (x) from the given function g (x) . The basic function can
be a polynomial, reciprocal, exponential, trigonometric, inverse trigonometric or
logarithmic.
2.
Identify the various translations, dilations and reflections that have been applied
to f (x) .
This stage can often be difficult and requires writing the given function g (x) in standard
transformations form. (In particular, identifying the horizontal translation can be tricky
and will usually require the algebraic manipulation shown above).
3.
Apply the various transformations in the correct order.
ALTERNATIVE METHOD: USE MAPPING
f ( x) ⎯Transforma
⎯ ⎯ ⎯tions
⎯→ g ( x) = Af ( Bx − C ) + D = Af ( B[ x − CB ]) + D = Af ( B[ x − δ ]) + D
1.
Dilation of factor A from the x-axis, ( x, y ) → ( x, Ay ) , if A < 0  Reflection in the x-axis.
2.
Dilation of factor
1
x

from the y-axis, ( x, Ay ) →  , Ay  , if B < 0  Reflection in the
B
B

y-axis.
x
 x

, Ay  →  + δ , Ay  .
B
 B

3.
Horizontal translation of δ units, 
4.
Vertical translation of D units, 
x
 x

+ δ , Ay  →  + δ , Ay + D  .
B
 B

(See Questions 36 and 37)
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COMPLEX NUMBERS
This topic involves:
•
C, the set of numbers z of the form z = x + yi, where x, y are real numbers and
•
Real and imaginary parts; complex conjugates.
•
Equality, addition, subtraction, multiplication and division.
•
Use of an Argand diagram to represent complex numbers.
•
Polar form (modulus and argument); multiplication and division in polar form, including
their geometric representation and interpretation.
•
de Moivre’s theorem and its use to find powers and roots of complex numbers in polar
form, including their geometric representation and interpretation. Proof for integral
powers
•
nth root of unity and other complex numbers and their location in the Complex Plane.
•
Factors over C of polynomials with integer coefficients and of z4 - a4 and z6 - a6 for
small integer values of a; and the fundamental theorem of algebra, including its
application to factorisation of polynomial functions of a single variable over C.
•
Solution over C of corresponding polynomial equations by completing the square, use
of the quadratic formula and using factorised form; occurrence of non-real roots in
conjugate pairs (conjugate root theorem). Informal consideration of the Fundamental
theorem of Algebra.
•
Factorisation of polynomial functions of a single variable over C for example
z 8 + 1, z 2 − i, z 3 − (2 − i ) z 2 + z − 2 + i
THE IMAGINARY NUMBER
i 2 = −1
By convention: i =
−1
(See Question 38)
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OPERATIONS INVOLVING PURELY
IMAGINARY NUMBERS
Any expression in the form bi (where b ≠ 0 ) is called a pure imaginary number.
The rules of arithmetic and algebra for real numbers also apply to pure imaginary numbers
i.e. imaginary numbers may be added, subtracted, multiplied and divided in the same way as
real numbers.
For example:
(a)
3i + 7i = 10i
(b)
7i − 2i = 5i
(c)
2i × −i × 4i = −8i 3 = −8(i 2 )i = −8( −1)i = 8i
(d)
( 2i ) 4 = 16(i ) 4 = 16(i 2 ) 2 = 16( −1) 2 = 16
(e)
( a 2 + ac )i ai ( a + c )
=
=a+c
ai
ai
Note: Every integral power of i can be expressed as one of i , − i, 1, − 1 .
For example:
i 2 = −1
i 3 = i 2i = ( −1)i = −i
i 4 = (i 2 ) 2 = ( −1) 2 = 1
i 5 = (i 4 )i = (1)i = i
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PROPERTIES OF COMPLEX NUMBERS
•
Expressions of the form are referred to as complex numbers.
•
The general form of a complex number is z = x + yi .
•
x represents the real part of z , called Re( z ).
•
y represents the imaginary part of z , called Im( z ).
•
The letter C is used to denote the set of complex numbers: {z : z = x + yi , x, y ∈ R} .
•
If y = 0 then z is a real number. The real numbers are therefore a subset of the
complex numbers: R ⊂ C .
EQUALITY
Two complex numbers are equal if and only if their real parts are equal and their imaginary
parts are equal. For Example:
If z = a + bi and w = c + di then z = w if (if and only if) a = c and b = d .
ADDITION AND SUBTRACTION
To add/subtract complex numbers, we add or subtract like terms. We add the imaginary
components together, and then the real components together.
( a + bi ) ± ( c + di ) = ( a ± c ) + (b ± d )i
Note:
•
The sum or difference of two complex numbers is itself a complex number.
•
If z1 = a + bi and z 2 = c + di then z1 + z 2 = z 2 + z1 .
•
If z1 = a + bi and z 2 = c + di and z 3 = e + fi then (z1 + z 2 ) + z 3 = z1 + (z 2 + z3 ) .
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