Transcript Higher Outcome 4 - Mathsrevision.com

Higher Unit 3
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Higher
Outcome 4
What is a Wave Function
Connection with Trig Identities Earlier
Maximum and Minimum Values
Solving Equations involving the Wave Function
Exam Type Questions
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The Wave Function
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Higher
Heart beat
E x p r e ssin g
Outcome 4
a co s x  b sin x in th e f o r m
k co s( x   ) o r k sin ( x   ).
Many wave shapes, whether occurring as
sound, light, water or electrical waves,
can be described mathematically as a combination
of sine and cosine waves.
Spectrum Analysis
Electrical
General shape for y = sinx + cosx
The Wave Function
Y
Higher
1. Like y = sin(x) shifted left
4 right
2. Like y =Outcome
cosx shifted
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3. Vertical height different
y = sin(x)+cos(x)
y = sin(x)
y = cos(x)
The Wave Function
Outcome 4
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Higher
Whenever a function is formed by adding cosine and sine
functions the result can be expressed as a related cosine
or sine function. In general:
a cos x  b sin x  k cos( x   )
a , b , k a n d  a re co n sta n ts
or
a cos x  b sin x  k sin( x   )
G iv e n a a n d b w e c a n c a c u la te k a n d 
With these constants the expressions on
the right hand sides
=
those on the left hand side
FOR ALL VALUES OF x
The Wave Function
Higher
Worked Example:
Outcome 4
W rite 4 cos x  3 sin x in the form k cos( x   ) , w here 0    360
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o
4 cos x  3 sin x
 k cos( x   )
 k  cos x cos   sin x sin 
4 cos x  3 sin x

o
o
R e m e m b e r !!!!!
cos( x   )
 cos x cos   sin x sin 
Re-arrange
 k cos  cos x  k sin  sin x
The left and right hand sides must be equal
for all values of x.
So, the coefficients of cos x and sin x must be equal:
k cos   4
k sin   3
A pair of simultaneous equations to be solved
The Wave Function
Outcome 4
Higher
Find tan ratio
Square and
add
note:
sin(+) and cos(+)
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k cos   4
k sin   3
k cos   k sin   4  3
k
2
2
2
 cos
2
2
sin 
co s 
2
2
  sin    16  9
k  25
2
2
2
k 5
 tan   tan
1
3
o
  = 3 6 .9
4
The Wave Function
k cos   4
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Note:
sin(+) and cos(+)
Outcome 4
Higher
k sin   3
k 5
  3 6 .9
o

2
90o
  is in the first quadrant
4 cos x  3 sin x
4 cos x  3 sin x
 k cos( x   )
 180
o
 5 cos( x  36.9)
o
o
S
A
T
C
270o
3
2
0o
The Wave
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Higher
Outcome 4
Example
W rite cos x 
cos x 
3 sin x in th e fo rm R cos( x   ), w h e re 0    2 
 R cos x cos   R sin x sin 
3 sin x
R cos   1
R sin  
Square and add
2
2
2
R 4
2
R2
 3
 3
1
o
 tan   
  tan 
 = 60
cos 
 1 
 1 
sin 

Find tan ratio
note:

sin(+) and cos(+)
2
90o
3
R cos   R sin   1  3
2
Expand and
Function equate
coefficients
 is in the first quadrant
 180
o
S
A
T
C
270o
3
2
0o
The Wave Function
Outcome 4
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Higher
  tan
180
1
 3
o

 = 60
 1 
 is in the first quadrant
= 
o
60 
o

3
 

3
Finally:
co s x 
3 sin x
 

 2 co s  x  
3 

The Wave
Higher
Expand and
equate
Function
coefficients
Outcome 4
Example
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W rite 5 cos 2 x  12 sin 2 x in the form k sin(2 x   ), w here 0    360
o
5 co s 2 x  1 2 sin 2 x  k sin 2 x co s   k co s 2 x sin 
Square and add
k cos   12
k sin   5
k co s   k sin   1 2  5
2
2
k  13
2
sin 
co s 

2
2
2
2
2
k  13
 tan   tan
1

Find tan ratio
2
noting sign ofo
90
sin(+) and cos(+)
 180
 5 
o

 = 2 2 .6
 12 
 is in the first quadrant
o
S
A
T
C
270o
3
2
0o
The Wave Function
Outcome 4
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Higher
  tan
1
Finally:
 5 
o

2
2
.6


 12 
 is in the first quadrant
5 cos 2 x  12 sin 2 x
 13 sin  2 x  22.6 
o
Maximum and Minimum Values
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Higher
Worked Example:
Outcome 4
a) W rite sin x  cos x in the form k cos( x   )
b) Hence find:
i)
Its maximum value and the value of x at
which this maximum occurs.
ii)
Its minimum value and the value of x at which
this minimum occurs.
Maximum and Minimum
Outcome 4
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Higher
 k cos( x   )  k cos x cos   k sin x sin 
sin x  cos x
k cos    1
Square and add
k sin   1
k
2
 cos

2
k 
sin 
co s 

  sin    1  1
2

2
2
Find tan ratio
o
90
note:
sin(+) and cos(-)
2
 tan  
1
   tan
1
1
 is in the 2
   135
o
Expand and
Values
equate
coefficients
nd
quadrant
1   4 5
o
 180
o
S
A
T
C
270o
3
2
0o
Maximum and Minimum Values
Outcome 4
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Higher
Maximum, we have: sin x  cos x

2 cos( x  135)
o ccu rrs w h e n
x  1 35  0

m a xim u m o f
the m axim um of
y 
x  13 5
2 cos( x  135)
y  cos x
th e m a xim u m o f
o
y 
o
o
is 1 w hen x  0
2 c os x
o
is
2
o
o
Maximum and Minimum Values
Outcome 4
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Higher
Minimum, we have:
sin x  cos x

2 cos( x  135)
o ccu rrs w h e n
x  135  180

m in im u m o f
the m inim um of
y 
y  cos x
th e m in im u m o f
2 cos( x  135)
o
y 
o
is -1 w hen x  180
2 cos x
o
is 
2
o
x  315
o
o
o
Maximum and Minimum Values
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Higher
Example
Outcome 4
A synthesiser adds two sound waves together to make a new sound.
The first wave is described by V = 75sin to and the second by
V = 100cos to, where V is the amplitude in decibels and t is the
time in milliseconds.
Find the minimum value of the resultant wave
and the value of t at which it occurs.
a)
E xpress the resultant w ave in the form
V result  75 sin t  100 cos t
V result  25  3 sin t  4 cos t 
3 sin t  4 cos t
 K sin( t   )
 25 k sin( t   )
 k sin( t   )
k sin( t   )
o
For later,
remember K = 25k
Maximum and Minimum
Outcome 4
Higher
 k sin( t   )
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3 sin t  4 cos t
 k sin t cos   k cos t sin 
3 sin t  4 cos t
k cos   3
Square and add
k sin    4
k
2
 cos
sin 
  sin    3  4
2
2
 tan   tan
1
 is in the 4
th
co s 

2
Expand and
Values
equate
coefficients
Find tan ratio
note:

sin(-) and cos(+)
2
90o
k 5
2
4
o

5
3
.1
 
3
quadrant
 180
o
S
A
T
C
270o
3
  360  53.1  306.9
o
2
0o
Maximum and Minimum Values
Outcome 4
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Higher
V result  125 sin( t  306.9)
remember K = 25k =25 x 5 = 125
The minimum value of sin is -1 and it occurs where the
angle is 270o
Therefore, the minimum value of Vresult is -125
T h e m in u m u m o ccu rs w h e re t  306.9  270
t
270  306.9
 576.9
t
576.9  360
 216.9
t  216.9
o
o
o
o
Adding or
subtracting 360o
leaves the sin
unchanged
Maximum and Minimum Values
Outcome 4
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Higher
Minimum, we have:
V result  75 sin t  100 cos t
m inim um of
the m inim um of
y  125 sin( x  216.9)
y  sin x
the m inim um of
 125 sin( t  216.9 )
o
o
o
is -1 w hen x  270
y  125 si n x
o
is  125
o ccu rrs w h e n x  216.9
o
o
Solving Trig Equations
Higher
S o lv e
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Outcome 4
Worked Example:
Step 1:
True for ALL
x means
coefficients
equal.
2 fo r 0  x  2 
3 cos x  sin x 
3 cos x  sin x in th e fo rm k cos( x   )
W rite
3 cos x  sin x  k cos( x   ).
3 cos x  sin x  k cos x cos   k sin x sin 
Compare Coefficients:
k cos  
3
k sin  
1
Square
&Add
k co s   k sin   3  1
2
2
2
2
k  4
2
k 2
Find tan ratio
Solving Trig Equations
note:
Outcome 4
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Higher
k cos  
3
k sin  
1
sin 
 tan   tan
cos 
180

o
o
1
 30 

 1 
o

30


 3
= 
o
sin(+) and cos(+)

6
 

3 co s x  sin x  2 co s  x  
6 

2
90o
 180
o
S
A
T
C
270o
3
2
0o
Solving Trig Equations
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Higher
Outcome 4
Step 2:
Re-write the trig. equation using your result from step 1,
then solve.
3 cos x  sin x 
 

2 co s  x   
6 

2

2
90o
2
 
1

co s  x   
6 
2

 180
o
 

1  1 
x


cos




6


 2 
 

o
0
x


4
5
a
n
d
3
1
5


6 

S
A
T
C
270o
3
  
7

x


a
n
d


6  4
4

2
0o
Solving Trig Equations
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Higher
Outcome 4
Step 2:
x

6


x
or
4


6
x
5
12
7
x
4
o
(7 5 )

4
or
x


x
or
6
2 3
12
7
4
o
(3 4 5 )


6
Solving Trig
Higher
Expand and
Equations equate
coefficients
Outcome 4
Example
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a ) E xp re ss 2 cos 2 x  3 sin 2 x in th e fo rm k sin(2 x   )
b)
H e n ce so lv e
2 cos 2 x  3 sin 2 x  1
fo r
0  x  360

Find tan ratio
note:2
o
sin(-) and90
cos(-)
 3 sin 2 x  2 cos 2 x  k sin(2 x   )
 k sin 2 x co s   k co s 2 x sin 
k sin    2 

k cos    3 
Square and add
 180
k   113
3
2
sin 
co s 
 tan   tan
1
o
o
o
S
A
T
C
270o
2
o
   3 3 .7
3
  1 8 0  3 3 .7  2 1 3 .7
o
3
2
o
0o
Solving Trig Equations
Outcome 4
Higher
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b ) W e n o w h a v e 2 cos 2 x  3 sin 2 x 
13 sin(2 x  213.7 )
0
2 cos 2 x  3 sin 2 x  1
W e so lv e
b y so lv in g
13 sin(2 x  213.7 )  1
sin(2 x  213.7 ) 
1
st
In th e 1
q u a d ra n t
13
sin
1
 1 
0

16.1


 13 
2x – 213.7 = 16.1o , (180-16.1o),(360+16.1o),(360+180-16.1o)
2x – 213.7 = 16.1o , 163.9o,
2x = 229.8o , 310.2o,
x = 114.9o ,
188.8o,
376.1o,
589.9o,
294.9o,
523.9o, ….
670.2o, ….
368.8o, ….
Solving Trig Equations
(From a past paper)
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Higher
Outcome 4
Example
A builder has obtained a large supply of 4 metre rafters. He wishes
to use them to build some holiday chalets. The planning department
insists that the gable end of each chalet should be in the form of an
isosceles triangle surmounting two squares, as shown in the diagram.
a) If θo is the angle shown in the diagram and A
is the area m2 of the gable end, show that
A  8  2  sin   2 cos 
o
b) Express 8  2  sin   2 cos 
o
o

o

in the form k sin     
c) Find algebraically the value of θo for which
the area of the gable end is 30m2.
4
o

o
4
4
Solving Trig Equations
(From a past paper)
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Higher
4
s
Outcome 4
Part (a)

s
Let the side of the square frames be s.
Use the cosine rule in the isosceles triangle:
a  b  c  2 bc cos A
2
2
2
 2 s   4  4  2  4  4 cos 
2
2
2
4 s  32 1  cos 
2

s  8 1  cos 
2
This is the area of one of the squares.
The formula for the area of a triangle is
A rea of T riangle 
1
2
 4  4  sin 

1
2
a b sin C
 8 sin 
Total area = Triangle + 2 x square:
A  8 sin   2  8 1  cos 

 8  2  sin   2 cos 


Solving Trig Equations
(From a past paper)
Higher
Outcome 4
Part (b)
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E xpress 8 sin   16 cos  in the form k sin    
C onsider 8  sin   2 cos    8 t sin    
S olve sin   2 cos   t sin    

and


Find tan ratio
rem em ber k  8 t .
note:

sin(+) and cos(+)
2
sin   2 co s   t sin  co s   t co s  sin 
t cos   1 

t sin   2 
sin 
co s 
t 5 t 5
2
 tan   tan
Finally:
90o
Square and add
1
2
 
1
  63.4
 180
o
8 sin   16 cos   8 5 sin    63.4 
o
S
A
T
C
270o
o
3
2
0o
Solving Trig Equations
(From a past paper)
Higher
Outcome 4
Part (c)
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Find algebraically the value of θo for which the area is the 30m2
A  8  2  sin 
From diagram
o < 90o
 2 cos    16  8 sin   16 θ
cos

ignore 2nd quad
o
A  16  8 5 sin    63.4 

30  16  8 5 sin    63.4 
14  8 5 sin    63.4 
sin    63.4  
  63.4  sin
   114.9
o
1
o
2
90o
o
7
 180
4 5
7 5


20


 51.5 a n d 128.5
o
o
o
S
A
T
C
270o
3
2
0o
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Higher Maths
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The Wave Function
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The Wave Function
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Higher
Part of the graph of y = 2 sin x + 5 cos x is shown
in the diagram.
a) Express y = 2 sin x + 5 cos x in the form k sin (x + a)
b)
where k > 0 and 0  a  360
Find the coordinates of the minimum turning point P.
Expand ksin(x + a):
k sin( x  a )  k sin x cos a  k cos x sin a
Equate coefficients:
k cos a  2
Square and add
Dividing:
k  2 5
2
2
tan a 
5
2
k sin a  5
k 
2
acute
Put together:
2 sin x  5 cos x 
Minimum when:
( x  68  )  270  
P has coords.
Previous
(202  , 
29
a  68 
a is in 1st quadrant
(sin and cos are +)
a  68
29 sin( x  68  )
x  202 
Hint
29 )
Quit
Quit
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Higher
a)
Write sin x - cos x in the form k sin (x - a) stating the values of k and a where
b)
k > 0 and 0  a  2
Sketch the graph of sin x - cos x for 0  a  2 showing clearly the graph’s
maximum and minimum values and where it cuts the x-axis and the y-axis.
Expand k sin(x - a):
k sin( x  a )  k sin x cos a  k cos x sin a
Equate coefficients:
k cos a  1
k 1 1
2
Square and add
2
sin x  cos x 
Put together:
Sketch Graph
m ax
m ax at

x
2
3
4
m in
m in at
Table of exact values
k 
2
tan a  1
Dividing:
Previous
k sin a  1
acute
2 sin( x 
2
a

4
Quit
a is in 1st quadrant
4
(sin and cos are +)
k 
)
2
a
a 

4

4
2
 2
x

7
4
Hint
2
Quit
Next
Maths4Scotland
Higher
Express 8 co s x  6 sin x in the form k cos( x  a ) w here k  0 and 0  a  360
Expand kcos(x + a):
k cos( x  a )  k cos x cos a  k sin x sin a
Equate coefficients:
k cos a  8
Square and add
k 8 6
Dividing:
tan a 
Put together:
2
2
k sin a  6
2
6
8
k  10
acute
a  37 
a is in 1st quadrant
(sin and cos are +)
a  37
8 cos x  6 sin x  10 cos( x  37  )
Hint
Previous
Quit
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Maths4Scotland
Higher
Find the maximum value of co s x  sin x and the value of x for which it occurs in
the interval 0  x  2.
R cos( x  a )  R cos x cos a  R sin x sin a
Express as Rcos(x - a):
Equate coefficients:
R cos a  1
Square and add
R  1 1
Dividing:
Put together:
Max value:
2
2
tan a   1
2
when
Previous
Table of exact values
R 
2
acute
cos x  sin x 
R sin a   1
a
2

a is in 4th quadrant
4
(sin is - and cos is +)


  0, x 
2 cos x 

x
Quit
7
4
a
7
4
7
4
Quit
7
Hint
4
Next
Maths4Scotland
Express
Higher
in the form k sin( x   )  , 0    360 and k  0
2 sin x   5 co s x 
Expand ksin(x - a):
k sin( x  a )  k sin x cos a  k cos x sin a
Equate coefficients:
k cos a  2
Square and add
k  2 5
2
5
2
Dividing:
tan a 
Put together:
2 cos x  5 sin x 
2
acute
k sin a  5
2
k 
29
a  68 
a is in 1st quadrant
a  68
(sin and cos are both +)
29 sin  x  68  
Hint
Previous
Quit
Quit
Next
Maths4Scotland
Higher
The diagram shows an incomplete graph of
 

y  3 sin  x   , for 0  x  2 
3

Find the coordinates of the maximum stationary point.
Max for sine occurs
(...) 

2
Coordinates of max s.p.
5
6
Sine takes values between 1 and -1
Max value of sine function:
Max value of function:
x
,
3

5
6
,
3

Hint
Previous
Quit
Quit
Next
Maths4Scotland
Higher
f ( x )  2 cos x   3 sin x 
a)
Express f (x) in the form k cos( x   ) 
b)
Hence solve algebraically f ( x )  0.5
Expand kcos(x - a):
k cos a  2
Square and add
k  2 3
Put together:
2
tan a 
a cu te
 x  56   82
Previous
3
2
2
acute
2 cos x  3 sin x 
Solve equation.
for
k 0
and
0    360
0  x  360
k cos( x  a )  k cos x cos a  k sin x sin a
Equate coefficients:
Dividing:
w here
k sin a  3
2
k 
13
a  56 
a  56
(sin and cos are both + )
13 cos  x  56  
co s  x  5 6   
13 cos  x  56    0.5
Cosine +, so 1st & 4th quadrants
Quit
a is in 1st quadrant
Quit
0 .5
13
x  138  or x  334 
Next
Hint
Maths4Scotland
Higher
Solve the simultaneous equations
k sin x   5
where k > 0 and 0  x  360
k co s x   2
Use tan A = sin A / cos A
Divide
tan x 
Find acute angle
5
2
acute
Determine quadrant(s)
x  68 
Sine and cosine are both + in original equations
Solution must be in 1st quadrant
State solution
x  68
Hint
Previous
Quit
Quit
Next
Maths4Scotland
Higher
Solve the equation 2 sin x   3 co s x   2 .5
R cos( x  a )  R cos x cos a  R sin x sin a
Use R cos(x - a):
R cos a   3
Equate coefficients:
R  2   3
2
Square and add
tan a  
Dividing:
Put together:
 x  146  
x  192 
or
Previous
2
3
2
acute
R sin a  2
2
R 
a  34 
13
a is in 2nd quadrant
13 cos  x  146  
co s  x  1 4 6   
13 cos  x  146    2.5
46
x  460 
a  146
(sin + and cos - )
2 sin x  3 cos x 
Solve equation.
a cu te
in the interval 0  x  360.
2 .5
13
Cosine +, so 1st & 4th quadrants
(out of range, so subtract 360°)
Quit
x  100 
Quit
or
x  192 
Hint
Next
Maths4Scotland
Higher
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Are you on Target !
Outcome 4
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Higher
•
Update you log book
•
Make sure you complete and correct
ALL of the Wave Function
questions in the past paper booklet.