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Higher Maths
Strategies
The Wave Function
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Maths4Scotland
Higher
The following questions are on
The Wave Function
Non-calculator questions will be indicated
You will need a pencil, paper, ruler and rubber.
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Maths4Scotland
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)
where k > 0 and 0  a  360
b) 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  22  52
tan a 
5
2
k sin a  5
k  29
acute
a  68
Put together:
2sin x  5cos x  29 sin( x  68)
Minimum when:
( x  68)  270  x  202
P has coords.
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a is in 1st quadrant
(sin and cos are +)
a  68
Hint
(202,  29)
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a)
b)
Higher
Write sin x - cos x in the form k sin (x - a) stating the values of k and a where
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 ksin(x - a):
k sin( x  a)  k sin x cos a  k cos x sin a
Equate coefficients:
k cos a  1
k 2  12  12
Square and add
tan a  1
Dividing:
k 2
acute
a


a is in 1st quadrant
4
(sin and cos are +)

2 a
4
sin x  cos x  2 sin( x  )
Put together:
4
Sketch Graph
max
max at
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k sin a  1
 2
x
3
4
min
min at
Table of exact values
2
 2
x
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7
4
k

4
Hint
2
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a
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Higher
Express 8 cos x  6 sin x in the form k cos( x  a) where 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 2  82  62
Dividing:
Put together:
tan a 
6
8
k sin a  6
k  10
acute
a  37
a is in 1st quadrant
(sin and cos are +)
a  37
8cos x  6sin x  10cos( x  37)
Hint
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Higher
Find the maximum value of cos x  sin x and the value of x for which it occurs in
the interval 0  x  2.
Express as Rcos(x - a):
R cos( x  a)  R cos x cos a  R sin x sin a
Equate coefficients:
R cos a  1
Square and add
R2  12  12
Dividing:
Put together:
Max value:
tan a  1
acute
R sin a  1
R 2
a

a is in 4th quadrant
4
(sin is - and cos is +)
 
  0, x 
cos x  sin x  2 cos x 
2
when
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Table of exact values
x 
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7
4
a
7
4
7
4
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7
4
Hint
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Higher
Express 2 sin x  5 cos x in the form k sin( x   ), 0    360 and k  0
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  22  52
5
2
k sin a  5
k  29
a  68
a is in 1st quadrant
Dividing:
tan a 
Put together:
2cos x  5sin x  29 sin  x  68
acute
(sin and cos are both +)
a  68
Hint
Previous
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Maths4Scotland
Higher
The diagram shows an incomplete graph of


y  3sin  x   , for 0  x  2
3

Find the coordinates of the maximum stationary point.
Max for sine occurs
(...) 
Coordinates of max s.p.
2
x
,
5
6
Sine takes values between 1 and -1
Max value of sine function:
Max value of function:

3

5
,3
6

Hint
Previous
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Higher
f ( x)  2 cos x  3sin 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  22  32
Put together:
tan a 
 x  56  82
Previous
and
0    360
0  x  360
3
2
acute
k sin a  3
k  13
a  56
a is in 1st quadrant
(sin and cos are both + )
a  56
2cos x  3sin x  13 cos  x  56
Solve equation.
acute
for
k 0
k cos( x  a)  k cos x cos a  k sin x sin a
Equate coefficients:
Dividing:
where
cos  x  56 
13 cos  x  56  0.5
Cosine +, so 1st & 4th quadrants
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0.5
13
x  138 or x  334
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Hint
Maths4Scotland
Higher
Solve the simultaneous equations
k sin x  5
where k > 0 and 0  x  360
k cos 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
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Higher
Solve the equation 2 sin x  3 cos x  2.5 in the interval 0  x  360.
R cos( x  a)  R cos x cos a  R sin x sin a
Use Rcos(x - a):
R cos a  3
Equate coefficients:
R 2  2 2   3 
Square and add
tan a  
Dividing:
Put together:
 x 146  46
x  192
or
Previous
acute
R  13
a  34
a is in 2nd quadrant
a  146
(sin + and cos - )
2sin x  3cos x  13 cos  x  146 
Solve equation.
acute
2
3
2
R sin a  2
cos  x  146 
13 cos  x 146  2.5
2.5
13
Cosine +, so 1st & 4th quadrants
x  460 (out of range, so subtract 360°)
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x  100
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or
x  192
Hint
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Maths4Scotland
Higher
You have completed all 9 questions in this presentation
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Higher
Table of exact values
sin
cos
tan
Return
30°
45°
60°

6
1
2

4

3
1
2
1
2
3
2
1
2
3
2
1
3
1
3