The Wave Function
Heart beat
Electrical
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
Expressing a cos x + b sin x in the form k cos(x ± ) or
k sin(x ± )
General shape for y = sin x + cos x
1. Like y = sin x shifted left
2. Like y = cos x shifted right
3. Vertical height (amplitude) different
y = sin x
y = cos x
y = sin x +cos x
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:
With these constants the expressions on the right hand sides = those on the left hand side
FOR ALL VALUES OF x
a cos x + b sin x = k cos(x ± ) or = k sin(x ± )
Where a, b, k and are constants
Given a and b, we can calculate k and .
Write 4 cos xo + 3 sin xo in the form k cos(x – )o, where 0 ≤ ≤ 360
cos(x – ) = cos x cos + sin x sin
k cos(x – )o = k cos xo cos + k sin xo sin 4 cos xo
Now equate with 4 cos xo + 3 sin xo+ 3 sin xo
It follows that : k cos = 4 and k sin = 3
cos2 x + sin2 x = 1
k2 = 42 + 32
k = √25k = 5
tan = ¾
= tan-1 0∙75 = 36∙9
4 cos xo + 3 sin xo = 5 cos(x – 36∙9)o
(kcos )2 + (ksin )2 = k2tan = sin
cos
Write cos x – √3 sin x in the form R cos(x + ), where 0 ≤ ≤ 2π
cos(x + ) = cos x cos – sin x sin
R cos(x + ) = R cos x cos – R sin x sin cos x – √3 sin x
It follows that : R cos = 1 and R sin = √3
cos2 x + sin2 x = 1
R2 = 12 + (√3)2
R = √4R = 2
tan = √3/1
= tan-1 √3
= π/3 (60)
cos x – √3 sin x = 2 cos(x + π/3)
(Rcos )2 + (Rsin )2 = R2 tan = sin cos
Write 5 cos 2x + 12 sin x in the form k sin(2x + ), where 0 ≤ ≤ 360
sin(2x + ) = sin 2x cos + cos 2x sin
k sin(2x + ) = k sin 2xo cos + k cos 2xo sin 12 sin 2xo + 5 cos 2xo
It follows that : k cos = 12 and k sin = 5
cos2 x + sin2 x = 1
k2 = 122 + 52
k = √169k = 13
tan = 5/12
= tan-1 0∙417 = 22∙6
5 cos 2x + 12 sin 2x = 13 sin(2x + 22∙6)
(kcos )2 + (ksin )2 = k2 tan = sin cos
Maximum and Minimum Values
MAX k cos (x ± ) is kk when (x ± ) = 0 or 360 (0 or 2π)
MIN k cos (x ± ) is – k when (x ± ) = 180 (π)
MAX k sin (x ± ) is kk when (x ± ) = 90 (π/2)
MIN k sin (x ± ) is – kk when (x ± ) = 270 (3π/2)
MAX cos x = 1
when x = 0o or 360o
MAX sin x = 1
when x = 90o
MIN cos x = –1
when x = 180o
MIN sin x = –1
when x = 270o
Write f(x) = sin x – cos x in the form k cos (x – ) and find the maximum of f(x) and the value of x at which occurs.
k cos(x – )o = k cos xo cos + k sin xo sin sin xo – cos xo
k cos = –1 k sin = 1
k2 = (–1)2 + 12
k = √2
AS
T C
cos –ve
sin +ve
tan = sin cos
tan = – 1
= (180 – 45)o angle = tan-1 1 = 45o
= 135o
f(x) = √2 cos (x – 135)o
MAX f(x) = √2
When angle = 0
x – 135 = 0
x = 135o
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.
sin( ) a) Express the resultant wave in the form ok t
Find the minimum value of the resultant wave and the value of t at which it occurs.
75sin 100cos sin( ) resultV t t K t
25 3sin 4cos 25 sin( ) resultV t t k t
3sin 4cos sin( ) t t k t For later, remember K = 25k
Maximum and Minimum Values
3sin 4cos sin( ) t t k t
3sin 4cos sin cos cos sin t t k t k t
cos 3
sin 4
k
k
2 2 2 2 2cos sin 3 4 k 5 k1sin 4
tan tan 53.1cos 3
o
th is in the 4 quadrant
360 53.1 306.9 o
Expand and equate
coefficients
C
AS
T0o180o
270o
90o
3
2
2
cos is +ve
sin is –ve
306.9 270 The minumum occurs where ot
125sin( 306.9)resultV t
The minimum value of sin is -1 and it occurs where the angle is 270o
Therefore, the minimum value of Vresult is – 125
270 306.9 576.9 ot
576.9 360 216.9 ot
216.9 ot
Adding or subtracting
360o leaves the sin unchanged
remember K = 25k =25 × 5 = 125
Minimum, we have:
125sin( 216.9) minimum of oy x
sin 1 270 the minimum of is - when o oy x x
125si 125n the minimum of is oy x
216.9occurrs when ox
75sin 100cos 125sin( 216.9 ) oresultV t t t
Solving Trig Equations
3 cos sin 2 0 2 Solve for x x x
3 cos sin cos( ).x x k x
cos 3
sin 1
k
k
2 2 2 2cos sin 3 1 k k2 4 k
2 k
3 cos sin cos cos sin sin x x k x k x
3 cos sin cos( ) Write in the form x x k x
1sin 1tan tan 30
cos 3
o
3 cos sin 2cos6
x x x
C
AS
T0o180
o
270o
90o
3
2
2
cos 3
sin 1
k
k
306
o
o
180 =
o
cos is +ve
sin is +ve
3 cos sin 2 x x
Re-write the trig. equation using your result from step 1, then solve.
2cos 26
x
1cos
6 2
x
C
AS
T0o180
o
270o
90o
3
2
2
1 1
cos6 2
x
6
o 0 45 and 315x7
6
and 4 4
x
cos is +ve
7
4 6 4 6
or x x
5 23
12 12
o o (75 ) or (345 ) x x
5 23
12 12
o o (75 ) or (345 ) x x
2cos2 3sin 2 sin(2 )
2cos2 3sin 2 1 0 360
a) Express in the form
b) Hence solve for ox x k x
x x x
sin 2
cos 3
k
k
2 13 k 13 k
1sin 2tan tan 33.7
cos 3
o
180 33.7 213.7o o o
C
AS
T0o180
o
270o
90o
3
2
2
sinxcoskcosxsink)xsin(k 222 xcosxsin)xsin(k 22232
cos is –ve
sin is –ve
02cos2 3sin 2 13sin(2 213.7)x x x b) We now have
2cos2 3sin 2 1
13sin(2 213.7) 1
x x
x
We solve
by solving
1sin(2 213.7)
13x 1 01
sin 16.113
st In the 1 quadrant
2x – 213.7 = 16.1o , (180-16.1o),(360+16.1o),(360+180-16.1o)
2x – 213.7 = 16.1o , 163.9o, 376.1o, 523.9o, ….
2x = 229∙8o , 310∙2o, 589∙9o, 670∙2o, ….
x = 114∙9o , 188∙8o, 294∙9o, 368∙8o, ….
Part of the graph of y = 2 sin x + 5 cos x is shown
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):
Equate coefficients:
Square and add
Dividing:
Put together:
Minimum when:
P has coords.
asinxcoskacosxsink)axsin(k xcosxsiny 52
2acosk 5asink222 52 k 29k
2
5atan 6852acute 1 tana
sin + , cos +
68a
)xsin(xcosxsin 682952
27068 x 202 x 1s inMin
),( 29202
2
2
Expand k sin(x - a): sin( ) sin cos cos sink x a k x a k x a
Equate coefficients: cos 1 sin 1k a k a
Square and add2 2 21 1 2k k
Dividing:
Put together:4 4
sin cos 2 sin( ) 2x x x k a
Sketch Graph
a) 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
b) 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.
max min2 2
3 7max at min at
4 4x x
tan 1a acute4
a
a is in 1st quadrant
(sin and cos are +)
4a