The Wave Function Heart beat Electrical Many wave shapes, whether occurring as sound, light, water...

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