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