## Thursday, January 4, 2007

### Precalculus 4.5b - Translations of Sine and Cosine Curves

4.5b Translations of Sine and Cosine Curves

Given the functions:

y = a sin (bxc) + d and y = a cos (bxc) + d creates horizontal and vertical translations of the basic sine and cosine curves.

The following explains what happens:

1. a is the amplitude
2. (2π)/b is the period
3. bc is the horizontal shift
4. d is the vertical shift

One complete cycle interval – the left and right endpoints can be determined by:

Left endpoint is bxc = 0
Right endpoint is bxc = 2π

Example 1: y = 2 sin (3x – π/6)

Amplitude = 2

bx – c = 0
3x – π/6 = 0
3x = π/6
x = π/18

bx – c = 2π
3x – π/6 = 2π
3x = 13π/6
x = 13π/18

So therefore [π/18, 13π/18 ] corresponds to one cycle.

13π/18 - π/18 = (12π/18)/ 4 = 3π/18

π/18 + 3π/18 = 4π/18 + 3π/18 = 7π/18 + 3π/18 = 10π/18 + 3π/18 = 13π/18

Therefore 5 key points are:

(π/18 , 0), (4π/18 , 2), (7π/18 , 0), (10π/18 , -2), (13π/18 , )

Example 2: y = 4 cos (2x + π/2)

Amplitude = 4

bx – c = 0
2x + π/2= 0
2x = - π/2
x =- π/4

bx – c = 2π
2x + π/2= 2π
2x = 3π/2
x = 3π/4

So therefore [-π/4, 3π/4 ] corresponds to one cycle.

3π/4 - -π/4 = (4π/4)/ 4 = π/4

-π/4 + π/4 = 0 + π/4 = π/4 + π/4 = 2π/4 + π/4 = 3π/4

Therefore 5 key points are:

(-π/4 , 4), (0 , 0), (π/4 , -4), (π/2 , 0), (3π/4 , 4 )

Example 3: y = 2 sin (3x - π/4 ) + 1

Amplitude = 2

bx – c = 0
3x - π/4 = 0
3x = π/4
x = π/12

bx – c = 2π
3x - π/4 = 2π
3x = 9π/4
x = 9π/12 = 3π/4

So therefore [π/12, 3π/4 ] corresponds to one cycle.

3π/4 - π/12 = (8π/12)/ 4 = π/6

π/12 + π/6 = π/4 + π/6 = 5π/12 + π/6 = 7π/12 + π/6 = 3π/4

Do the horizontal shift first, then the vertical. Therefore 5 key points are:
(π/12 , 0), (π/4 , 2), (5π/12 , 0), (7π/12 , -2), (3π/4 , 0 )

Now do the vertical shift up one so the points become
(π/12 , 1), (π/4 , 3), (5π/12 , 1), (7π/12 , -1), (3π/4 , 1 )
Mathematical Modeling:

Sine and Cosine curves can be used to model many real-life situations, including electric currents, musical tones, radio waves, tides, and weather patterns.

Example: Chicago’s high temperatures are the following. Find a trigonometric model.

t= { 1 ,2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
C ={29, 33.5, 45.8, 58.6, 70.1, 79.6, 83.7, 81.8, 74.8, 63.3, 48.4, 34.0 }

Using your graphing calculator, place the t-values in List One and the C-values in List Two.

Next using sin-reg you will find the function:

y = a sin (bx + c) + d

y = 28.58 sin (.47x – 1.81) + 55.71 rounding to the second decimal place. By graphing the original equation and plotting the points, how well did the model fit?

Homework #32: pg. 330; #15, 21, 23, 33, 45, 47, 53 – 61 odd, 63, 79, 80
pg. 320 #52, 80, 91, 93