jdlogo

jdlogo

jdlogo

jdlogo

jdlogo

Home

Radians & Degrees

Angles in Standard Position Revisited

Trig. Functions & Graphs

Transformations

Trig. Identities

Trig. Equations

Summary&Test

LESSON 4:TRANSFORMATIONS  HOMEWORK QUESTIONS

 

jdsmathnotes

 


Quick Review:

 

x

(radians)

 

0

 

 

x

(degrees)

 

0

 

30

 

60

 

90

 

120

 

150

 

180

 

210

 

240

 

270

 

300

 

330

 

360

sin x

(exact)

 

0

 

1

 

0

 

-1

 

0

sin x

(approx.)

 

0

 

0.5

 

0.87

 

1

 

0.87

 

0.5

 

0

 

-0.5

 

-0.87

 

-1

 

-0.87

 

-0.5

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

x

(radians)

 

0

 

 

x

(degrees)

 

0

 

30

 

60

 

90

 

120

 

150

 

180

 

210

 

240

 

270

 

300

 

330

 

360

cos x

(exact)

 

1

 

0

 

-1

 

0

 

1

cos x

(approx.)

 

1

 

0.5

 

0.87

 

1

 

-0.5

 

-0.87

 

-1

 

-0.87

 

-0.5

 

0

 

0.5

 

0.87

 

1

 

 

 

 

 

                       

Text Box: In summary, to graph y = a sin [k(x – d)] + c from the graph of y = sin(x), follow these ideas:

·	If a < 0, we have a reflection in the x-axis
·	If k < 0, we have a reflection in the y-axis
·	If  | a | < 1, we have a vertical compression , factor | a |
·	If  | a | > 1, we have a vertical stretch, factor | a |
·	
·	If  | k | < 1, we have a horizontal stretch, factor 1/k
·	If  | k | > 1, we have a horizontal compression, factor 1/k
·	The value of d gives the horizontal translation (phase shift)
·	The value of c gives the vertical translation (shift)
 

 

 

 

 

 

 

 

 

 

 

 

 

 

Homework questions cont’d:  (Solutions below)

 

1.  State the amplitude, period, phase shift, domain and range for each of the following trigonometric functions.  Draw the graph of at least

one complete period for each.

 

2.  The graph below shows a sine function of the form  y = a sin k(x – d) + c.  find the values of the parameters a,k, d, c.

 

 

 

3.  The graph below shows a sine function of the form  y = a cos k(x – d) + c.  find the values of the parameters a,k, d, c.

 

 

 

 

Solutions:

 

 

 

 

x0

0

90

180

270

360

y

0

1

0

-1

0

 

 

 

        

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

x0

0

90

180

270

360

y

0

1

0

-1

0

 

 

 

 

     

 

 

 

 

 

 

 

 

 

 

 

 

x0

0

90

180

270

360

y

1

0

-1

0

1

 

 

      

 

 

In radians, the points (see table at top of page) would be:

 

 

 

Amplitude:   |a| = 2

 

Domain is given as:

 

Range: The maximum value of y is 3 and the minimum value is –1

           

 

 

 

 

 

 

 

x0

0

90

180

270

360

y

0

1

0

-1

0

 

 

 

 

 

        (x, y) -----------------------à (½ x + 22.50,  3y + 1)

        Using the points from the table above we get points for the transformed graph

        (x, y) -----------------------------à (½ x + 22.50,  3y +1)

        (0, 0) ------------------------à (22.50, 1)

        (900, 1) ----------------------à (67.50, 4)

        (1800, 0) ---------------------à (112.50, 1)

        (2700 , -1) --------------------à (157.50, -2)

        (3600, 0) ----------------------à (202.50, 1)    [red graph at left]

 

       

 

 

 

 

 

 

 

 

 

 

 

In radians, the points (see table at top of page) would be:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Amplitude:   |a| = 3

 

Domain is given as:

 

 

Range: The maximum value of y is 4 and the minimum value is – 2

 

 

2.  The graph below shows a sine function of the form  y = a sin k(x – d) + c.  find the values of the parameters a,k, d, c.

 

 

Solution:

 

 

 

Recall the basic sine curve starts at (0, 0).  If this point is shifted right 450 and down 1, we end up on the given curve.

Hence d = 450 and c = -1, yielding the equation:

                                                           

 

 

 

 

3.  The graph below shows a sine function of the form  y = a cos k(x – d) + c.  find the values of the parameters a,k, d, c.

 

 

Solution:

 

 

 

 

 

Recall the basic cosine curve starts at (0, 1).  If this point is shifted right 300 and up 0.5, we end up on the given curve.

Hence d = 300 and c = 0.5, yielding the equation:

 

 

                                                           

 

  

           

Return to top of page

Click here to return to lesson

Click here to return to homework questions page 1