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G RAPHING S INE AND C OSINE F UNCTIONS

Learn how to graph sine and cosine functions with vertical and horizontal shifts, reflections, and translations. Understand transformations, amplitudes, and periods. Practice graphing with examples and solutions. Check your graphs for accuracy and master circular motion modeling with Ferris wheel height calculations.

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G RAPHING S INE AND C OSINE F UNCTIONS

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  1. GRAPHING SINE AND COSINE FUNCTIONS In previous chapters you learned that the graph ofy=a•f (x–h) +k is related to the graph of y= |a| •f (x) by horizontal and vertical translations and by a reflection when a is negative. This also applies to sine, cosine, and tangent functions.

  2. GRAPHING SINE AND COSINE FUNCTIONS TRANSFORMATIONS OF SINE AND COSINE GRAPHS To obtain the graph of y = a sin b(x – h) + k or y = a cos b(x – h) + k Transform the graph of y = | a | sin bx or y = | a | cos bx as follows.

  3. GRAPHING SINE AND COSINE FUNCTIONS TRANSFORMATIONS OF SINE AND COSINE GRAPHS VERTICAL SHIFT Shift the graph k units vertically. k y=a•sinbx+k y=a•sinbx

  4. GRAPHING SINE AND COSINE FUNCTIONS TRANSFORMATIONS OF SINE AND COSINE GRAPHS HORIZONTAL SHIFT Shift the graph h units Vertically. h y=a•sinb(x–h) y=a•sinbx

  5. GRAPHING SINE AND COSINE FUNCTIONS TRANSFORMATIONS OF SINE AND COSINE GRAPHS y=a•sinbx+k REFLECTION If a < 0, reflect the graph in the line y = k after any vertical and horizontal shifts have been performed. y=–a•sinbx+k

  6. Graphing a Vertical Translation Because the graph is a transformation of the graph of y = 3 sin 4x, the amplitude is 3 and the period is = .  2 2 4   3  8 4 2 8 Graph y = – 2 + 3 sin 4x. SOLUTION By comparing the given equationto the general equation y = a sin b(x – h) + k, you can see that h = 0, k= – 2, and a > 0. Therefore translate the graph of y = 3 sin 4xdown two units.

  7. Graphing a Vertical Translation     3 3   8 8 4 4 2 2 8 8 y=– 2 Graph y = – 2 + 3 sin 4x. The graph oscillates 3 units up and down from its center line y=– 2. SOLUTION Therefore, the maximum value of the function is – 2 + 3= 1 and the minimum value of the function is– 2 – 3= –5

  8. Graphing a Vertical Translation   On y = k: (0, 2); , – 2 ; , – 2 4 2  Maximum: , 1 8   3  3 8 4 2 8 8 Minimum: , – 5 Graph y = – 2 + 3 sin 4x. SOLUTION The five key points are:

  9. Graphing a Vertical Translation CHECK   3  8 4 2 8 Graph y = – 2 + 3 sin 4x. You can check your graph with a graphing calculator. Use theMaximum, Minimum and Intersect features to check the key points.

  10. Graphing a Vertical Translation  2 Graph y = 2 cos x – . 3 4 Because the graph is a transformation of the graph ofy = 2 cosx, the amplitude is 2 and the period is = 3 . 2 3 2 2 3 SOLUTION

  11. Graphing a Vertical Translation 2 π Graph y = 2 cos x – . 3 4 By comparing the given equation to the general equation y = a cos b(x–h) + k, you can see that h = , k = 0, and a > 0.  4 Therefore, translate the graph of y = 2 cos x right unit. 2 3  4 Notice that the maximum occurs unit to the right ofthe y-axis.  4 SOLUTION

  12. Graphing a Horizontal Translation  2 Graph y = 2 cos x – . 3 4     1 On y = k: • 3+ , 0 = (, 0); 4 4 4 4 4 5 3 • 3+ , 0 = , 0 4 2  Maximum: 0 +, 2 = , 2 ; 4 13 3+, 2 = , 2 ; 4  1 7 Minimum: • 3+, – 2= , – 2 2 4 4 SOLUTION The five key points are:

  13. Graphing a Reflection Graph y = – 3 sin x. Because the graph is a reflection of the graph of y = 3 sin x, the amplitude is 3 and the period is 2. SOLUTION When you plot the five points on the graph, note that the intercepts are the same as they are for the graph of y= 3 sin x.

  14. Graphing a Reflection Graph y = – 3 sin x. SOLUTION However, when the graph is reflected in the x-axis, the maximum becomes a minimum and the minimum becomes a maximum.

  15. Graphing a Reflection 1 • 2, 0 = (, 0) 2  1 Minimum:•2, – 3 = , – 3 2 4 3 3 Maximum: •2, 3 = , 3 2 4 Graph y = – 3 sin x. SOLUTION The five key points are: On y = k:(0, 0); (2, 0);

  16. Modeling Circular Motion  h = 25 sin t – 7.5 + 30 15 FERRIS WHEEL You are riding a Ferris wheel. Your height h (in feet) above the ground at any time t (in seconds) can be modeled by the following equation: The Ferris wheel turns for 135 seconds before it stops to let the first passengers off. Graph your height above the ground as a function of time. What are your minimum and maximum heights above the ground?

  17. Modeling Circular Motion 2  The amplitude is 25 and the period is = 30. h = 25 sin t – 7.5 + 30  15 15 130 The wheel turns = 4.5 times in 135 seconds, so the graph shows 4.5 cycles. 30 SOLUTION

  18. Modeling Circular Motion  h = 25 sin t – 7.5 + 30 15 SOLUTION The key five points are (7.5, 30), (15, 55), (22.5, 30), (30, 5) and (37.5, 30).

  19. Modeling Circular Motion  h = 25 sin t – 7.5 + 30 15 SOLUTION Since the amplitude is 25 and the graph is shifted up 30units, the maximum height is 30+ 25 = 55 feet. The minimum height is 30– 25 = 5 feet.

  20. TRANSFORMATIONS OF TANGENT GRAPHS To obtain the graph of y = atan b(x – h) + ktransform the graph of y= a tanbx as follows. | | GRAPHING TANGENT FUNCTIONS • Shift the graph k units vertically and h units horizontally. • Then, if a < 0, reflect the graph in the line y = k.

  21. Combining a Translation and a Reflection  Graph y = – 2 tan x + . 4 SOLUTION The graph is a transformation of the graph of y = 2 tan x, so the period is .

  22. Combining a Translation and a Reflection  Graph y = – 2 tan x + . 4 By comparing the given equation to y = a tan b(x– h) + k, you can see that h = – , k = 0, and a < 0.  4 Therefore translate the graph of y= 2 tan x left unit and then reflect it in the x-axis.  4 SOLUTION

  23. Combining a Translation and a Reflection  Graph y = – 2 tan x + . 4   3    x = – – = – ; x = – = 2•1 4 4 2•1 4 4  (h, k) = – , 0 4      – – , 2 = – , 2 ; – , – 2 = (0, – 2) 4•1 4 2 4•1 4 Asymptotes: On y = k: Halfway points:

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