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MATH 201 - Week 9 MATH 201 - Week 9 Ferenc Balogh Concordia University 2008 Winter Based on the textbook J. Stuart, L. Redlin, S. Watson, Precalculus - Mathematics for Calculus, 5th Edition, Thomson All figures and videos are made using MAPLE 11 and ImageMagick-convert. MATH 201 - Week 9 Overview The Unit Circle (Section 5.1) The Unit Circle Terminal Points on the Unit Circle The Reference Number MATH 201 - Week 9 Overview The Unit Circle (Section 5.1) The Unit Circle Terminal Points on the Unit Circle The Reference Number Trigonometric Functions of Real Numbers (Section 5.2) The Trigonometric Functions Values of Trigonometric Functions Fundamental Identities MATH 201 - Week 9 Overview The Unit Circle (Section 5.1) The Unit Circle Terminal Points on the Unit Circle The Reference Number Trigonometric Functions of Real Numbers (Section 5.2) The Trigonometric Functions Values of Trigonometric Functions Fundamental Identities Trigonometric Graphs (Section 5.3) Graphs of the Sine and Cosine Functions Graphs of Transformations of Sine and Cosine MATH 201 - Week 9 The Unit Circle The unit circle is the circle of radius 1 centered at the origin. The equation of the unit circle is x 2 + y 2 = 1. MATH 201 - Week 9 The Unit Circle The unit circle is the circle of radius 1 centered at the origin. The equation of the unit circle is x 2 + y 2 = 1. Example. The points √ P(1, 0), Q 3 1 , 2 2 ! are on the unit circle because 12 + 02 √ !2 2 3 1 + 2 2 2 2 1 1 √ + −√ 2 2 = 1 = 1 = 1 , R 1 1 √ , −√ 2 2 MATH 201 - Week 9 The Unit Circle Example. Find the x-coordinate of the point P on the unit circle located in the second quadrant whose y -coordinate is equal to 12 . MATH 201 - Week 9 The Unit Circle Example. Find the x-coordinate of the point P on the unit circle located in the second quadrant whose y -coordinate is equal to 12 . Solution. The point P(x0 , y0 ) is on the unit circle therefore its coordinates satisfy the equation x 2 + y 2 = 1. Since y0 = 12 , x = x0 solves the equation 2 1 x + = 1. 2 2 2 1 1 3 x =1− =1− = . 2 4 4 2 Since P is in the second quadrant, we have to take the negative solution, so r √ 3 3 x0 = − =− . 4 2 MATH 201 - Week 9 The Unit Circle Terminal Points on the Unit Circle Let t be an arbitrary real number. The terminal point corresponding to the number t on the unit circle is given by the following procedure: MATH 201 - Week 9 The Unit Circle Terminal Points on the Unit Circle Let t be an arbitrary real number. The terminal point corresponding to the number t on the unit circle is given by the following procedure: If t = 0 then the corresponding terminal point is (1, 0). MATH 201 - Week 9 The Unit Circle Terminal Points on the Unit Circle Let t be an arbitrary real number. The terminal point corresponding to the number t on the unit circle is given by the following procedure: If t = 0 then the corresponding terminal point is (1, 0). If t > 0 then we take the endpoint of the (possibly self-overlapping) arc starting at (1, 0) of length t along the circumference of the unit circle in positive (counterclockwise) direction. MATH 201 - Week 9 The Unit Circle Terminal Points on the Unit Circle Let t be an arbitrary real number. The terminal point corresponding to the number t on the unit circle is given by the following procedure: If t = 0 then the corresponding terminal point is (1, 0). If t > 0 then we take the endpoint of the (possibly self-overlapping) arc starting at (1, 0) of length t along the circumference of the unit circle in positive (counterclockwise) direction. If t < 0 then we take the endpoint of the (possibly self-overlapping) arc starting at (1, 0) of length |t| along the circumference of the unit circle in negative (clockwise) direction. MATH 201 - Week 9 The Unit Circle Terminal Points on the Unit Circle Let t be an arbitrary real number. The terminal point corresponding to the number t on the unit circle is given by the following procedure: If t = 0 then the corresponding terminal point is (1, 0). If t > 0 then we take the endpoint of the (possibly self-overlapping) arc starting at (1, 0) of length t along the circumference of the unit circle in positive (counterclockwise) direction. If t < 0 then we take the endpoint of the (possibly self-overlapping) arc starting at (1, 0) of length |t| along the circumference of the unit circle in negative (clockwise) direction. Remark. The notion of terminal point is similar to the notion of terminal direction for angles. MATH 201 - Week 9 The Unit Circle Terminal Points on the Unit Circle Example. Find the terminal points for t = π, t= π , 2 t = −3π. MATH 201 - Week 9 The Unit Circle Terminal Points on the Unit Circle Example. Find the terminal points for t = π, t= π , 2 t = −3π. Solution. First of all, the unit circle has a perimeter of 2π. MATH 201 - Week 9 The Unit Circle Terminal Points on the Unit Circle Example. Find the terminal points for t = π, t= π , 2 t = −3π. Solution. First of all, the unit circle has a perimeter of 2π. The terminal point for t = π is (−1, 0) since π is exactly the half of the perimeter. MATH 201 - Week 9 The Unit Circle Terminal Points on the Unit Circle Example. Find the terminal points for t = π, t= π , 2 t = −3π. Solution. First of all, the unit circle has a perimeter of 2π. The terminal point for t = π is (−1, 0) since π is exactly the half of the perimeter. The terminal point for t = the total length. π 2 is (0, 1) since π 2 is one-fourth of MATH 201 - Week 9 The Unit Circle Terminal Points on the Unit Circle Example. Find the terminal points for t = π, t= π , 2 t = −3π. Solution. First of all, the unit circle has a perimeter of 2π. The terminal point for t = π is (−1, 0) since π is exactly the half of the perimeter. The terminal point for t = the total length. π 2 is (0, 1) since π 2 is one-fourth of The terminal point for t = −3π is (−1, 0) since we have to measure a distance of one and a half of the total perimeter in the negative direction on the circumference. MATH 201 - Week 9 The Unit Circle Terminal Points on the Unit Circle Here is a list of some special terminal points: t terminal point 0 0) (1, √ 3 1 π , 6 2 2 π 1 √ , √1 4 2 √ 2 3 π 1 3 2, 2 π 2 π 2π (0, 1) (−1, 0) (1, 0) MATH 201 - Week 9 The Unit Circle Terminal Points on the Unit Circle Here is a list of some special terminal points: t terminal point 0 0) (1, √ 3 1 π , 6 2 2 π 1 √ , √1 4 2 √ 2 3 π 1 3 2, 2 π 2 π 2π (0, 1) (−1, 0) (1, 0) Remark. Notice that different values of t can give the same terminal point on the unit circle (one could call these coterminal numbers). MATH 201 - Week 9 The Unit Circle The Reference Number The reference number t̄ associated with a real number t is the shortest distance along the circumference of the unit circle between the terminal point determined by t and the x-axis. This is the analogue of the reference angle. MATH 201 - Week 9 The Unit Circle The Reference Number The reference number t̄ associated with a real number t is the shortest distance along the circumference of the unit circle between the terminal point determined by t and the x-axis. This is the analogue of the reference angle. Examples. The reference number of π is 0 since the terminal point (−1, 0) is on the x-axis. The reference number of π2 is π2 since the distance of its terminal point (0, 1) from the x-axis is π2 along the unit circle. π The reference number of 2π 3 is 3 since the distance of its terminal point from the x-axis is 2π 3 along the unit circle. MATH 201 - Week 9 The Unit Circle The Reference Number Example. Find the reference numbers for 3 t = − π, 2 t= 7π , 8 t=− 5π . 6 MATH 201 - Week 9 The Unit Circle The Reference Number Example. Find the reference numbers for 3 t = − π, 2 t= 7π , 8 t=− 5π . 6 Solution. For t = − 32 π the terminal point is (0, 1). The corresponding reference number is t̄ = π2 . MATH 201 - Week 9 The Unit Circle The Reference Number Example. Find the reference numbers for 3 t = − π, 2 t= 7π , 8 t=− 5π . 6 Solution. For t = − 32 π the terminal point is (0, 1). The corresponding reference number is t̄ = π2 . If t = 7π 8 the terminal point is in the second quadrant and it is closer to the negative half of the x-axis. The reference π number is π − 7π 8 = 8. MATH 201 - Week 9 The Unit Circle The Reference Number Example. Find the reference numbers for 3 t = − π, 2 t= 7π , 8 t=− 5π . 6 Solution. For t = − 32 π the terminal point is (0, 1). The corresponding reference number is t̄ = π2 . If t = 7π 8 the terminal point is in the second quadrant and it is closer to the negative half of the x-axis. The reference π number is π − 7π 8 = 8. √ 3 1 For t = − 5π , the terminal point is given by − , − 6 2 2 . π Therefore the reference number is 6 . MATH 201 - Week 9 The Unit Circle The Reference Number How to find terminal points using reference numbers? Suppose that t is a given real number. 1 Find the reference number corresponding to t̄. 2 Find the terminal point determined by the reference number t̄ of the form (a, b). 3 The terminal point determined by t is of the form (±a, ±b) where the signs are fixed by the quadrant of the terminal point of t. MATH 201 - Week 9 The Unit Circle The Reference Number Example. Find the terminal points using reference numbers for 3 7π π 2π , t = π, t = , t=− . t= 3 4 6 4 MATH 201 - Week 9 The Unit Circle The Reference Number Example. Find the terminal points using reference numbers for 3 7π π 2π , t = π, t = , t=− . t= 3 4 6 4 Solution. π For t = 2π 3 , the reference number is t̄ = 3 . √ The terminal point of t̄ is 12 , 23 . The terminal point of t lies in the second quadrant therefore the signs of the x-coordinate and the y -coordinate are negative and positive respectively. √ 3 1 is − , So the terminal point of t = 2π 3 2 2 . MATH 201 - Week 9 The Unit Circle The Reference Number Example. Find the terminal points using reference numbers for 3 7π π 2π , t = π, t = , t=− . t= 3 4 6 4 Solution. π For t = 2π 3 , the reference number is t̄ = 3 . √ The terminal point of t̄ is 12 , 23 . The terminal point of t lies in the second quadrant therefore the signs of the x-coordinate and the y -coordinate are negative and positive respectively. √ 3 1 is − , So the terminal point of t = 2π 3 2 2 . the reference number is t̄ = π4 . The terminal point of t̄ is √12 , √12 . The terminal point of t lies in the second quadrant therefore the signs of the x-coordinate and the y -coordinate are negative and positive respectively. √1 , √1 . So the terminal point of t = 3π is − 4 2 2 For t = 3π 4 , MATH 201 - Week 9 The Unit Circle The Reference Number Example. Find the terminal points using reference numbers for 3 7π π 2π , t = π, t = , t=− . t= 3 4 6 4 Solution. π For t = 2π 3 , the reference number is t̄ = 3 . √ The terminal point of t̄ is 12 , 23 . The terminal point of t lies in the second quadrant therefore the signs of the x-coordinate and the y -coordinate are negative and positive respectively. √ 3 1 is − , So the terminal point of t = 2π 3 2 2 . the reference number is t̄ = π4 . The terminal point of t̄ is √12 , √12 . The terminal point of t lies in the second quadrant therefore the signs of the x-coordinate and the y -coordinate are negative and positive respectively. √1 , √1 . So the terminal point of t = 3π is − 4 2 2 For t = 3π 4 , MATH 201 - Week 9 The Unit Circle The Reference Number Solution (Continued). π For t = 7π 6 , the reference number √ is t̄ = 6 . The terminal point of t̄ is 23 , 12 . The terminal point of t lies in the third (−, −). √ quadrant: 3 1 7π So the terminal point of t = 6 is − 2 , − 2 . MATH 201 - Week 9 The Unit Circle The Reference Number Solution (Continued). π For t = 7π 6 , the reference number √ is t̄ = 6 . The terminal point of t̄ is 23 , 12 . The terminal point of t lies in the third (−, −). √ quadrant: 3 1 7π So the terminal point of t = 6 is − 2 , − 2 . For t = − π4 , the reference number is t̄ = π4 . The terminal point of t̄ is √12 , √12 . The terminal point of t lies in the fourth quadrant: (+, −). π 1 1 √ √ So the terminal point of t = − 4 is ,− 2 . 2 MATH 201 - Week 9 The Unit Circle The Reference Number Solution (Continued). π For t = 7π 6 , the reference number √ is t̄ = 6 . The terminal point of t̄ is 23 , 12 . The terminal point of t lies in the third (−, −). √ quadrant: 3 1 7π So the terminal point of t = 6 is − 2 , − 2 . For t = − π4 , the reference number is t̄ = π4 . The terminal point of t̄ is √12 , √12 . The terminal point of t lies in the fourth quadrant: (+, −). π 1 1 √ √ So the terminal point of t = − 4 is ,− 2 . 2 MATH 201 - Week 9 Trigonometric Functions of Real Numbers Definition of Trigonometric Functions of Real Numbers Let t be a real number and let P(x, y ) denote the terminal point of t. The trigonometric functions of t are defined to be sin t = y csc t = 1 y cos t = x tan t = y x 1 x cot t = x y sec t = Remark. Note the similarities and differences between the expressions defining the trigonometric functions for angles and for real numbers. MATH 201 - Week 9 Trigonometric Functions of Real Numbers Example. Find the trigonometric functions of t = 5π 6 . MATH 201 - Week 9 Trigonometric Functions of Real Numbers Example. Find the trigonometric functions of t = 5π 6 . π Solution. The reference number of t is t̄ = π − 5π 6 = 6 therefore √ the terminal point is (− 3 1 2 , 2 ). 5π 6 5π cos 6 5π tan 6 5π csc 6 5π sec 6 5π cot 6 sin √ So x = − = y= 1 2√ 3 2 y 1 = −√ x 3 1 =2 y 1 2 = −√ x 3 √ x =− 3 y = x =− = = = = 3 2 and y = 21 . MATH 201 - Week 9 Trigonometric Functions of Real Numbers The domains of Trigonometric Functions Function Domain sin R cos R tan R \ π2 + kπ | k is an integer csc R \ {kπ π | k is an integer} sec R \ 2 + kπ | k is an integer R \ {kπ | k is an integer} cot Range [−1, 1] [−1, 1] R (−∞, −1] ∪ [1, ∞) (−∞, −1] ∪ [1, ∞) R MATH 201 - Week 9 Trigonometric Functions of Real Numbers Values of Trigonometric Functions Special values and signs of the trigonometric functions t 0 π 6 π 4 π 3 π 2 sin t cos t tan t csc t sec t cot t 0 1 0 − 1 − √ √ 3 1 2 √1 √ 2 3 2 2 3 √ √3 √1 √1 1 2 2 1 2 √2 √ 3 1 2 √ √1 3 2 2 2 3 3 1 0 − 1 − 0 MATH 201 - Week 9 Trigonometric Functions of Real Numbers Values of Trigonometric Functions Special values and signs of the trigonometric functions t 0 π 6 π 4 π 3 π 2 sin t cos t tan t csc t sec t cot t 0 1 0 − 1 − √ √ 3 1 2 √1 √ 2 3 2 2 3 √ √3 √1 √1 1 2 2 1 2 √2 √ 3 1 2 √ √1 3 2 2 2 3 3 1 0 − 1 − 0 Quad. sin t cos t tan t csc t sec t cot t I + + + + + + I + − − + − − III − − + − − + IV − + − − + − MATH 201 - Week 9 Trigonometric Functions of Real Numbers Values of Trigonometric Functions Example. Use the known special values to find the exact values of π 20π 4π , csc . sin , cot − 3 6 3 MATH 201 - Week 9 Trigonometric Functions of Real Numbers Values of Trigonometric Functions Example. Use the known special values to find the exact values of π 20π 4π , csc . sin , cot − 3 6 3 Solution. π 4π t̄ = 4π 3 − π = 3 . The terminal point of 3 is in the third quadrant so √ 3 π 4π = − sin = − . sin 3 3 2 MATH 201 - Week 9 Trigonometric Functions of Real Numbers Values of Trigonometric Functions Example. Use the known special values to find the exact values of π 20π 4π , csc . sin , cot − 3 6 3 Solution. π 4π t̄ = 4π 3 − π = 3 . The terminal point of 3 is in the third quadrant so √ 3 π 4π = − sin = − . sin 3 3 2 t̄ = π6 . The terminal point of − π6 is in the fourth quadrant so π √ π cot − = − cot = − 3. 6 6 MATH 201 - Week 9 Trigonometric Functions of Real Numbers Values of Trigonometric Functions Example. Use the known special values to find the exact values of π 20π 4π , csc . sin , cot − 3 6 3 Solution. π 4π t̄ = 4π 3 − π = 3 . The terminal point of 3 is in the third quadrant so √ 3 π 4π = − sin = − . sin 3 3 2 t̄ = π6 . The terminal point of − π6 is in the fourth quadrant so π √ π cot − = − cot = − 3. 6 6 t̄ = π3 . The terminal point of csc 20π 3 is in the second quadrant so 20π π 2 = csc = √ . 3 3 3 MATH 201 - Week 9 Trigonometric Functions of Real Numbers Values of Trigonometric Functions Fundamental Identities Reciprocial Identities csc t = 1 sin t sec t = tan t = sin t cos t 1 cos t cot t = cot t = 1 tan t cos t sin t Pythagorean Identities sin2 t + cos2 t = 1 tan2 t + 1 = sec2 t 1 + cot2 t = csc2 t MATH 201 - Week 9 Trigonometric Functions of Real Numbers Values of Trigonometric Functions Example. Use the calculator to find the approximate values of sin 22.4, cot (−1.3) , csc 31. MATH 201 - Week 9 Trigonometric Functions of Real Numbers Values of Trigonometric Functions Example. Use the calculator to find the approximate values of sin 22.4, cot (−1.3) , csc 31. Solution. First of all, make sure that the calculator is in rad mode. sin 22.4 ≈ −0.3976. MATH 201 - Week 9 Trigonometric Functions of Real Numbers Values of Trigonometric Functions Example. Use the calculator to find the approximate values of sin 22.4, cot (−1.3) , csc 31. Solution. First of all, make sure that the calculator is in rad mode. sin 22.4 ≈ −0.3976. cot (−1.3) = 1 ≈ −0.2776 tan (−1.3) MATH 201 - Week 9 Trigonometric Functions of Real Numbers Values of Trigonometric Functions Example. Use the calculator to find the approximate values of sin 22.4, cot (−1.3) , csc 31. Solution. First of all, make sure that the calculator is in rad mode. sin 22.4 ≈ −0.3976. cot (−1.3) = csc 31 = 1 ≈ −0.2776 tan (−1.3) 1 ≈ −2.475 sin 31 MATH 201 - Week 9 Trigonometric Functions of Real Numbers Values of Trigonometric Functions Reminder. A function f is said to be even if f (−x) = f (x) for all x in its domain. A function f is said to be odd if f (−x) = −f (x) for all x in its domain. Even-Odd Properties sin(−t) = − sin t cos(−t) = cos t tan(−t) = − tan t csc(−t) = − csc t sec(−t) = sec t cot(−t) = − cot t MATH 201 - Week 9 Trigonometric Functions of Real Numbers Values of Trigonometric Functions Example. Express sin t in terms of cos t where t is chosen from the fourth quadrant. MATH 201 - Week 9 Trigonometric Functions of Real Numbers Values of Trigonometric Functions Example. Express sin t in terms of cos t where t is chosen from the fourth quadrant. Solution. In the fourth quadrant, the sine function is negative. Using the basic Pythagorean identity we get sin2 t + cos2 t = 1 sin2 t = 1 − cos2 t p sin t = ± 1 − cos2 t Since sin t is negative in the fourth quadrant and the square root is non-negative we have p sin t = − 1 − cos2 t. MATH 201 - Week 9 Trigonometric Functions of Real Numbers Values of Trigonometric Functions Example. Express sin t in terms of cos t where t is chosen from the fourth quadrant. Solution. In the fourth quadrant, the sine function is negative. Using the basic Pythagorean identity we get sin2 t + cos2 t = 1 sin2 t = 1 − cos2 t p sin t = ± 1 − cos2 t Since sin t is negative in the fourth quadrant and the square root is non-negative we have p sin t = − 1 − cos2 t. Remark. If t is in the third quadrant, for example, then the + sign is valid. MATH 201 - Week 9 Trigonometric Functions of Real Numbers Values of Trigonometric Functions Example. If sec t = 3 and t is from the fourth quadrant, find the values of the other trigonometric functions. MATH 201 - Week 9 Trigonometric Functions of Real Numbers Values of Trigonometric Functions Example. If sec t = 3 and t is from the fourth quadrant, find the values of the other trigonometric functions. Solution. It is convenient to find the sine and cosine first: 1 1 cos t = = . sec t 3 Using the expression from the previous page we have r r √ p 1 8 8 2 sin t = − 1 − cos t = − 1 − = − =− . 9 9 3 From these, it is easy to generate the other trigonometric functions: √ sin t =− 8 tan t = cos t 1 3 csc t = = −√ sin t 8 cos t 1 cot t = = −√ sin t 8 MATH 201 - Week 9 Trigonometric Graphs Graphs of the Sine and Cosine Functions A function f is said to be periodic if there is a number p such that f (t + p) = f (t) for all t. The least such positive p (if this exists) is called the period of f . MATH 201 - Week 9 Trigonometric Graphs Graphs of the Sine and Cosine Functions A function f is said to be periodic if there is a number p such that f (t + p) = f (t) for all t. The least such positive p (if this exists) is called the period of f . Periodicity of sin and cos The sine and cosine functions are periodic and their period is 2π: sin(t + 2π) = sin t cos(t + 2π) = cos t. MATH 201 - Week 9 Trigonometric Graphs Graphs of the Sine and Cosine Functions A function f is said to be periodic if there is a number p such that f (t + p) = f (t) for all t. The least such positive p (if this exists) is called the period of f . Periodicity of sin and cos The sine and cosine functions are periodic and their period is 2π: sin(t + 2π) = sin t cos(t + 2π) = cos t. This fact is very important: it is enough to find the shape of their graphs in the interval between 0 and 2π and then ’make copies’. MATH 201 - Week 9 Trigonometric Graphs Graphs of the Sine and Cosine Functions We can use the special values of the trigonometric values to sketch the graphs of the sine and cosine functions: The graph of the sine function The graph of the cosine function MATH 201 - Week 9 Trigonometric Graphs Graphs of the Sine and Cosine Functions Vertical and Horizontal Strechings and Shrinkings The graph of a transformed sine function y = A sin(kx) is obtained from the original graph y = sin x using 1 horizontal stretching/shrinking depending on the parameter k 2 vertical stretching/shrinking depending on the parameter A MATH 201 - Week 9 Trigonometric Graphs Graphs of the Sine and Cosine Functions Vertical and Horizontal Strechings and Shrinkings The graph of a transformed sine function y = A sin(kx) is obtained from the original graph y = sin x using 1 horizontal stretching/shrinking depending on the parameter k 2 vertical stretching/shrinking depending on the parameter A The period of A sin(kx) is 2π k : 2π A sin k x + = A sin (kx + 2π) = A sin (kx) . k MATH 201 - Week 9 Trigonometric Graphs Graphs of the Sine and Cosine Functions Vertical and Horizontal Strechings and Shrinkings The graph of a transformed sine function y = A sin(kx) is obtained from the original graph y = sin x using 1 horizontal stretching/shrinking depending on the parameter k 2 vertical stretching/shrinking depending on the parameter A The period of A sin(kx) is 2π k : 2π A sin k x + = A sin (kx + 2π) = A sin (kx) . k The magnitude of A sin(kx) is given by the number |A|: this is called the amplitude. MATH 201 - Week 9 Trigonometric Graphs Graphs of the Sine and Cosine Functions Additional Horizontal Shifts The graph of a transformed sine function y = A sin(k(x − b)) is obtained from the original graph y = sin x using 1 horizontal stretching/shrinking depending on the parameter k 2 horizontal shifting given by b 3 vertical stretching/shrinking depending on the parameter A MATH 201 - Week 9 Trigonometric Graphs Graphs of the Sine and Cosine Functions Additional Horizontal Shifts The graph of a transformed sine function y = A sin(k(x − b)) is obtained from the original graph y = sin x using 1 horizontal stretching/shrinking depending on the parameter k 2 horizontal shifting given by b 3 vertical stretching/shrinking depending on the parameter A The number b is called the phase shift. MATH 201 - Week 9 Trigonometric Graphs Graphs of the Sine and Cosine Functions Example. Consider the function f (x) = −3 sin(2(x − 5)). What is the amplitude, period and phase shift of f (x)? Sketch the graph of f . MATH 201 - Week 9 Trigonometric Graphs Graphs of the Sine and Cosine Functions Example. Consider the function f (x) = −3 sin(2(x − 5)). What is the amplitude, period and phase shift of f (x)? Sketch the graph of f . Solution. The amplitude is 3, the period is 2π 2 = π and the phase shift is 5. The graph is obtained from the graph of y = sin x using the following sequence of elementary transformations: 1 2 1 horizontal shrinking of factor 2 horizontal shifting to the right of 5 units 3 vertical stretching of factor 3 4 reflection to the x-axis. MATH 201 - Week 9 Trigonometric Graphs Graphs of the Sine and Cosine Functions