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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. I The Unit Circle (Section 5.1) I I I I Trigonometric Functions of Real Numbers (Section 5.2) I I I I The Unit Circle Terminal Points on the Unit Circle The Reference Number The Trigonometric Functions Values of Trigonometric Functions Fundamental Identities Trigonometric Graphs (Section 5.3) I I Graphs of the Sine and Cosine Functions Graphs of Transformations of Sine and Cosine 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 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 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: I If t = 0 then the corresponding terminal point is (1, 0). I 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. I 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. 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. I The terminal point for t = π is (−1, 0) since π is exactly the half of the perimeter. I The terminal point for t = the total length. I 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. π 2 is (0, 1) since π 2 is one-fourth of 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). 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. I The reference number of π is 0 since the terminal point (−1, 0) is on the x-axis. I 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. I 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. Example. Find the reference numbers for 3 t = − π, 2 t= 7π , 8 t=− 5π . 6 Solution. I For t = − 32 π the terminal point is (0, 1). The corresponding reference number is t̄ = π2 . I 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 . I 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. Example. Find the terminal points using reference numbers for t= 2π , 3 3 t = π, 4 t= 7π , 6 π t=− . 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 So the terminal point of t = 2π is − , 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. 3π 1 √1 √ So the terminal point of t = 4 is − 2 , 2 . For t = 3π 4 , 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 . 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. Example. Find the trigonometric functions of t = 5π 6 . 5π Solution. The reference number of t is t̄ = π − 6 = π6 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 . The domains of Trigonometric Functions Function sin cos tan csc sec cot Domain R R R \ π2 + kπ | k is an integer R \ {kπ | k is an integer} R \ π2 + kπ | k is an integer R \ {kπ | k is an integer} Range [−1, 1] [−1, 1] R (−∞, −1] ∪ [1, ∞) (−∞, −1] ∪ [1, ∞) R 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 1 2 √ √ 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 − + − − + − Example. Use the known special values to find the exact values of π 4π 20π sin , cot − . , csc 3 6 3 Solution. I t̄ = 4π − π = π . The terminal point of 4π is in the third 3 3 3 quadrant so √ π 3 4π = − sin = − . sin 3 3 2 I t̄ = π6 . The terminal point of − π6 is in the fourth quadrant so π √ π = − cot = − 3. cot − 6 6 I t̄ = π3 . The terminal point of csc 20π 3 is in the second quadrant so 20π π 2 = csc = √ . 3 3 3 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 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. I sin 22.4 ≈ −0.3976. I cot (−1.3) = I csc 31 = 1 ≈ −0.2776 tan (−1.3) 1 ≈ −2.475 sin 31 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 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. 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 8 8 1 =− . sin t = − 1 − cos2 t = − 1 − = − 9 9 3 From these, it is easy to generate the other trigonometric functions: √ sin t tan t = =− 8 cos t 1 3 csc t = = −√ sin t 8 cos t 1 cot t = = −√ sin t 8 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’. 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 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. 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. 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. horizontal shrinking of factor 1 2 2. horizontal shifting to the right of 5 units 3. vertical stretching of factor 3 4. reflection to the x-axis.