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Lecture Notes 4.3 Sine Waves.Pdf 4.3 The Sine Wave The sine function has domain the set of all real numbers: (−∞; 1) but the range is just [−1; 1] since all y-coordinates on the unit circle must be between −1 and 1. Similarly, the domain of cosine is (−∞; 1) and the range is [−1; 1]: 4.3.1 Symmetries of sine and cosine Let's consider the definition of sine and cosine on the unit circle and ask about symmetries. Are either of these functions even? odd? We assume that a positive angle θ involves a counterclockwise rotation while a negative value of θ means we move clockwise. Is it clear that moving clockwise instead of counterclockwise does not change the sign of the x-value of the point P (x; y)? That is, for any angle θ, cos(−θ) = cos(θ) and so cosine is an even function. However, if we begin to move clockwise around the origin, beginning on the x-axis at (1; 0) then the y-value of the point P (x; y) immediately becomes negative instead of positive. Reversing the direction of rotation reverses the sign of the y-value and so sin(−θ) = − sin(θ): Therefore the sine function is odd. Here is the graph of the sine function. Figure 13. The graph of f(θ) = sin(θ): Notice the rotational symmetry about the origin. But the sine function has much more symmetry than just rotational symmetry about the origin. It is in fact periodic with period 2π (≈ 6:28:) The reasons for this are obvious: since 2π radians makes a complete revolution of the circle then sin(θ + 2π) = sin(θ): Some worked problems p 3 1. Describe the set of all the angles θ that satisfy the trig equation sin θ = − 2 : 160 Solution. If the sine of an angle is negative then it must be in the third of fourth quadrants. From 4π 5π p our knowledge of 30-60-90 triangles, we see that θ = and θ = are angles whose sine is − 3 : 3 3 2 But then since the sine function is periodic with period 2π we know that 4π 5π θ = + 2πk and θ = + 2πk, 3 3 where k is an integer, will also be solutions. 2. Solve each of the following equations. (a) cos x = 1 (b) 2 cos x = 1 (c) (2 cos x − 1)(cos x − 1) = 0: Solutions. (a) Since cos(0) = 1 then cos x = 1 means that x is either 0 or 0 plus some multiple of 2π: We can write this all in the form f2πk : k 2 Zg: π 1 π π (b) Since cos 3 = 2 then x = 3 is a solution to 2 cos x = 1. So is x = − 3 : (Remember, f(x) = cos x is an even function!) Since the period of cosine is 2π then our set of all solutions is π π f 3 + 2πk : k 2 Zg [ {− 3 + 2πk : k 2 Zg: (c) Any solution to (2 cos x − 1)(cos x − 1) = 0 is either a solution to cos x − 1 = 0 or a solution to 2 cos x − 1 = 0: We have already solved these equations in parts (a) and (b). All of the solutions to parts (a) and (b) are solutions to part (c). So our answer is π π f2πk : k 2 Zg [ f 3 + 2πk : k 2 Zg [ {− 3 + 2πk : k 2 Zg: 4.3.2 Amplitude, period and phase shift In practical applications many periodic functions are tranformations of the sine function. A transfor- mation of the sine function is often called a sine wave or a sinusoid. In general, sine waves will have form f(θ) = a sin(b(θ − c)) + d: (12) From our earlier discussion of transformations, we see that one can transform the graph of sin(θ) into the graph of f(θ) = a sin(b(θ + c)) + d by the following steps (in this order!): 1. Shift right by c, 2. Shrink horizontally by a factor of b, 3. Expand vertically by a factor of a, 4. Shift up by d. 161 In regards to the sine function, some of these translations are associated with particular terms. We will revisit the concept of period and introduce new terms frequency, amplitude and phase shift. Period of a sine wave Since the sine function has period 2π then the sine wave given by the function f(θ) = a sin(b(θ−c))+d 2π will have period : (We use an absolute value sign here since we want the period to be positive and it jbj is possible that b is negative.) Frequency of a sine wave The period of a sine wave tells us how many units of the input variable are required before the function repeats. The frequency a sine wave is the number of times the wave repeats within a single unit of the input variable θ; this is the reciprocal of the period. Thus the frequency of the standard sine wave sin(x) 1 jbj is and so the frequency of f(θ) = a sin(b(θ − c)) + d is : 2π 2π Electronic transmissions involve the sine wave. The frequency of the transmission represents the number of copies of the sine wave which occur within a single unit of time (often one second.) For example, an electromagnetic wave with frequency 4:3 × 1014 oscillates 430; 000; 000; 000; 000 (4 hundred million million) times in one second and is perceived by our eyes as the color red. Scientists often use the term \hertz" to represent \cycles per second" and so we say that frequency of red light is 4:3 × 1014 hertz. A light wave of lower frequency will not be visible to our eyes; waves of higher frequency will show up as orange, yellow, and so on. (For the frequencies of various colors, see the Wikipedia article on color.) Amplitude of a sine wave The height of the standard sine wave oscillates between a maximum of 1 and minimum of −1. If we consider the midpoint of this wave, then the wave rises 1 unit above and then drops 1 unit below this midpoint. This variation from the \average" height is the amplitude of the sine wave. For the standard sine wave the amplitude is 1. The amplitude of the sine wave f(θ) = a sin(b(θ − c)) + d is just jaj: (Again, we use absolute value because we want the amplitude to be positive.) Phase shift of a sine wave The graph of the standard sine wave sin(θ) passes through the origin (0; 0). A sine wave might be shifted to the right by an amount c; this is the phase shift of the sine wave f(θ) = a sin(b(θ − c)) + d: Note that the phase shift can be negative. A negative phase shift means that the graph of sin θ is being shifted by a certain amount to the left. The graph of cosine. The graph of cosine function has a very similar wave pattern to that of sine. Here is a graph of the cosine function. 162 Figure 14. The graph of f(x) = cos(θ): Just as we did with sine waves, we may consider graphs of g(θ) = a cos(b(θ − c)) + d: There is no significant difference in meaning for the period, frequency, amplitude or phase shift when 2π jbj discussing the cosine function; here the function g(θ) has period p = ; frequency f = , amplitude jbj 2π jaj and phase shift c. We tend to concentrate on the sine wave and ignore the cosine function. This is merely because the graph of cosine function is really a shift of the graph of sine! A careful examination of the graphs of these functions (or an examination of the definitions of cosines and sines on the unit circle) demonstrate that π the graph of cos(θ) is the graph of sin(θ) shifted to the left by : Therefore 2 π cos(θ) = sin(θ + ): (13) 2 π We could think of the cosine function as a sine wave with phase shift − . 2 4.3.3 The symmetries of the six trig functions Since the sine function is odd and the cosine function is even then sin(−θ) − sin(θ) tan(−θ) = = = − tan(θ) cos(−θ) cos(θ) and so the tangent function is odd. Here is a graph of the tangent function: 163 Figure 15. The graph of f(x) = tan(θ): y If the central angle θ gives the point P (x; y) on the unit circle then the tangent of θ is . The tangent x −y of θ + π will then be and since the minus signs will cancel we see that −x y tan(θ + π) = = tan(θ): x So the tangent function has period p = π, not 2π! The reciprocals of cosine, sine and tangent with have the same \parity" (even/odd property) as the original function. So the secant function is even while cosecant and cotangent are both odd. Just like cosine and sine, the secant and cosecant functions have period 2π: The cotangent function, like the tangent function, has period π: Worked problems with sine waves 1. Describe the transformations necessary to change the graph of y = sin x into the graph of π y = −5 sin(2(x − 4 )) + 1 Solution. (These must be done in exactly this order. Any other order is incorrect.) π (a) Shift right by 4 . (b) Shrink horizontally by a factor of 2. (c) Expand vertically by a factor of 5 and reflect across the x-axis.
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