The Sine Wave The sine function has domain the set of all real - - PowerPoint PPT Presentation

Elementary Functions

Part 4, Trigonometry Lecture 4.3a, The Sine Wave

• Dr. Ken W. Smith

Sam Houston State University

2013

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The Sine Wave

The sine function has domain the set of all real numbers: (−∞, ∞) but the range is just [−1, 1] since all y-coordinates on the unit circle must be between −1 and 1.

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Cosine

Similarly, the domain of cosine is (−∞, ∞) and the range is [−1, 1].

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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 starting at the point P(1, 0) and moving above the x-axis, while a negative value of θ means we move clockwise, starting at the point P(1, 0) and moving below the x-axis, 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.

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Symmetries of sine and cosine

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.

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Central Angles and Arcs

Here is the graph of the sine function. 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.) Since 2π radians makes a complete revolution of the circle then sin(θ + 2π) = sin(θ).

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Some worked problems

1 Describe the set of all the angles θ that satisfy the trig equation

sin θ = −

√ 3 2 .

• Solution. If the sine of an angle is negative then it must be in the

third of fourth quadrants. From our knowledge of 30-60-90 triangles, we see that θ = 4π 3 (240◦) and θ = 5π 3 (300◦) are angles whose sine is −

√ 3 2 .

Since the sine function is periodic these are not the only solutions to this equation! Since the sine function is periodic with period 2π we know that θ = 4π 3 + 2πk and θ = 5π 3 + 2πk, (where k is an integer) will also be solutions.

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The Sine Wave

Solve each of the following equations.

1 cos x = 1 2 2 cos x = 1 3 (2 cos x − 1)(cos x − 1) = 0.

Solutions.

1 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 0 + 2πk (where k is an integer) or {2πk : k ∈ Z}.

2 Since cos π 3 = 1 2 then x = π 3 is a solution to 2 cos x = 1. So also is

x = − π

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

3 + 2πk : k ∈ Z} ∪ {− π 3 + 2πk : k ∈ Z}.

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The Sine Wave

3 Solve the equation (2 cos x − 1)(cos x − 1) = 0.

• Solution. Any solution to (2 cos x − 1)(cos x − 1) = 0 is either a

solution to 2 cos x − 1 = 0 or a solution to cos x − 1 = 0. We have already solved these equations in the previous two problems. All of the solutions the previous two problems are solutions to this

• problem. So our answer is

{2πk : k ∈ Z} ∪ { π

3 + 2πk : k ∈ Z} ∪ {− π 3 + 2πk : k ∈ Z}.

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Amplitude, period and phase shift

In practical applications many periodic functions are transformations of the sine function. A transformation 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. (1) 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 (about the line x = c), 3 Expand vertically by a factor of a 4 Shift up by d.

Some of these translations are associated with particular terms. We revisit the concept of period and introduce new terms frequency, amplitude and phase shift.

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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 will have period 2π |b| . (We use an absolute value sign here since we want the period to be positive and it is possible that b is negative.)

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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) is 1 2π and so the frequency of f(θ) = a sin(b(θ − c)) + d is |b| 2π.

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Frequency

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 (430 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, green, and so on.

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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 |a|. (Again, we use absolute value because we want the amplitude to be positive.)

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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.

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The sine wave

In the next presentation, we will continue to look further at sine waves (sinusoids) including the cosine function. (End)

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Elementary Functions

Part 4, Trigonometry Lecture 4.3b, The Cosine and other Sinusoids

• Dr. Ken W. Smith

Sam Houston State University

2013

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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.

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Central Angles and Arcs

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 discussing the cosine function. The function g(θ) has period p = 2π |b| , frequency f = |b| 2π, amplitude |a| 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 demonstrate that the graph of cos(θ) is the graph of sin(θ) shifted to the left by π 2 . cos(θ) = sin(θ + π 2 ). We could think of the cosine function as a sine wave with phase shift −π 2 .

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The symmetries of the six trig functions

Since the sine function is odd and the cosine function is even then tan(−θ) = sin(−θ) cos(−θ) = − sin(θ) cos(θ) = − tan(θ) and so the tangent function is odd. Here is a graph of the tangent function:

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The symmetries of the six trig functions

If the central angle θ gives the point P(x, y) on the unit circle then the tangent of θ is y

• x. The tangent of θ + π will then be −y

−x and since the minus signs will cancel tan(θ + π) = y x = tan(θ). So the tangent function has period p = π, not 2π!

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Symmetries of six trig functions

The reciprocals of cosine, sine and tangent have the same “parity” (even/odd-ness) as the original function. So the secant function is even while cosecant and cotangent are both

• dd.

Just like cosine and sine, the secant and cosecant functions have period 2π. The cotangent function, like the tangent function, has period π.

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Worked problems with sine waves

1 Describe the transformations necessary to change the graph of

y = sin x into the graph of y = −5 sin(2(x − π

4 )) + 1

• Solution. (These must be done in exactly this order. Any other order

is incorrect.)

1 Shift right by π

4 .

2 Shrink horizontally by a factor of 2 (about the line x = π

4 ).

3 Expand vertically by a factor of 5 and reflect across the x-axis. 4 Shift up 1.

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The Sine Wave

2 Give the amplitude, period and phase shift for the sine wave

y = −5 sin(2(x − π

4 )) + 1.

Solution.

1 In the previous problem we began by shifting right by π

4 . This is the

phase shift.

2 Then we shrunk the graph horizontally by a factor of 2 so the period is

2π 2 = π.

3 Then we stretched the graph vertically by a factor of 5 and turned it

• ver. The amplitude should always be positive (it represents a

deviation from the mean) and so the amplitude is 5.

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Examples of Sine Waves

3 A person’s blood pressure follows a sine wave corresponding to the

beats of the heart. A particular individual’s blood pressure at time t (measured in minutes) is p(t) = 20 sin(160πt) + 100. What does this tell us about the person’s heart rate and blood pressure?

• Solution. To transform the sine wave f(θ) = sin(θ) into the graph of

p(t) = 20 sin(160πt) + 100 we first shrink the graph horizontally by a factor of 160π so that the wave has period

2π 160π = 1 80.

Then stretch the graph vertically by a factor of 20 and then shift it up 100. This means that a single heartbeat occurs in

1 80th of a minute and

that the frequency is 80 beats per minute. The amplitude of this sine wave is 20; the blood pressure varies from a maximum of 100 + 20 = 120 to a minimum of 100 − 20 = 80. So the person’s blood pressure is 120/80. The phase shift is zero.

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The Sine Wave

4 Data from The Weather Channel summarizes Houston weather

averages. One approximation to the average highs in Houston is the equation H(m) = 15 sin( π

6 (m − 4.5)) + 78

where m = 1 represents the month of January, etc. (See orange curve below.)

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The Sine Wave

Given the equation, H(m) = 15 sin( π

6 (m − 4.5)) + 78

answer the following questions.

1 What is the period of this function? 2 What is its amplitude? 3 What does the phase shift m = 4.5 say about the month of April? 4 Use this model to estimate the average high in January, April, July

and October. How do those numbers compare with our chart?

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The Sine Wave

H(m) = 15 sin( π

6 (m − 4.5)) + 78

• Solution. To transform the sine wave f(θ) = sin(θ) into the graph of

H(m) = 15 sin( π

6 (m − 4.5)) + 78 we shift the sine wave right by 4.5 and

then shrink the graph horizontally by a factor of π/6. Since we shrunk the graph horizontally by a factor of π/6, the period is now 2π π/6 = 12. Not surprisingly, this tells us that the graph repeats every 12 months. We then stretch the graph vertically by a factor of 15 (so that the amplitude is 15) and then shift it up 78 so that if varies from a high of 93 to a low of 63. The phase shift of m = 4.5 tells us that the month of April is close to the annual average; it is 3 months after the lowest temperatures (in January) and 3 months before the highest temperatures (in July/August.) Notice that our model equation for Houston average high monthly temperatures, does not quite fit the Weather Channel data, but is certainly close.

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The Sine Wave

In the next presentation, we will look trigonometry on right triangles. (End)

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