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The sine has domain the set of all real numbers: (−∞, ∞) but the range is just [−1, 1] since all y-coordinates on the must be between −1 and 1. Elementary Functions Part 4, Lecture 4.3a, The

Dr. Ken W. Smith

Sam Houston State University

2013

Smith (SHSU) Elementary Functions 2013 1 / 29 Smith (SHSU) Elementary Functions 2013 2 / 29 Cosine Symmetries of sine and cosine

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

Smith (SHSU) Elementary Functions 2013 3 / 29 Smith (SHSU) Elementary Functions 2013 4 / 29 Symmetries of sine and cosine Central 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 However, if we begin to move clockwise around the origin, beginning on symmetry about the origin. It is in fact periodic with period 2π (≈ 6.28.) the x-axis at (1, 0) then the y-value of the point P (x, y) immediately Since 2π makes a complete revolution of the circle then becomes negative instead of positive. sin(θ + 2π) = sin(θ).

Reversing the direction of rotation reverses the sign of the y-value and so

sin(−θ) = − sin(θ).

Therefore the sine function is odd.

Smith (SHSU) Elementary Functions 2013 5 / 29 Smith (SHSU) Elementary Functions 2013 6 / 29 Some worked problems The Sine Wave

Solve each of the following . 1 Describe the set of all the angles θ that satisfy the trig √ 1 3 cos x = 1 sin θ = − 2 . 2 2 cos x = 1 Solution. If the sine of an angle is negative then it must be in the 3 (2 cos x − 1)(cos x − 1) = 0. third of fourth quadrants. From our knowledge of 30-60-90 , 4π 5π we see that θ = (240◦) and θ = (300◦) are angles whose sine Solutions. 3 3 √ 1 Since cos(0) = 1 then cos x = 1 means that x is either 0 or 0 plus is − 3 . 2 some multiple of 2π. We can write this all in the form 0 + 2πk (where k is an ) or 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 {2πk : k ∈ Z}. know that π 1 π 2 Since cos 3 = 2 then x = 3 is a solution to 2 cos x = 1. So also is 4π 5π π θ = + 2πk and θ = + 2πk, x = − 3 . (Remember, f(x) = cos x is an even function!) Since the 3 3 period of cosine is 2π then our set of all solutions is (where k is an integer) will also be solutions. π π { 3 + 2πk : k ∈ Z} ∪ {− 3 + 2πk : k ∈ Z}.

Smith (SHSU) Elementary Functions 2013 7 / 29 Smith (SHSU) Elementary Functions 2013 8 / 29 The Sine Wave , period and 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 will have form 3 Solve the equation (2 cos x − 1)(cos x − 1) = 0. f(θ) = a sin(b(θ − )) + d. (1) 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. From our earlier discussion of transformations, we see that one can We have already solved these equations in the previous two problems. 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!): All of the solutions the previous two problems are solutions to this 1 Shift right by c, problem. So our answer is 2 Shrink horizontally by a factor of b (about the line x = c), π π {2πk : k ∈ Z} ∪ { 3 + 2πk : k ∈ Z} ∪ {− 3 + 2πk : k ∈ Z}. 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 , amplitude and phase shift. Smith (SHSU) Elementary Functions 2013 9 / 29 Smith (SHSU) Elementary Functions 2013 10 / 29 Period of a sine wave 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. Since the sine function has period 2π then the sine wave given by the 2π function f(θ) = a sin(b(θ − c)) + d will have period . The frequency a sine wave is the number of times the wave repeats within |b| a single unit of the input variable θ; this is the reciprocal of the period. (We use an absolute value sign here since we want the period to be positive and it is possible that b is negative.) 1 Thus the frequency of the standard sine wave sin(x) is and so the 2π |b| frequency of f(θ) = a sin(b(θ − c)) + d is . 2π

Smith (SHSU) Elementary Functions 2013 11 / 29 Smith (SHSU) Elementary Functions 2013 12 / 29 Frequency Amplitude of a sine wave

Electronic transmissions involve the sine wave. The frequency of the The height of the standard sine wave oscillates between a maximum of 1 transmission represents the number of copies of the sine wave which occur and minimum of −1. within a single unit of time (often one second.)

14 If we consider the midpoint of this wave, then the wave rises 1 unit above For example, an electromagnetic wave with frequency 4.3 × 10 oscillates and then drops 1 unit below this midpoint. 430, 000, 000, 000, 000 (430 million million) times in one second and is perceived by our eyes as the color red. This variation from the “average” height is the amplitude of the sine wave. For the standard sine wave the amplitude is 1. Scientists often use the term “” to represent “cycles per second” and 14 so we say that frequency of red light is 4.3 × 10 hertz. 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 A light wave of lower frequency will not be visible to our eyes; waves of positive.) higher frequency will show up as orange, yellow, green, and so on.

Smith (SHSU) Elementary Functions 2013 13 / 29 Smith (SHSU) Elementary Functions 2013 14 / 29 Phase shift of a sine wave The sine wave

The graph of the standard sine wave sin(θ) passes through the origin (0, 0). In the next presentation, we will continue to look further at sine waves (sinusoids) including the cosine function. 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. (End) 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.

Smith (SHSU) Elementary Functions 2013 15 / 29 Smith (SHSU) Elementary Functions 2013 16 / 29 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. Elementary Functions Part 4, Trigonometry Lecture 4.3b, The Cosine and other Sinusoids

Dr. Ken W. Smith

Sam Houston State University

2013

Smith (SHSU) Elementary Functions 2013 17 / 29 Smith (SHSU) Elementary Functions 2013 18 / 29 Central Angles and Arcs The symmetries of the six trig functions

Just as we did with sine waves, we may consider graphs of Since the sine function is odd and the cosine function is even then sin(−θ) − sin(θ) g(θ) = a cos(b(θ − c)) + d. tan(−θ) = = = − tan(θ) cos(−θ) cos(θ) There is no significant difference in meaning for the period, frequency, amplitude or phase shift when discussing the cosine function. and so the function is odd. Here is a graph of the tangent function: 2π |b| The function g(θ) has period p = , frequency f = , amplitude |a| |b| 2π 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 π left by . 2 π cos(θ) = sin(θ + ). 2 π We could think of the cosine function as a sine wave with phase shift − . Smith (SHSU) Elementary Functions 2013 192 / 29 Smith (SHSU) Elementary Functions 2013 20 / 29 The symmetries of the six trig functions Symmetries of six trig functions

If the central angle θ gives the point P (x, y) on the unit circle then the y −y tangent of θ is . The tangent of θ + π will then be and since the x −x minus signs will cancel y The reciprocals of cosine, sine and tangent have the same “parity” tan(θ + π) = = tan(θ). x (even/odd-ness) as the original function. So the tangent function has period p = π, not 2π! 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.

Smith (SHSU) Elementary Functions 2013 21 / 29 Smith (SHSU) Elementary Functions 2013 22 / 29 Worked problems with sine waves The Sine Wave

2 Give the amplitude, period and phase shift for the sine wave 1 Describe the transformations necessary to change the graph of y = −5 sin(2(x − π )) + 1. π 4 y = sin x into the graph of y = −5 sin(2(x − 4 )) + 1 Solution. Solution. (These must be done in exactly this order. Any other order π 1 In the previous problem we began by shifting right by 4 . This is the is incorrect.) phase shift. π 1 Shift right by 4 . 2 Then we shrunk the graph horizontally by a factor of 2 so the period is 2 Shrink horizontally by a factor of 2 (about the line x = π ). 2π 4 2 = π. 3 Expand vertically by a factor of 5 and reflect across the x-axis. 3 Then we stretched the graph vertically by a factor of 5 and turned it 4 Shift up 1. over. The amplitude should always be positive (it represents a deviation from the mean) and so the amplitude is 5.

Smith (SHSU) Elementary Functions 2013 23 / 29 Smith (SHSU) Elementary Functions 2013 24 / 29 Examples of Sine Waves The Sine Wave

4 Data from The Weather Channel summarizes Houston weather 3 A person’s blood pressure follows a sine wave corresponding to the averages. beats of the heart. A particular individual’s blood pressure at time t One approximation to the average highs in Houston is the equation (measured in minutes) is H(m) = 15 sin( π (m − 4.5)) + 78 p(t) = 20 sin(160πt) + 100. 6 where m = 1 represents the month of January, etc. (See orange curve What does this tell us about the person’s heart rate and blood below.) 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 2π 1 factor of 160π so that the wave has period 160π = 80 . Then stretch the graph vertically by a factor of 20 and then shift it up 100. 1 This means that a single heartbeat occurs in 80 th 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. TheSmith phase (SHSU) shift is zero. Elementary Functions 2013 25 / 29 Smith (SHSU) Elementary Functions 2013 26 / 29 The Sine Wave The Sine Wave

Given the equation, H(m) = 15 sin( π (m − 4.5)) + 78 π 6 H(m) = 15 sin( 6 (m − 4.5)) + 78 answer the following questions. Solution. To transform the sine wave f(θ) = sin(θ) into the graph of π 1 What is the period of this function? H(m) = 15 sin( 6 (m − 4.5)) + 78 we shift the sine wave right by 4.5 and 2 What is its amplitude? then shrink the graph horizontally by a factor of π/6. Since we shrunk the 2π 3 What does the phase shift m = 4.5 say about the month of April? graph horizontally by a factor of π/6, the period is now = 12. 4 Use this model to estimate the average high in January, April, July π/6 Not surprisingly, this tells us that the graph repeats every 12 months. and October. How do those numbers compare with our chart? 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

Smith (SHSU) Elementary Functions 2013 27 / 29 temperatures,Smith (SHSU)does not quite fitElementary the Weather Functions Channel data, but is2013 28 / 29 certainly close. This is typical in scientific modeling – we want a good first approximation and then we explore reasons that the model does not quite fit real life. The Sine Wave

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

(End)

Smith (SHSU) Elementary Functions 2013 29 / 29