How can we model systems that vary in a smooth, wavelike cycle, rising and falling again and again?
How can we model the shape of waves in water, sound waves, radio waves, the motion of the tides in and out over the course of a day, the shaking of an earthquake, or the varying time of sunrise over the course of a year?
You probably first learned about sines, cosines, and tangents when you were studying triangles. However, these functions are amazingly useful in an enormous variety of contexts. These functions are so handy that scientists and mathematicians always keep them in mind as part of our standard toolbox. We use sines and cosines whenever we see anything that varies in a smooth wave cycle, going up and down by the same amount, again and again on a regular basis.
A tall water tower is swaying back and forth in the wind. Using an ultrasonic ranging device, we measure the distance from our device to the tower (in centimeters) every two seconds with these measurements recorded below.
| Time (sec) | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 | 20 |
| Distance (cm) | 30.9 | 23.1 | 14.7 | 12.3 | 17.7 | 26.7 | 32.3 | 30.1 | 21.8 | 13.9 | 12.6 |
Use the coordinate plane below to create a graph of these data points.
What is the water tower's maximum distance away from the device?
What is the smallest distance measured from the tower to the device?
If the water tower was sitting still and no wind was blowing, what would be the distance from the tower to the device? We call this the tower's equilibrium position.
What is the maximum distance that the tower moves away from its equilibrium position? We call this the amplitude of the oscillations.
How much time does it take the water tower to sway back and forth in a complete cycle? We call this the period of oscillation.
Sines, cosines, and tangents are very useful when studying triangles. The input into each of these functions is an angle, and the output tells us the ratio of the lengths of the sides of the triangle. There are two commonly used units for measuring angles, degrees and radians, and so there are two commonly used versions of the trigonometric functions. There's \(\sin x\) where \(x\) is in degrees, and there's \(\sin x\) where \(x\) is in radians. Calculus is a lot easier if we measure angles in radians, so that's what we'll use throughout this course. If you ever have trouble getting the right numbers from your calculator, you may want to double check that your calculator is in radian mode.
So, what is a radian?
A radian is a measure of angle which is defined so that if we have an angle with a size of one radian on a unit circle (with a radius \(r= 1\)), then the length of the arc along the circumference of the circle is also equal to one, as we see in Figure 2. Because the circumference of a circle is \(2 \pi r\text{,}\) this means that for one complete circle,
\begin{equation*} 360^\circ = 2 \pi \text{ radians. } \end{equation*}Similarly half a circle is
\begin{equation*} 180^\circ = \pi \text{ radians } \end{equation*}and a right angle is
\begin{equation*} 90^\circ = \frac{\pi}{2} \text{ radians. } \end{equation*}So one radian is
\begin{equation*} 57.3^\circ \approx \frac{180^\circ}{\pi} = 1 \text{ radian } . \end{equation*}Because we define the radian in this way, this means that the arc length \(s\) along the circumference of a circle with radius \(r\) over angle \(\theta\) can be calculated as \(s = r \theta\) as long as the angle \(\theta\) is measured in radians.
Let's draw a unit circle with its center at the origin and think about a point moving along the circumference of this circle. We start with the point on the \(x\) axis with coordinates (1,0), as shown in the figure below, and define this location to be an angle of \(\theta = 0\) radians. Then, we let this point move up, so that our point is at an angle \(\theta\) above the \(x\) axis. The sine and cosine functions are defined so that they give us the coordinates of our point:
\begin{equation*} x = \cos \theta \quad \text{ and } \quad y = \sin \theta \end{equation*}This means that an angle of \(\theta = 2 \pi\) carries us around one full circle and brings us back to our starting point on the \(x\) axis again, with coordinates (1,0). That means that \(\sin 2 \pi = \sin 0 = 0\text{.}\) Similarly \(\theta = 4 \pi\) carries us around two full circles, and \(\theta = 6 \pi\) carries us around three full circles.
Next we can use the Pythagorean Theorem, and remember that our hypotenuse is equal to one to see that
\begin{equation*} \sin^2 \theta + \cos^2 \theta = 1 \end{equation*}This is a very useful relationship!
In this activity we will review the trigonometry of the special angles \(0^\circ\), \(30^\circ\), \(45^\circ\), and their multiples.
Use the fact that \(180^\circ\) is the same as \(\pi\) radians, convert each of the following angle measurements to radians.
| Degrees | \(0^\circ\) | \(30^\circ\) | \(45^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(135^\circ\) | \(150^\circ\) | \(180^\circ\) |
| Radians | \(0\) | \(\pi\) | |||||||
| Degrees | \(210^\circ\) | \(225^\circ\) | \(240^\circ\) | \(270^\circ\) | \(300^\circ\) | \(315^\circ\) | \(330^\circ\) | \(360^\circ\) | |
| Radians | |||||||||
In part (a) of this problem there are several patterns that can help in remembering the radian conversions for certain angles. For example, you should have found that \(30^\circ\) converts to \(\frac{\pi}{6}\) radians. Therefore, \(60^\circ\) should be twice \(\frac{\pi}{6}\) which indeed it is: \(60^\circ = \frac{\pi}{3}\) radians. What other similar patterns can you find? What is the minimum number of radian measures that you need to memorize?
The sides of a \(30-60-90\) triangle follow well-known ratios. Consider the equilateral triangle on the left of the figure below. Fill in the rest of the sides and angles on the figure and use them to determine the trigonometric values of \(30^\circ\) and \(60^\circ\).
| Angle (Degrees) | Angle (Radians) | Sine | Cosine | Tangent |
| \(30^{\circ}\) | ||||
| \(60^{\circ}\) |
The sides of a \(45-45-90\) triangle also follow well-known ratios. Consider the isosceles triangle on the right of the figure below. Fill in the rest of the sides and angles on the figure and use them to determine the trigonometric values of \(45^\circ\).
| Angle (Degrees) | Angle (Radians) | Sine | Cosine | Tangent |
| \(45^{\circ}\) |
Finally, we can organize all of the information about the special right
triangles on a well-known organizational tool: the unit circle. The \(x\)
coordinate of each point is the cosine of the angle and the \(y\) coordinate of
each point is the sine of the angle.
To get beyond trigonometry, rather than using the angle \(\theta\) as the input to our sine and cosine functions, instead we will put our function input on the \(x\) axis. Then we can plot the output of on the \(y\) axis. This produces the following graph:
Here we can see that the range (output) of the sine function is the interval from \(-1\) to \(+1\text{.}\) (\(\sin x\) can never equal 2!) The domain (input) of the sine function is extends from \(-\infty\) to \(+\infty\text{,}\) but the cycle repeats every \(2 \pi\) along the \(x\) axis. (Do you understand how the definition of sine on the unit circle makes both of these facts true?)
The following terms will be very important when we describe functions like this:
The period of a trig function is how far along the \(x\) axis it takes to complete one full cycle. The units of the period are seconds per cycle.
The frequency of a trig function is the number of complete cycles completed in 1 unit. The units of the frequency are cycles per second.
The amplitude of a trig function is how far it goes on the \(y\) axis above and below its average value.
The function \(f(x) = \sin x\) has a period of \(2 \pi\) and an amplitude of \(1\text{.}\) If we plot both the sine and the cosine functions together we see the following graph: From this we see that the function \(g(x) = \cos x\) also has a period of \(2 \pi\) and an amplitude of \(1\text{.}\) The difference is that \(\sin 0 = 0\text{,}\) so the sine function starts at its average value, halfway between a peak and a trough. On the other hand \(\cos 0 = 1\text{,}\) so the cosine function starts at a peak. This means that we can turn a cosine function into a sine function and visa versa simply by shifting:
\begin{equation*} \sin(x) = \cos \left( x - \frac{\pi}{2} \right) \text{ and } \cos(x) = \sin \left( x + \frac{\pi}{2} \right). \end{equation*}To model real data with the sine or cosine function, we must be able to change the amplitude, the period, and the average value of our wave, to get what we call a sinusoidal function. Every sinusoidal function can be written either of the two following forms. Note well that the sine function could be replaced by the cosine function.
\(|A|\) is the amplitude (the distance from the centerline of the wave to the maximum or minimum).
\(B\) is the angular frequency, which determines the period, with \(B = \frac{2 \pi}{\mbox{Period} }\text{.}\)
\(C\) is the average value.
\(t_0\) is the shift along the \(t\) axis, a time when \(f\) is at an average value and increasing
\(\phi\) is the shift in radians, the angle at which the oscillations begin
The parameter \(B\) can be a little surprising. Because \(B\) is inversely related to the period, this means that larger values of \(B\) result in a shorter period, and smaller values of \(B\) result in a longer period, as we see in the graphs below:
When modeling with sinusoidal functions it is sometimes more convenient to use sine versus cosine or visa versa.Suppose we measure the temperature every hour throughout a day and find that \(T\) varies in a smooth sinusoidal pattern. We find that the average temperature is \(60^\circ\text{,}\) the amplitude is \(20^\circ\text{,}\) and the period is 24 hours. The minimum temperature is at 4am, the maximum temperature is at 4pm, and so it is at the average temperature and increasing at 10am. How would we model the temperature as a sinusoidal function?
The figure below gives us the voltage produced by an electrical circuit as a function of time.
What is the amplitude of the oscillations?
What is the period of the oscillations?
What is the average value of the voltage?
What is the shift along the \(t\) axis, \(t_0\)?
What is a formula for this function?
Suppose the following sinusoidal function models the water level on a pier in the ocean as it changes due to the tides during a certain day.
\begin{equation*} w(t) = 4.3 \sin \left( 0.51 t + 0.82 \right) + 10.6 \end{equation*}Using the formula above, make a table showing the water level every two hours for a 24 hour period starting at midnight.
| time (hours) | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 | 20 | 22 | 24 |
| water level (ft) | |||||||||||||
Sketch a graph of this function using the data from your table in part (a).
What is the period of oscillation of this function?
What time is high tide?
The tangent function has a completely different shape than the sine and cosine functions because it is defined to be
\begin{equation*} \tan \theta = \frac{\sin \theta}{\cos \theta} \end{equation*}as long as \(\cos \theta \neq 0\text{,}\) so that we don't have a divide-by-zero problem. This means that the tangent function is undefined at \(\pm \pi/2, \pm 3 \pi/2, \cdots\text{,}\) and the function has vertical asymptotes at these points. Because the tangent function blows up to infinity, we don't often fit tangent functions to real world data.
Suppose we want to find the value of \(x\) so that \(\sin x = 0.75\text{.}\) To find this, we use an inverse sine function, written as either \(\arcsin\) or \(\sin^{-1}\text{.}\) In this case \(\sin^{-1} 0.75 \approx 0.848\text{,}\) because \(\sin 0.848 \approx 0.75\text{.}\) The range of the sine function only goes from \(-1\) to \(+1\text{,}\) so that means the domain of \(\arcsin\) only goes from \(-1\) to \(+1\text{.}\) It is impossible to find a solution to \(\sin x = 3\text{,}\) because the sine function doesn't go that high!
It is important to remember that the sine function repeats itself in the same pattern again and again, every \(2\pi\text{,}\) so \(x = 0.848\) is not the only solution to \(\sin x = 0.75\text{.}\) Another solution is \(x = 0.848 + 2 \pi \approx 7.13\text{.}\) Another solution is \(x = 0.848 + 4 \pi \approx 13.4.\text{.}\) And of course we could subtract multiples of \(2 \pi\) as well to get \(x = 0.848 - 2 \pi \approx -5.44\text{.}\) As a result, we have the \(\arcsin\) function output the value between \(-\pi/2\) and \(+\pi/2\text{.}\) This means that the arcsine function has a very limited domain and range, only existing for \(-1 \leq x \leq 1\) and \(-\pi/2 \leq y \leq \pi/2\text{.}\)
Warning: Even though we write the inverse sine function as \(\sin^{-1} x\text{,}\) it is a completely different thing than \(1 / \sin x\text{.}\)
\begin{equation*} \sin^{-1}x \neq \frac{1}{\sin x} \end{equation*}We can define a similar inverse function for the cosine, which we call \(\arccos\) or \(\cos^{-1}\text{.}\) The domain of this function is \(-1 \leq x \leq 1\text{,}\) and we choose the range to be \(0 \leq y \leq \pi\text{.}\)
Recall that the tangent function has a range going all the way from negative infinity to positive infinity, as \(x\) goes from \(-\pi/2\) to \(+\pi/2\text{.}\) As a result, the inverse tangent function has a domain of \(-\infty \lt x \lt \infty\text{,}\) and a range of \(-\pi/2 \leq y \leq \pi/2\text{.}\) The three primary inverse trigonometric functions are shown in Figure 7.
Trigonometric functions are utilized to model periodic behavior such as tides, sound waves, or voltage through an electrical circuit.
Converting between radian measure and degree measure can be achieved by remembering that
\begin{equation*} 2\pi \text{ radians } = 360^\circ \end{equation*}for \(f(t) = A \sin( B ( t - t_0) ) + C\) or \(f(t) = A \sin( B t + \phi) + C\)
\(A\) is the amplitude.
\(B\) is the angular frequency, which determines the period, with \(B = \frac{2 \pi}{\mbox{Period} }\text{.}\)
\(C\) is the average value.
\(t_0\) is the shift along the \(t\) axis, a time when \(f\) is at an average value and increasing
\(\phi\) is the shift in radians, the angle at which the oscillations begin
What is the formula for a sinusoidal function that has a minimum at coordinates \((3,6)\) followed by a maximum at \((5,9)\text{?}\)
For each of the following trigonometric functions find the amplitude, period, and phase shift.
\begin{align*} a(x) \amp = -\sqrt{19} \sin\left( 7 \pi x + 18 \right)\\ b(x) \amp = -15 \cos\left( 5\left( x+\frac{\pi}{2} \right) \right)\\ c(x) \amp = -2 \cos\left( \frac{x}{7} \right)\\ d(x) \amp = 5 \cos\left( 5x-\frac{\pi}{8} \right) \end{align*}Write the equation of the trigonometric function shown in each plot below.
The number of hours of daylight varies sinusoidally throughout the year. The maximum occurs on the summer solstice, June 21, when we have 15 hours and 50 minutes of daylight. The minimum occurs on the winter solstice, December 21, when we have only 8 hours and 33 minutes of daylight. Find the formula for a function to describe this. The input to your function should be \(d\text{,}\) the number of days since the beginning of the year, so that \(d = 5\) on January 5. The output of your function should be the amount of daylight, in minutes. Assume that this is not a leap year. Hint: Because we know the date of maximum, it is easier to write this in terms of a cosine function.
