7.3 Graphs of Sine and Cosine

Vanessa Coffelt

7.3 Graphs of Sine and Cosine

In Section 7.2, we discussed how to interpret the sine and cosine of real numbers. To review, we identify a real number $t$ with an oriented angle $\theta$ measuring $t$ radians^[1] and define $\sin(t) = \sin(\theta)$ and $\cos(t) = \cos(\theta)$ . Every real number can be identified with one and only one angle $\theta$ this way, therefore the domains of the functions $f(t) = \sin(t)$ and $g(t) = \cos(t)$ are all real numbers, $(-\infty, \infty)$ .

When it comes to range, recall that the sine and cosine of angles are coordinates of points on the Unit Circle and hence, each fall between $-1$ and $1$ inclusive. As the real number line,^[2] when wrapped around the Unit Circle completely covers the circle, we can be assured that every point on the Unit Circle corresponds to at least one real number. Putting these two facts together, we conclude the range of $f(t) = \sin(t)$ and $g(t) = \cos(t)$ are both $[-1,1]$ . We summarize these two important facts below.

Theorem 7.5 Domain and Range of the Cosine and Sine Functions

The function
- has a domain of $(-\infty, \infty)$
- has a range of $[-1,1]$
The function
- has a domain of $(-\infty, \infty)$
- has a range of $[-1,1]$

Our aim in this section is to become familiar with the graphs of $f(t) = \sin(t)$ and $g(t) = \cos(t)$ . To that end, we begin by making a table and plotting points. We’ll start by graphing $f(t) = \sin(t)$ by making a table of values and plotting the corresponding points. We’ll keep the independent variable ` $t$ ‘ for now and use the default ` $y$ ‘ as our dependent variable.^[3] Note in the graph below, on the right, the scale of the horizontal and vertical axis is far from 1:1. (We will present a more accurately scaled graph shortly.)

$\begin{array}{ccc}\begin{array}{|r||r|r|} \hline t & \sin(t) & (t,\sin(t)) \\ \hline 0 & 0 & (0, 0) \\ [1pt] \hline \frac{\pi}{4} & \frac{\sqrt{2}}{2} & \left(\frac{\pi}{4}, \frac{\sqrt{2}}{2}\right) \\ [1pt] \hline \frac{\pi}{2} & 1 & \left(\frac{\pi}{2}, 1\right) \\ [1pt] \hline \frac{3\pi}{4} & \frac{\sqrt{2}}{2} & \left(\frac{3\pi}{4}, \frac{\sqrt{2}}{2}\right) \\ [1pt] \hline \pi & 0 & (\pi, 0) \\ [1pt] \hline \end{array} & \qquad \qquad & \begin{array}{|r||r|r|} \hline t & \sin(t) & (t,\sin(t)) \\ \hline \frac{5\pi}{4} & -\frac{\sqrt{2}}{2} & \left(\frac{5\pi}{4}, -\frac{\sqrt{2}}{2}\right) \\ [1pt] \hline \frac{3\pi}{2} & -1 & \left(\frac{3\pi}{2}, -1 \right) \\ [1pt] \hline \frac{7\pi}{4} & -\frac{\sqrt{2}}{2} & \left(\frac{7\pi}{4}, -\frac{\sqrt{2}}{2}\right) \\ [1pt] \hline 2\pi & 0 & (2\pi, 0) \\ [1pt] \hline \end{array} \\ \end{array}$

The graph of y equals sine of t. The graph has a horizontal axis which counted by pi/4 and goes from 0 to 2 pi. The y axis counts by 1 and goes from -1 to 1. The curve starts at (0,0) and then increases to 1 at pi over 2, back to 0 at pi, down to -1 at 3 pi/2 and back up to 0 at 2 pi. The curve is smooth and continuous. — One Cycle of the Sine Function

If we plot additional points, we soon find that the graph repeats itself. This shouldn’t come as too much of a surprise considering Theorem 7.3. In fact, in light of that theorem, we expect the function to repeat itself every $2\pi$ units. Below is a more accurately scaled graph highlighting the portion we had already graphed above. The graph is often described as having a `wavelike’ nature and is sometimes called a sine wave or, more technically, a sinusoid.

A sine wave on a cartesian plane. The graph extends in both as t approaches negative infinity and as t approaches positive infinity. The graph is a wave that slowly bounces between negative one and positive one. The portion of the graph from the origin to when t is 2 pi is highlighted. — A More Accurately Scaled Graph of f(t) = sin (t)

Note that by copying the highlighted portion of the graph and pasting it end-to-end, we obtain the entire graph of $f(t) = \sin(t)$ . We give this `repeating’ property a name.

Definition 7.4 Periodic Functions

A function $f$ is said to be periodic if there is a real number $c$ so that $f(t+c) = f(t)$ for all real numbers $t$ in the domain of $f$ . The smallest positive number $p$ for which $f(t+p) = f(t)$ for all real numbers $t$ in the domain of $f$ , if it exists, is called the period of $f$ .

We have already seen a family of periodic functions in Section 1.3.1: the constant functions. However, despite being periodic a constant function has no period. (We’ll leave that odd gem as an exercise for you.)

Returning to $f(t) = \sin(t)$ , we see that by Definition 7.4, $f$ is periodic as $\sin(t + 2\pi) = \sin(t)$ . To determine the period of $f$ , we need to find the smallest real number $p$ so that $f(t+p) = f(t)$ for all real numbers $t$ or, said differently, the smallest positive real number $p$ such that $\sin(t+p) = \sin(t)$ for all real numbers $t$ .

We know that $\sin(t + 2\pi) = \sin(t)$ for all real numbers $t$ but the question remains if any smaller real number will do the trick. Suppose $p>0$ and $\sin(t + p) = \sin(t)$ for all real numbers $t$ . Then, in particular, $\sin(0+p) = \sin(0)=0$ so that $\sin(p) = 0$ . From this we know $p$ is a multiple of $\pi$ . Given $\sin\left(\frac{\pi}{2} \right) \neq \sin\left(\frac{\pi}{2} + \pi \right)$ , we know $p \neq \pi$ . Hence, $p = 2\pi$ so the period of $f(t) = \sin(t)$ is $2\pi$ .

Having period $2\pi$ essentially means that we can completely understand everything about the function $f(t) = \sin(t)$ by studying one interval of length $2\pi$ , say $[0,2\pi]$ .^[4] For this reason, when graphing sine (and cosine) functions, we typically restrict our attention to graphing these functions over the course of one period to produce one cycle of the graph.

Not surprisingly, the graph of $g(t) = \cos(t)$ exhibits similar behavior as $f(t) = \sin(t)$ .^[5]

$\begin{array}{|r||r|r|} \hline t & \cos(t) & (t,\cos(t)) \\ \hline 0 & 1 & (0, 1) \\ [2pt] \hline \frac{\pi}{4} & \frac{\sqrt{2}}{2} & \left(\frac{\pi}{4}, \frac{\sqrt{2}}{2}\right) \\ [2pt] \hline \frac{\pi}{2} & 0 & \left(\frac{\pi}{2}, 0\right) \\ [2pt] \hline \frac{3\pi}{4} & -\frac{\sqrt{2}}{2} & \left(\frac{3\pi}{4}, -\frac{\sqrt{2}}{2}\right) \\ [2pt] \hline \pi & -1 & (\pi, -1) \\ [2pt] \hline \frac{5\pi}{4} & -\frac{\sqrt{2}}{2} & \left(\frac{5\pi}{4}, -\frac{\sqrt{2}}{2}\right) \\ [2pt] \hline \frac{3\pi}{2} & 0 & \left(\frac{3\pi}{2}, 0 \right) \\ [2pt] \hline \frac{7\pi}{4} & \frac{\sqrt{2}}{2} & \left(\frac{7\pi}{4}, \frac{\sqrt{2}}{2}\right) \\ [2pt] \hline 2\pi & 1 & (2\pi, 1) \\ [2pt] \hline \end{array}$

The graph of y equals cosine of t. The graph has a horizontal axis which counted by pi/4 and goes from 0 to 2 pi. The y axis counts by 1 and goes from -1 to 1. The curve starts at (0,1) and then decreases to 0 at pi over 2, down to -1 at pi, and back up to 0 at 3 pi/2 and returns to 1 at 2 pi. The curve is smooth and continuous. — One Cycle of the Cosine Function

Like $f(t)=\sin(t)$ , $g(t) = \cos(t)$ is a wavelike curve with period $2\pi$ . Moreover, the graphs of the sine and cosine functions have the same shape – differing only in what appears to be a horizontal shift. As we’ll prove in Section 8.2, $\sin\left(t + \frac{\pi}{2}\right) = \cos(t)$ , which means we can obtain the graph of $y=\cos(t)$ by shifting the graph of $y=\sin(t)$ to the left $\frac{\pi}{2}$ units.^[6]

A cosine wave on a cartesian plane. The graph extends in both as t approaches negative infinity and as t approaches positive infinity. The graph is a wave that slowly bounces between negative one and positive one. The portion of the graph from the t equals 0 to 2 pi is highlighted. — A More Accurately Scaled Graph of f(t) = cos (t)

While arguably the most important property shared by $f(t) = \sin(t)$ and $g(t) = \cos(t)$ is their periodic `wavelike’ nature,^[7] their graphs suggest these functions are both continuous and smooth. Recall from Section 2.2 that, like polynomial functions, the graphs of the sine and cosine functions have no jumps, gaps, holes in the graph, vertical asymptotes, corners or cusps.

Moreover, the graphs of both $f(t) = \sin(t)$ and $g(t) = \cos(t)$ meander and never `settle down’ as $t \rightarrow \pm \infty$ to any one real number. So even though these functions are `trapped’ (or bounded) between $-1$ and $1$ , neither graph has any horizontal asymptotes.

Lastly, the graphs of $f(t) = \sin(t)$ and $g(t) = \cos(t)$ suggest each enjoy one of the symmetries introduced in Section 2.2. The graph of $y = \sin(t)$ appears to be symmetric about the origin while the graph of $y = \cos(t)$ appears to be symmetric about the $y$ -axis. Indeed, as we’ll prove in Section 8.2, $f(t) = \sin(t)$ is, in fact, an odd function: ^[8] that is, $\sin(-t) = -\sin(t)$ and $g(t) = \cos(t)$ is an even function, so $\cos(-t) = \cos(t)$ .

We summarize all of these properties in the following result.

Theorem 7.6 Properties of the Sine and Cosine Functions

The function
- has domain $(-\infty, \infty)$
- has range $[-1,1]$
- is continuous and smooth
- is odd
- has period $2 \pi$
The function
- has domain $(-\infty, \infty)$
- has range $[-1,1]$
- is continuous and smooth
- is even
- has period $2\pi$
Conversion formulas: $\sin\left(t + \frac{\pi}{2}\right) = \cos(t)$ and $\cos\left(t - \frac{\pi}{2} \right) = \sin(t)$

Now that we know the basic shapes of the graphs of $y = \sin(t)$ and $y = \cos(t)$ , we can use the results of Section 1.6 to graph more complicated functions using transformations. As mentioned already, the fact that both of these functions are periodic means we only have to know what happens over the course of one period of the function in order to determine what happens to all points on the graph. To that end, we graph the `fundamental cycle‘, the portion of each graph generated over the interval $[0, 2\pi]$ , for each of the sine and cosine functions below.

Two graphs side by side. The first is the fundamental cycle for the sine function and the second is the fundamental cycle of the cosine function. The main points at t equals 0, pi over 2, pi, three pi over 2 and 2 pi. The first graph is labeled "The fundamental cycle of y = sin (t)." The second is labeled "The fundamental cycle of y = cos (t)." — Graphs of the Fundamental Cycles of Sine and Cosine Functions

In working through Section 1.6, it was very helpful to track `key points’ through the transformations. The `key points’ we’ve indicated on the graphs above correspond to the quadrantal angles and generate the zeros and the extrema of functions. Due to the fact that the quadrantal angles divide the interval $[0,2\pi]$ into four equal pieces, we shall refer to these angles henceforth as the `quarter marks.’

It is worth noting that because the transformations discussed in Section 1.6 are linear,^[9] the relative spacing of the points before and after the transformations remains the same.^[10] In particular, wherever the interval $[0, 2\pi]$ is mapped, the quarter marks of the new interval correspond to the quarter marks of $[0, 2\pi]$ . (Can you see why?) We will exploit this fact in the following example.

Example 7.3.1

Example 7.3.1.1

Graph one cycle of the following functions. State the period of each.

$f(t) = 3 \sin(2t)$

Solution:

Graph one cycle of $f(t) = 3 \sin(2t)$ .

One way to proceed is to use Theorem 1.12 and follow the procedure outlined there.

Starting with the fundamental cycle of $y= \sin(t)$ , we divide each $t$ -coordinate by $2$ and multiply each $y$ -coordinate by 3 to obtain one cycle of $y = 3 \sin(2t)$ .

Rendered by QuickLaTeX.com

As a result of one cycle of $y=f(t)$ being completed over the interval $[0, \pi]$ , the period of $f$ is $\pi$ .

Example 7.3.1.2

Graph one cycle of the following functions. State the period of each.

$g(t) = 2 \cos\left(t +\frac{\pi}{2} \right) +1$

Solution:

Graph one cycle of $g(t) = 2 \cos\left(t +\frac{\pi}{2} \right) +1$ .

Starting with the fundamental cycle of $y = \cos(t)$ and using Theorem 1.12, we subtract $\frac{\pi}{2}$ from each of the $t$ -coordinates, then multiply each $y$ -coordinate by $2$ , and add $1$ to each $y$ -coordinate.

We find one cycle of $y=g(t)$ is completed over the interval $\left[ -\frac{\pi}{2}, \frac{3\pi}{2} \right]$ , the period is $\frac{3\pi}{2} - \left(- \frac{\pi}{2} \right) = 2\pi$ .

Rendered by QuickLaTeX.com

As previously mentioned, the curves graphed in Example 7.3.1 are examples of sinusoids. A sinusoid is the result of taking the graph of $y = \sin(t)$ or $y = \cos(t)$ and performing any of the transformations mentioned in Section 1.6. We graph one cycle of a generic sinusoid below. Sinusoids can be characterized by four properties: period, phase shift, vertical shift (or `baseline’), and amplitude.

A generic sinusoid graph without a cartesian plane. The graph contains a dashed line from the starting point and ending point of the one cycle. The dashed line is labeled "baseline". There is a vertical arrowed line between the dashed baseline and the high point of the sinusoid. The arrowed line is labeled amplitude. Below the graph marking the distance between the start and stop of the one cycle is a line labeled period. — Characteristics of a Sinusoid

We have already discussed the period of a sinusoid. If we think of $t$ as measuring time, the period is how long it takes for the sinusoid to complete one cycle and is usually represented by the letter $T$ . The standard period of both $\sin(t)$ and $\cos(t)$ is $2\pi$ , but horizontal scalings will change this.

In Example 7.3.1, for instance, the function $f(t) = 3 \sin(2t)$ has period $\pi$ instead of $2\pi$ because the graph is horizontally compressed by a factor of $2$ as compared to the graph of $y = \sin(t)$ . However, the period of $g(t) = 2 \cos\left(t +\frac{\pi}{2} \right) +1$ is the same as the period of $\cos(t)$ , $2\pi$ , because there are no horizontal scalings.

The phase shift of the sinusoid is the horizontal shift. Again, thinking of $t$ as time, the phase shift of a sinusoid can be thought of as when the sinusoid `starts’ as compared to $t=0$ . Assuming there are no reflections across the $y$ -axis, we can determine the phase shift of a sinusoid by finding where the value $t=0$ on the graph of $y = \sin(t)$ or $y=\cos(t)$ is mapped to under the transformations.

For $f(t) = 3 \sin(2t)$ , the phase shift is ` $0$ ‘ because the value $t=0$ on the graph of $y = \sin(t)$ remains stationary under the transformations. Loosely speaking, this means both $y=\sin(t)$ and $y=3\sin(2t)$ `start’ at the same time. The phase shift of $g(t) = 2 \cos\left(t +\frac{\pi}{2} \right) +1$ is $-\frac{\pi}{2}$ or ` $\frac{\pi}{2}$ to the left‘ because the value $t = 0$ on the graph of $y=\cos(t)$ is mapped to $t = -\frac{\pi}{2}$ on the graph of $y= 2 \cos\left(t +\frac{\pi}{2} \right) +1$ . Again, loosely speaking, this means $y=2 \cos\left(t +\frac{\pi}{2} \right) +1$ starts $\frac{\pi}{2}$ time units earlier than $y=\cos(t).$

The vertical shift of a sinusoid is exactly the same as the vertical shifts in Section 1.6 and determines the new `baseline’ of the sinusoid. Thanks to symmetry, the vertical shift can always be found by averaging the maximum and minimum values of the sinusoid. For $f(t) = 3 \sin(2t)$ , the vertical shift is $0$ whereas the vertical shift of $g(t) = 2 \cos\left(t +\frac{\pi}{2} \right) +1$ is $1$ or ` $1$ up.’

The amplitude of the sinusoid is a measure of how `tall’ the wave is, as indicated in the figure below. Said differently, the amplitude measures how much the curve gets displaced from its `baseline. ‘ The amplitude of the standard cosine and sine functions is $1$ , but vertical scalings can alter this.

In Example 7.3.1, the amplitude of $f(t) = 3 \sin(2t)$ is $3$ , owing to the vertical stretch by a factor of $3$ as compared with the graph of $y = \sin(t)$ . In the case of $g(t) = 2 \cos\left(t +\frac{\pi}{2} \right) +1$ , the amplitude is $2$ due to its vertical stretch as compared with the graph of $y = \cos(t)$ . Note that the ` $+1$ ‘ here does not affect the amplitude of the curve; it merely changes the `baseline’ from $y=0$ to $y=1$ .

The following theorem shows how these four fundamental quantities relate to the parameters which describe a generic sinusoid. The proof follows from Theorem 1.12 and is left to the reader in Exercise 31.

Theorem 7.7

For $B > 0$ , the graphs of

$S(t) = A \sin(B t + C) + D \quad \text{and} \quad E(t) = A \cos(B t + C) + D$

have frequency of $B$
have period $T = \dfrac{2\pi}{B}$
have amplitude $|A|$
have phase shift $-\dfrac{C}{B}$
have vertical shift or `baseline’ $D$

We put Theorem 7.7 to good use in the next example.

Example 7.3.2

Example 7.3.2.1

Use Theorem 7.7 to determine the frequency, period, phase shift, amplitude, and vertical shift of each of the following functions and use this information to graph one cycle of each function.

$f(t) = 3 \cos\left(\frac{\pi t - \pi}{2}\right) + 1$

Solution:

Determine the frequency, period, phase shift, amplitude, and vertical shift of $f(t) = 3 \cos\left(\frac{\pi t - \pi}{2}\right) + 1$ and then graph on cycle of the function.

To use Theorem 7.7, we first need to rewrite $f(t)$ in the form prescribed by Theorem 7.7. To that end, we rewrite: $f(t) = 3 \cos\left(\frac{\pi t - \pi}{2}\right) + 1 = 3\cos\left(\frac{\pi}{2} t + \left(-\frac{\pi}{2}\right)\right) + 1$ .

From this, we identify $A = 3$ , $B = \frac{\pi}{2}$ , $C = -\frac{\pi}{2}$ and $D = 1$ .

According to Theorem 7.7, the frequency is $B = \frac{\pi}{2}$ , the period is $T=\frac{2\pi}{B} = \frac{2\pi}{\pi/2} = 4$ , the phase shift is $-\frac{C}{B} = -\frac{-\pi/2}{\pi/2} = 1$ (indicating a shift to the right $1$ unit), the amplitude is $|A| = |3| = 3$ , and the vertical shift is $D = 1$ (indicating a shift up $1$ unit.)

To graph $y = f(t)$ , we know one cycle begins at $t=1$ (the phase shift.) Because the period is $4$ , we know the cycle ends $4$ units later at $t=1+4 = 5$ . If we divide the interval $[1,5]$ into four equal pieces, each piece has length $\frac{4}{4} = 1$ . Hence, we to get our quarter marks, we start with $t=1$ and add $1$ unit until we reach the endpoint, $t=5$ . Our new quarter marks are: $t=1$ , $t=2$ , $t=3$ , $t=4$ , and $t=5.$

We now substitute these new quarter marks into $f(t)$ to obtain the corresponding $y$ -values on the graph.^[11] We connect the dots in a `wavelike’ fashion to produce the graph on the right.

Note that we can (partially) spot-check our answer by noting the average of the maximum and minimum is $\frac{4+(-2)}{2} = 1$ (our vertical shift) and the amplitude, $4 - 1 = 1 - (-2) = 3$ is indeed $3$ .

$\begin{array}{|r||r|r|} \hline t & f(t) & (t,f(t)) \\ [2pt] \hline 1 & 4 & (1,4) \\ [2pt] \hline 2 & 1 & (2,1) \\ [2pt] \hline 3 & -2 & (3,-2) \\ [2pt] \hline 4 & 1 & (4,1) \\ [2pt] \hline 5 & 4 & (5,4) \\ [2pt] \hline \end{array}$

Rendered by QuickLaTeX.com

Thought not asked for, this example provides a nice opportunity to interpret the ordinary frequency: $f = \frac{1}{T} = \frac{1}{4}$ . Hence, $\frac{1}{4}$ of the sinusoid is traced out over an interval that is $1$ unit long.

Example 7.3.2.2

Use Theorem 7.7 to determine the frequency, period, phase shift, amplitude, and vertical shift of each of the following functions and use this information to graph one cycle of each function.

$g(t) = \frac{1}{2} \sin(\pi - 2t) + \frac{3}{2}$

Solution:

Determine the frequency, period, phase shift, amplitude, and vertical shift of $g(t) = \frac{1}{2} \sin(\pi - 2t) + \frac{3}{2}$ and then graph on cycle of the function.

Turning our attention now to the function $g$ , we first note that the coefficient of $t$ is negative. In order to use Theorem 7.7, we need that coefficient to be positive. Hence, we first use the odd property of the sine function to rewrite $\sin(\pi-2t)$ so that instead of a coefficient of $-2$ , $t$ has a coefficient of $2$ . We get

$\begin{array}{rcl} \sin(\pi-2t) &=& \sin(-2t+\pi) \\ &=& \sin(- (2t-\pi)) \\ &=& -\sin(2t-\pi) \end{array}$

Hence, $g(t) = -\frac{1}{2} \sin(2t + (-\pi)) + \frac{3}{2}.$

We identify $A = -\frac{1}{2}$ , $B = 2$ , $C = -\pi$ and $D = \frac{3}{2}$ .

The frequency is $B = 2$ , the period is $T=\frac{2\pi}{2} = \pi$ , the phase shift is $-\frac{-\pi}{2} = \frac{\pi}{2}$ (indicating a shift right $\frac{\pi}{2}$ units), the amplitude is $\left| - \frac{1}{2} \right| = \frac{1}{2}$ , and, finally, the vertical shift is up $\frac{3}{2}.$

Proceeding as before, we know one cycle of $g$ starts at $t = \frac{\pi}{2}$ and ends at $t = \frac{\pi}{2} + \pi = \frac{3\pi}{2}$ . Dividing the interval $\left[ \frac{\pi}{2}, \frac{3 \pi}{2} \right]$ into four equal pieces gives pieces of length $\frac{\pi}{4}$ units. Hence, to obtain our new quarter marks, we start at $t = \frac{\pi}{2}$ and add $\frac{\pi}{4}$ until we reach $t=\frac{3\pi}{2}$ . Our new quarter marks are: $t = \frac{\pi}{2}$ , $t = \frac{3\pi}{4}$ , $t = \pi$ , $t = \frac{5\pi}{4}$ , $t = \frac{3\pi}{2}$ . Substituting these values into $g$ gives us the points to plot to produce the graph below on the right.

Again, we can quickly check the vertical shift by averaging the maximum and minimum values: $\frac{2+1}{2} = \frac{3}{2}$ and verify the amplitude: $2 - \frac{3}{2} = \frac{3}{2} - 1 = \frac{1}{2}.$

$\begin{array}{|r||r|r|} \hline t & g(t) & (t,g(t)) \\ [2pt] \hline \frac{\pi}{2} & \frac{3}{2} & \left(\frac{\pi}{2}, \frac{3}{2}\right) \\ [2pt] \hline \frac{3\pi}{4} & 1 & \left(\frac{3\pi}{4} , 1 \right) \\ [2pt] \hline \pi & \frac{3}{2} & \left(\pi , \frac{3}{2} \right) \\ [2pt] \hline \frac{5\pi}{4} & 2 & \left(\frac{5\pi}{4} , 2 \right) \\ [2pt] \hline \frac{3\pi}{2} & \frac{3}{2} & \left(\frac{3\pi}{2}, \frac{3}{2} \right) \\ [2pt] \hline \end{array}$

Rendered by QuickLaTeX.com

Note that in this section, we have discussed two ways to graph sinusoids: using Theorem 1.12 from Section 1.6 and using Theorem 7.7. Both methods will produce one cycle of the resulting sinusoid, but each method may produce a different cycle of the same sinusoid.

For example, if we graphed the function $g(t) = \frac{1}{2} \sin(\pi - 2t) + \frac{3}{2}$ from Example 7.3.2 using Theorem 1.12, we obtain the following:

There is a a table and then two graphs. The rows of the table start with the headers t, g(t), and (t, g(t)). The table includes angles t of pi over 2, pi over 4, 0, negative pi over 2, and negative pi over 4. For the next column is the values of g(t) at each t. You get three halves, 2, three halves, one, and three halves, respectively. The last column is the ordered pairs. The first graph is one cycle of g(t) using the table values. The second graph is also of g(t), but the graph starts at pi over 2 and ends at three pi over 2. The graph drops to one at three pi over four and then increasing to two at five pi over 4 and then back to three halves at three pi over 2. — Graphs of g(t) using Theorems 1.12 and 7.7

Comparing this result with the one obtained in Example 7.3.2 side by side, we see that one cycle ends right where the other starts. The cause of this discrepancy goes back to using the odd property of sine.

Essentially, the odd property of the sine function converts a reflection across the $y$ -axis into a reflection across the $t$ -axis. (Can you see why?) For this reason, whenever the coefficient of $t$ is negative, Theorems 1.12 and 7.7 will produce different results.

In the Exercises, we assume the problems are worked using Theorem 7.7. If you choose to use Theorems 1.12 instead, your answer may look different than what is provided even though both your answer and the textbook’s answer represent one cycle of the same function.

In the next example, we use Theorem 7.7 to determine the formula of a sinusoid given the graph of one cycle. Note that in some disciplines, sinusoids are written in terms of sines whereas in others, cosines functions are preferred. To cover all bases, we ask for both.

Example 7.3.3

Example 7.3.3.1

Below is the graph of one complete cycle of a sinusoid $y=f(t)$ .

Rendered by QuickLaTeX.com

Write $f(t)$ in the form $E(t) = A \cos(Bt + C) + D$ , for $B > 0$ .

Solution:

Write $f(t)$ in the form $E(t) = A \cos(Bt + C) + D$ , for $B > 0$ .

One cycle is graphed over the interval $[-1,5]$ , thus its period is $T=5-(-1) = 6$ .

According to Theorem 7.7, $6 = T = \frac{2\pi}{B}$ , so that $B = \frac{\pi}{3}$ .

Next, we see that the phase shift is $-1$ , so we have $-\frac{C}{B} = -1$ , or $C= B = \frac{\pi}{3}$ .

To find the baseline, we average the maximum and minimum values: $D = \frac{1}{2}\left[ \frac{5}{2} + \left(-\frac{3}{2}\right)\right] = \frac{1}{2}(1) = \frac{1}{2}$ .

To find the amplitude, we subtract the maximum value from the baseline: $A = \frac{5}{2} - \frac{1}{2} = 2$ .

Putting this altogether, we obtain our final answer is $f(t) = 2 \cos\left(\frac{\pi}{3} t + \frac{\pi}{3} \right) + \frac{1}{2}$ .

Example 7.3.3.2

Below is the graph of one complete cycle of a sinusoid $y=f(t)$ .

Rendered by QuickLaTeX.com

Write $f(t)$ in the form $S(t) = A \sin(Bt + C) + D$ , for $B > 0$ .

Solution:

Write $f(t)$ in the form $S(t) = A \sin(Bt + C) + D$ , for $B > 0$ .

Because we have written $f(t)$ in terms of cosines, we can use the conversion from sine to cosine as listed in Theorem 7.6.

As $\cos(t) = \sin\left(t +\frac{\pi}{2} \right)$ , $\cos\left(\frac{\pi}{3} t + \frac{\pi}{3} \right) = \sin\left( \left[\frac{\pi}{3} t + \frac{\pi}{3}\right] + \frac{\pi}{2} \right)$ , so $\cos\left(\frac{\pi}{3} t + \frac{\pi}{3} \right) = \sin\left(\frac{\pi}{3} t + \frac{5\pi}{6} \right)$ .

Our final answer is $f(t) = 2 \sin\left(\frac{\pi}{3} t + \frac{5\pi}{6} \right) + \frac{1}{2}$ .

However, for the sake of completeness, we provide another solution strategy which enables us to write $f(t)$ in terms of sines without starting with our answer from part 1.

Note that we obtain the period, amplitude, and vertical shift as before: $B = \frac{\pi}{3}$ , $A = 2$ and $B = \frac{1}{2}$ . The trickier part is finding the phase shift.

To that end, we imagine extending the graph of the given sinusoid as in the figure below so that we can identify a cycle beginning at $\left(\frac{7}{2}, \frac{1}{2}\right)$ . Taking the phase shift to be $\frac{7}{2}$ , we get $-\frac{C}{B} = \frac{7}{2}$ , or $C= -\frac{7}{2} B = -\frac{7}{2}\left(\frac{\pi}{3}\right) = -\frac{7\pi}{6}.$

Hence, our answer is $f(t) = 2 \sin\left(\frac{\pi}{3} t - \frac{7\pi}{6}\right) + \frac{1}{2}.$

Rendered by QuickLaTeX.com

Note that each of the answers given in Example 7.3.3 is one choice out of many possible answers. For example, when fitting a sine function to the data, we could have chosen to start at $\left(\frac{1}{2}, \frac{1}{2}\right)$ taking $A = -2$ . In this case, the phase shift is $\frac{1}{2}$ so $C = -\frac{\pi}{6}$ for an answer of $f(t) = -2 \sin\left(\frac{\pi}{3} t - \frac{\pi}{6}\right) + \frac{1}{2}$ . The ultimate check of any solution is to graph the answer and check it matches the given data.

7.3.1 Applications of Sinusoids

In the same way exponential functions can be used to model a wide variety of phenomena in nature,^[12] the sine and cosine functions can be used to model their fair share of natural behaviors. Our first foray into sinusoidal motion revisits circular motion – in particular Equation 7.3.

Example 7.3.4

Recall from Exercise 55 in Section 7.1.2 that The Giant Wheel at Cedar Point is a circle with diameter 128 feet which sits on an 8 foot tall platform making its overall height 136 feet. It completes two revolutions in 2 minutes and 7 seconds. Assuming that the riders are at the edge of the circle, find a sinusoid which describes the height of the passengers above the ground $t$ seconds after they pass the point on the wheel closest to the ground.

Solution:

We sketch the problem situation below and assume a counter-clockwise rotation.^[13]

Rendered by QuickLaTeX.com

We know from the equations given on Circular Motion in Section 7.2.3 that the $y$ -coordinate for counter-clockwise motion on a circle of radius $r$ centered at the origin with constant angular velocity (frequency) $\omega$ is given by $y = r\sin(\omega t)$ . Here, $t=0$ corresponds to the point $(r,0)$ so that $\theta$ , the angle measuring the amount of rotation, is in standard position.

In our case, the diameter of the wheel is 128 feet, so the radius is $r = 64$ feet. Because the wheel completes two revolutions in 2 minutes and 7 seconds (which is $127$ seconds) the period $T = \frac{1}{2} (127) = \frac{127}{2}$ seconds. Hence, the frequency is $B = \frac{2\pi}{T} = \frac{4 \pi}{127}$ radians per second.

Putting these two pieces of information together, we have that $y = 64 \sin\left(\frac{4 \pi}{127} t\right)$ describes the $y$ -coordinate on the Giant Wheel after $t$ seconds, assuming it is centered at $(0,0)$ with $t=0$ corresponding to point $Q$ .

In order to find an expression for $h$ , we take the point $O$ in the figure as the origin. As the base of the Giant Wheel ride is $8$ feet above the ground and the Giant Wheel itself has a radius of $64$ feet, its center is $72$ feet above the ground. To account for this vertical shift upward,^[14] we add $72$ to our formula for $y$ to obtain the new formula $h = y + 72 = 64 \sin\left(\frac{4 \pi}{127} t\right) + 72.$

Next, we need to adjust things so that $t=0$ corresponds to the point $P$ instead of the point $Q$ . This is where the phase comes into play. Geometrically, we need to shift the angle $\theta$ in the figure back $\frac{\pi}{2}$ radians.

From the discussion on circular motion, we know $\theta = \omega t = \frac{4 \pi}{127} t$ , so we (temporarily) write the height in terms of $\theta$ as $h =64 \sin\left(\theta\right) + 72$ . Subtracting $\frac{\pi}{2}$ from $\theta$ gives

$\begin{array}{rcl} h(t) &=& 64 \sin\left(\theta - \frac{\pi}{2}\right) + 72 \\[6pt] &=& 64\sin\left(\frac{4 \pi}{127} t -\frac{\pi}{2} \right) + 72 \end{array}$

We can check the reasonableness of our answer by graphing $y = h(t)$ over the interval $\left[0, \frac{127}{2}\right]$ and visualizing the path of a person on the Big Wheel ride over the course of one rotation.

Rendered by QuickLaTeX.com

A few remarks about Example 7.3.4 are in order. First, note that the amplitude of $64$ in our answer corresponds to the radius of the Giant Wheel. This means that passengers on the Giant Wheel never stray more than $64$ feet vertically from the center of the Wheel, which makes sense. Second, the phase shift of our answer works out to be $\frac{\pi/2}{4\pi/127} = \frac{127}{8} = 15.875$ . This represents the `time delay’ (in seconds) we introduce by starting the motion at the point $P$ as opposed to the point $Q$ . Said differently, passengers which `start’ at $P$ take $15.875$ seconds to `catch up’ to the point $Q.$

7.3.2 Section Exercises

In Exercises 1 – 12, graph one cycle of the given function. State the period, amplitude, phase shift and vertical shift of the function.

$f(t) = 3\sin(t)$
$g(t) = \sin(3t)$
$h(t) = -2\cos(t)$
$f(t) = \cos \left( t - \frac{\pi}{2} \right)$
$g(t) = -\sin \left( t + \frac{\pi}{3} \right)$
$h(t) = \sin(2t - \pi)$
$f(t) = -\frac{1}{3}\cos \left( \frac{1}{2}t + \frac{\pi}{3} \right)$
$g(t) = \cos (3t - 2\pi) + 4$
$h(t) = \sin \left( -t - \frac{\pi}{4} \right) - 2$
$f(t) = \frac{2}{3} \cos \left( \frac{\pi}{2} - 4t \right) + 1$
$g(t) = -\frac{3}{2} \cos \left( 2t + \frac{\pi}{3} \right) - \frac{1}{2}$
$h(t) = 4\sin (-2\pi t + \pi)$

In Exercises 13 – 16, a sinusoid is graphed. Write a formula for the sinusoid in the form $S(t) = A \sin(B t + C) + B$ and $E(t) = A \cos(B t + C) + B$ . Select $B$ so $B > 0$ . Check your answer by graphing.

Use the graph of to graph each of the following functions. State the period of each.
1. $f(t) = | 4 \sin(t)|$
2. $g(t) = \sqrt{4 \sin(t)}$

In Exercises 18 – 23, use a graphing utility to graph each function and discuss the related questions with your classmates.

$f(t) = \cos(3t) + \sin(t)$ . Is this function periodic? If so, what is the period?
$f(t) = \frac{\sin(t)}{t}$ . What appears to be the horizontal asymptote of the graph?
$f(t) = t \sin(t)$ . Graph $y = \pm t$ . What do you notice?
$f(t) = \sin\left(\frac{1}{t}\right)$ . What’s happening as $t \rightarrow 0$ ?
$f(t) = e^{-0.1t} \left( \cos(2t) + \sin(2t)\right)$ . Graph $y = \pm e^{-0.1t}$ on the same set of axes. What do you notice?
$f(t) = e^{-0.1t} \left( \cos(2t) + 2\sin(t)\right)$ . Graph $y = \pm e^{-0.1t}$ on the same set of axes. What do you notice?
Show every constant function $f$ is periodic by explaining why $f(x + 117) = f(x)$ for all real numbers $x$ . Then show that $f$ has no period by showing that you cannot find a smallest number $p$ such that $f(x + p) = f(x)$ for all real numbers $x$ .Said differently, show that $f(x + p) = f(x)$ for all real numbers $x$ for ALL values of $p > 0$ , so no smallest value exists to satisfy the definition of `period’.
The sounds we hear are made up of mechanical waves. The note `A’ above the note `middle C’ is a sound wave with ordinary frequency $f = 440$ Hertz $= 440 \frac{\text{cycles}}{\text{second}}$ . Find a sinusoid which models this note, assuming that the amplitude is $1$ and the phase shift is $0$ .
The voltage $V$ in an alternating current source has amplitude $220 \sqrt{2}$ and ordinary frequency $f = 60$ Hertz. Find a sinusoid which models this voltage. Assume that the phase is $0$ .
The London Eye is a popular tourist attraction in London, England and is one of the largest Ferris Wheels in the world. It has a diameter of 135 meters and makes one revolution (counter-clockwise) every 30 minutes. It is constructed so that the lowest part of the Eye reaches ground level, enabling passengers to simply walk on to, and off of, the ride. Find a sinusoid which models the height $h$ of the passenger above the ground in meters $t$ minutes after they board the Eye at ground level.
In Section 7.2.3, we found the $x$ -coordinate of counter-clockwise motion on a circle of radius $r$ with angular frequency $\omega$ to be $x = r\cos(\omega t)$ , where $t=0$ corresponds to the point $(r,0)$ . Suppose we are in the situation of Exercise 27 above. Find a sinusoid which models the horizontal displacement $x$ of the passenger from the center of the Eye in meters $t$ minutes after they board the Eye. Here we take $x(t) > 0$ to mean the passenger is to the right of the center, while $x(t) < 0$ means the passenger is to the left of the center.
In Exercise 52 in Section 7.1.2, we introduced the yo-yo trick `Around the World’ in which a yo-yo is thrown so it sweeps out a vertical circle. As in that exercise, suppose the yo-yo string is 28 inches and it completes one revolution in 3 seconds. If the closest the yo-yo ever gets to the ground is 2 inches, find a sinusoid which models the height $h$ of the yo-yo above the ground in inches $t$ seconds after it leaves its lowest point.
Consider the pendulum below. Ignoring air resistance, the angular displacement of the pendulum from the vertical position, , can be modeled as a sinusoid.^[15]

The amplitude of the sinusoid is the same as the initial angular displacement, $\theta_{0}$ , of the pendulum and the period of the motion is given by

$T = 2\pi \sqrt{\dfrac{\ell}{g}}$

where $\ell$ is the length of the pendulum and $g$ is the acceleration due to gravity.
1. Find a sinusoid which gives the angular displacement $\theta$ as a function of time, $t$ . Arrange things so $\theta(0) = \theta_{0}.$
2. In Exercise 24 Section 4.1, you found the length of the pendulum needed in Jeff’s antique Seth-Thomas clock to ensure the period of the pendulum is $\frac{1}{2}$ of a second. Assuming the initial displacement of the pendulum is $15^{\circ}$ , find a sinusoid which models the displacement of the pendulum $\theta$ as a function of time, $t$ , in seconds.
Use Theorem 1.12 to prove Theorem 7.7.

Section 7.3 Exercise Answers can be found in the Appendix … Coming soon

which, you'll recall essentially `wraps the real number line around the Unit Circle ↵
in particular the interval $[0, 2\pi)$ ↵
Keep in mind that we're using ` $y$ ' here to denote the output from the sine function. It is a coincidence that the $y$ -values on the graph of $y=\sin(t)$ correspond to the $y$ -values on the Unit Circle. ↵
Technically, we should study the interval $[0,2\pi)$ , (In some advanced texts, the interval of choice is $[-\pi, \pi)$ ) as whatever happens at $t=2\pi$ is the same as what happens at $t=0$ . As we will see shortly, $t=2\pi$ gives us an extra `check' when we go to graph these functions. ↵
Here note that the dependent variable ` $y$ ' represents the outputs from $g(t) = \cos(t)$ which are $x$ -coordinates on the Unit Circle. ↵
Hence, we can obtain the graph of $y = \sin(t)$ by shifting the graph of $y = \cos(t)$ to the right $\frac{\pi}{2}$ units: $\cos\left(t - \frac{\pi}{2} \right) = \sin(t)$ . ↵
this is the reason they are so useful in the Sciences and Engineering ↵
The reader may wish to review Definitions 2.5 and 2.6 as needed. ↵
See the remarks at the beginning of Section 1.6. ↵
If we use a linear function $f(t) = mt + b$ to transform the inputs, $t$ , then $\Delta[f(t)] = m \Delta t$ . That is, the change after the transformation, $m \Delta t$ , is just a multiple of the change before the transformation, $\Delta t$ . ↵
Note when we substitute the quarter marks into $f(t)$ , the argument of the cosine function simplifies to the quadrantal angles. That is, when we substitute $t=1$ , the argument of cosine simplifies to $0$ ; when we substitute $t=2$ , the argument simplifies $\frac{\pi}{2}$ and so on. This provides a quick check of our calculations. ↵
See Section 5.7. ↵
Otherwise, we could just observe the motion of the wheel from the other side. ↵
We are readjusting our `baseline' from $y=0$ to $y=72$ . ↵
Provided $\theta$ is kept `small.' Carl remembers the `Rule of Thumb' as being $20^{\circ}$ or less. Check with your friendly neighborhood physicist to make sure. ↵