On page ??, 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] (which, you’ll recall essentially ‘wraps the real number line around the Unit Circle) and define \(\sin (t) = \sin (\theta )\) and \(\cos (t) = \cos (\theta )\). Since every real number can be identified with one and only one angle \(\theta \) this way, 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. Since the real number line, ^[2] (in particular the interval \([0, 2\pi )\)) 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.

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] (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.)

We use desmos to help us construct the graph below. Note the scale of the horizontal and vertical axis is far from 1:1, but you are invited to take advantage of desmos’s ‘Zoom Square’ feature to see a more accurate graph.

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 coterminalsamcosinesinethm. 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.

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.

We have already seen a family of periodic functions in Section ConstantandLinearFunctions: 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 periodic, \(f\) is periodic since \(\sin (t + 2\pi ) = \sin (t)\). 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 \(\pi \). Since \(\sin \left (\frac {\pi }{2} \right ) \neq \sin \left (\frac {\pi }{2} + \pi \right ) \), 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] (Technically, we should study the interval \([0,2\pi )\),¹since 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.) 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 \(f(t) = \sin (t)\) as seen below. ^[5] (Here note that the dependent variable ‘\(y\)’ represents the outputs from \(g(t) = \cos (t)\) which are \(x\)-coordinates on the Unit Circle.)

We start with a table of values.

Next, we use desmos to help us graph one cycle.

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 MoreTrigonometricIdentities, \(\sin \left (t + \frac {\pi }{2}\right ) = \cos (t)\), 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] (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: \(\cos \left (t - \frac {\pi }{2} \right ) = \sin (t)\).)

While arguably the most important property shared by \(f(t) = \sin (t)\) and \(g(t) = \cos (t)\) is their periodic ‘wavelike’ nature, ^[7] (this is the reason they are so useful in the Sciences and Engineering) their graphs suggest these functions are both continuous and smooth. Recall from Section GraphsofPolynomials 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.

Note the graphs of both \(f(t) = \sin (t)\) and \(g(t) = \cos (t)\) meander as \(t \rightarrow -\infty \) or as \( \rightarrow \infty \). Said differently, none of the limits \(\lim _{t \rightarrow -\infty } \sin (t)\), \(\lim _{t \rightarrow \infty } \sin (t)\), \(\lim _{t \rightarrow -\infty } \cos (t)\), or \(\lim _{t \rightarrow \infty } \cos (t)\) exist. 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 GraphsofPolynomials. 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 MoreTrigonometricIdentities, \(f(t) = \sin (t)\) is, in fact, an odd function: ^[8] (The reader may wish to review Definitions evenfunctiondefn and oddfunctiondefn as needed.) 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.

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 Transformations to graph more complicated functions using transformations. 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 sine and cosine:

In working through Section Transformations , 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. Since 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 Transformations are linear, ^[9] (See the remarks at the beginning of Section Transformations.) the relative spacing of the points before and after the transformations remains the same. ^[10] (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 the transformation, \(m \Delta t\), is just a multiple of the change before the transformation, \(\Delta t\).) 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.

As previously mentioned, the curves graphed in Example cosinesinegraphex1 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 Transformations. 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.

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 cosinesinegraphex1, 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 \(\cos (t)\), \(2\pi \), since 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 ‘\(0\)’ since 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’ since 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 Transformations 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 cosinesinegraphex1, 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 transformationsthm and is left to the reader in Exercise proofsinusoidformexercise.

The parameter \(\omega \) mentioned above is called the angular frequency, or more simply, the frequency of the sinusoid and is the number of cycles the sinusoid completes over an interval of length \(2\pi \). That is, \(\omega \) measures how ‘frequently’ the sinusoid repeats over an interval of length \(2\pi \). As we’ll see in the next example, we can always ensure \(\omega > 0\) using the even and odd properties of the cosine and sine functions, respectively. If \(t\) represents time, \(\omega \) as represents how fast the sinusoid is being generated in terms of radians per unit time. In essence, it is the angular speed of the curve.

A quantity closely related to the angular frequency of the sinusoid is the ordinary frequency of the sinusoid, usually denoted \(f\). The ordinary frequency of a sinusoid measures the number of cycles the sinusoid completes over an interval of length \(1\). Since the period, \(T\) represents the length of the interval required for a sinusoid to make one complete cycle, we have \(f =\frac {1}{T}\). Once again, if \(t\) represents time, the ordinary frequency measures how fast the sinusoid is being generated in terms of complete cycles per unit time. ^[11] (If \(t\) is time measured in seconds, then one cycle per second is 1 Hertz.)

Note that since \(T = \frac {2 \pi }{\omega }\), \(f = \frac {1}{T} = \frac {\omega }{2\pi }\). Rewriting, we get \(\omega = 2 \pi f\). To understand this equation in terms of units, recall \(1\) complete cycle (revolution) around the Unit Circle counts for \(2\pi \) radians. Hence, to get from \(f\), measured in cycles per unit time, to \(\omega \), measured in radians per unit time, we need to multiply by \(2\pi \).

If the concepts of period and frequency seem familiar, they should. In Section RadianMeasure, we discussed these very same ideas in the context of Example EarthRotationEx and revisited them again in Section TheCircularFunctionsSineandCosine in Equation equationsforcircularmotion. and Example Lakelandrotates. On the one hand, the notions presented here are more general, since they are not tied directly to circular motion. On the other hand, the stipulation in this section that \(\omega > 0\) means we are restricting our attention to angular speeds instead of the more general angular velocities. ^[12] (Recall that velocity is speed with a direction. In equationsforcircularmotion, \(\omega > 0\) indicated counter-clockwise motion while \(\omega < 0\) indicated clockwise motion.)

Last, but not least, the quantity \(\phi \) mentioned in Theorem sinusoidform is called the phase or phase angle of the sinusoid. The phase of a sinusoid is the angle in the argument which corresponds to \(t=0\), and is important in describing waves in fields such as physics and electronics. When graphing sinusoids, however, we focus our attention on the horizontal shift induced by \(\phi \), \(-\frac {\phi }{\omega }\).

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

Note that in this section, we have discussed two ways to graph sinusoids: using Theorem transformationsthm from Section Transformations and using Theorem sinusoidform. 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 cosinesinegraphex2 using Theorem transformationsthm, we obtain the following:

Comparing this result (in green) with the one obtained in Example cosinesinegraphex2 (in purple) 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 transformationsthm and sinusoidform will produce different results.

In the Exercises, we assume the problems are worked using Theorem sinusoidform. If you choose to use Theorems transformationsthm 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 sinusoidform 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.

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

1. Write \(f(t)\) in the form \(C(t) = A \cos (\omega t + \phi ) + B\), for \(\omega > 0\).
2. Write \(f(t)\) in the form \(S(t) = A \sin (\omega t + \phi ) + B\), for \(\omega > 0\).

1. Since one cycle is graphed over the interval \([-1,5]\), its period is \(T=5-(-1) = 6\). According to Theorem sinusoidform, \(6 = T = \frac {2\pi }{\omega }\), so that \(\omega = \frac {\pi }{3}\). Next, we see that the phase shift is \(-1\), so we have \(-\frac {\phi }{\omega } = -1\), or \(\phi = \omega = \frac {\pi }{3}\).

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

2. Since we have written \(f(t)\) in terms of cosines, we can use the conversion from sine to cosine as listed in Theorem cosinesinefunctionprops. Since \(\cos (t) = \sin \left (t +\frac {\pi }{2} \right )\), \(\cos \left (\frac {\pi }{3} t + \frac {\pi }{3} \right ) = \sin \left ( \left [\frac {\pi }{3} t + \frac {\pi }{3}\right ] + \frac {\pi }{2} \right )\), 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: \(\omega = \frac {\pi }{3}\), \(A = 2\) 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 \(-\frac {\phi }{\omega } = \frac {7}{2}\), or \(\phi = -\frac {7}{2} \omega = -\frac {7}{2}\left (\frac {\pi }{3}\right ) = -\frac {7\pi }{6}\). Hence, our answer is \(f(t) = 2 \sin \left (\frac {\pi }{3} t - \frac {7\pi }{6}\right ) + \frac {1}{2}\).

   

Note that each of the answers given in Example fitsinusoidtodata1 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 \(\phi = -\frac {\pi }{6}\) 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.

1 Applications of Sinusoids

In the same way exponential functions can be used to model a wide variety of phenomena in nature, ^[14] (See Section ??.) 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 ?? .

Recall from Exercise ?? in Section ?? 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.

We sketch the problem situation below and assume a counter-clockwise rotation. ^[15] (Otherwise, we could just observe the motion of the wheel from the other side.)

We know from the Equation ?? in Section ?? 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. Since 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 angular frequency is \(\omega = \frac {2\pi }{T} = \frac {4 \pi }{127}\) 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. Since 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, ^[16] (We are readjusting our ‘baseline’ from \(y=0\) to \(y=72\).) 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 motivating Equation ??, 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 \(h(t) = 64 \sin \left (\theta - \frac {\pi }{2}\right ) + 72 = 64\sin \left (\frac {4 \pi }{127} t -\frac {\pi }{2} \right ) + 72\).

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.

A few remarks about Example 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 \(Q\).

Our next example revisits the daylight data first introduced in Section ??, Exercise ??.

The scenario described in Example 5 is a typical example of where the circular functions are useful outside the context of angles or circular motion. Indeed, sine and cosine functions are used extensively to model a wide range of periodic phenomena including signal analysis, wave physics, and even quantum mechanics. We close this section discussing limits involving sine and cosine.

2 Limits involving Sine and Cosine

We’ve already stated as fact (but not proven) that \(f(t) = \sin (t)\) and \(g(t) = \cos (t)\) are continuous. Per Definition ??, this means that \(\lim _{t \rightarrow a} \sin (t) = \sin (a)\) and \(lim_{t \rightarrow a} \cos (t) = \cos (a)\) for all real numbers, \(a\). This means, for instance, we can compute \(\lim _{t \rightarrow \pi } \cos (t) = \cos (\pi ) = -1\). Per the discussion following Definition ??, so long as we avoid the usual domain pitfalls, we can also compute:

In the next example we investigate a few (similar looking) limits of sinusoids.