10.5 Graphs of the Trigonometric Functions

Graphs of the Cosine and Sine Functions

From Theorem in Section, we know that the domain of $f(t)=\cos (t)$ and of $g(t)=\sin (t)$ is all real numbers, $(-\mathrm{\infty },\mathrm{\infty }),$ and the range of both functions is $[-1,1]$. The Even / Odd Identities in Theorem tell us $\cos (-t)=\cos (t)$ for all real numbers $t$ and $\sin (-t)=-\sin (t)$ for all real numbers $t$. This means $f(t)=\cos (t)$ is an even function, while $g(t)=\sin (t)$ is an odd function.¹ Another important property of these functions is that for coterminal angles $\alpha$ and $\beta$, $\cos (\alpha )=\cos (\beta )$ and $\sin (\alpha )=\sin (\beta )$. Said differently, $\cos (t+2\pi k)=\cos (t)$ and $\sin (t+2\pi k)=\sin (t)$ for all real numbers $t$ and any integer $k$. This last property is given a special name.

We have already seen a family of periodic functions in Section: 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 the circular functions, we see that by Definition, $f(t)=\cos (t)$ is periodic, since $\cos (t+2\pi k)=\cos (t)$ for any integer $k$. 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 $\cos (t+p)=\cos (t)$ for all real numbers $t$. We know that $\cos (t+2\pi )=\cos (t)$ for all real numbers $t$ but the question remains if any smaller real number will do the trick. Suppose $p>0$ and $\cos (t+p)=\cos (t)$ for all real numbers $t$. Then, in particular, $\cos (0+p)=\cos (0)$ so that $\cos (p)=1$. From this we know $p$ is a multiple of $2\pi$ and, since the smallest positive multiple of $2\pi$ is $2\pi$ itself, we have the result. Similarly, we can show $g(t)=\sin (t)$ is also periodic with $2\pi$ as its period.² Having period $2\pi$ essentially means that we can completely understand everything about the functions $f(t)=\cos (t)$ and $g(t)=\sin (t)$ by studying one interval of length $2\pi$, say $[0,2\pi ]$.³ In some advanced texts, the interval of choice is $[-\pi ,\pi )$.

One last property of the functions $f(t)=\cos (t)$ and $g(t)=\sin (t)$ is worth pointing out: both of these functions are continuous and smooth. Recall from Section that geometrically this means the graphs of the cosine and sine functions have no jumps, gaps, holes in the graph, asymptotes, corners or cusps. As we shall see, the graphs of both $f(t)=\cos (t)$ and $g(t)=\sin (t)$ meander nicely and don't cause any trouble. We summarize these facts in the following theorem.

In the chart above, we followed the convention established in Section and used $x$ as the independent variable and $y$ as the dependent variable.⁴ This allows us to turn our attention to graphing the cosine and sine functions in the Cartesian Plane. To graph $y=\cos (x)$, we make a table as we did in Section using some of the `common values' of $x$ in the interval $[0,2\pi ]$. This generates a portion of the cosine graph, which we call the `fundamental cycle' of $y=\cos (x)$.

$$\begin{array}{ccc}& & \\ \hfill x& \hfill \cos (x)& \hfill (x,\cos (x))\\ & & \\ \hfill 0& \hfill 1& \hfill (0,1)\\ & & \\ \hfill \frac{\pi }{4}& \hfill \frac{\sqrt{2}}{2}& \hfill (\frac{\pi }{4},\frac{\sqrt{2}}{2})\\ & & \\ \hfill \frac{\pi }{2}& \hfill 0& \hfill (\frac{\pi }{2},0)\\ & & \\ \hfill \frac{3\pi }{4}& \hfill -\frac{\sqrt{2}}{2}& \hfill (\frac{3\pi }{4},-\frac{\sqrt{2}}{2})\\ & & \\ \hfill \pi & \hfill -1& \hfill (\pi ,-1)\\ & & \\ \hfill \frac{5\pi }{4}& \hfill -\frac{\sqrt{2}}{2}& \hfill (\frac{5\pi }{4},-\frac{\sqrt{2}}{2})\\ & & \\ \hfill \frac{3\pi }{2}& \hfill 0& \hfill (\frac{3\pi }{2},0)\\ & & \\ \hfill \frac{7\pi }{4}& \hfill \frac{\sqrt{2}}{2}& \hfill (\frac{7\pi }{4},\frac{\sqrt{2}}{2})\\ & & \\ \hfill 2\pi & \hfill 1& \hfill (2\pi ,1)\end{array}$$

A few things about the graph above are worth mentioning. First, this graph represents only part of the graph of $y=\cos (x)$. To get the entire graph, we imagine `copying and pasting' this graph end to end infinitely in both directions (left and right) on the $x$-axis. Secondly, the vertical scale here has been greatly exaggerated for clarity and aesthetics. Below is an accurate-to-scale graph of $y=\cos (x)$ showing several cycles with the `fundamental cycle' plotted thicker than the others. The graph of $y=\cos (x)$ is usually described as `wavelike' – indeed, many of the applications involving the cosine and sine functions feature modeling wavelike phenomena.

We can plot the fundamental cycle of the graph of $y=\sin (x)$ similarly, with similar results.

$$\begin{array}{ccc}& & \\ \hfill x& \hfill \sin (x)& \hfill (x,\sin (x))\\ & & \\ \hfill 0& \hfill 0& \hfill (0,0)\\ & & \\ \hfill \frac{\pi }{4}& \hfill \frac{\sqrt{2}}{2}& \hfill (\frac{\pi }{4},\frac{\sqrt{2}}{2})\\ & & \\ \hfill \frac{\pi }{2}& \hfill 1& \hfill (\frac{\pi }{2},1)\\ & & \\ \hfill \frac{3\pi }{4}& \hfill \frac{\sqrt{2}}{2}& \hfill (\frac{3\pi }{4},\frac{\sqrt{2}}{2})\\ & & \\ \hfill \pi & \hfill 0& \hfill (\pi ,0)\\ & & \\ \hfill \frac{5\pi }{4}& \hfill -\frac{\sqrt{2}}{2}& \hfill (\frac{5\pi }{4},-\frac{\sqrt{2}}{2})\\ & & \\ \hfill \frac{3\pi }{2}& \hfill -1& \hfill (\frac{3\pi }{2},-1)\\ & & \\ \hfill \frac{7\pi }{4}& \hfill -\frac{\sqrt{2}}{2}& \hfill (\frac{7\pi }{4},-\frac{\sqrt{2}}{2})\\ & & \\ \hfill 2\pi & \hfill 0& \hfill (2\pi ,0)\end{array}$$

As with the graph of $y=\cos (x)$, we provide an accurately scaled graph of $y=\sin (x)$ below with the fundamental cycle highlighted.

It is no accident that the graphs of $y=\cos (x)$ and $y=\sin (x)$ are so similar. Using a cofunction identity along with the even property of cosine, we have

$$\sin (x)=\cos \left(\frac{\pi }{2}-x\right)=\cos \left(-\left(x-\frac{\pi }{2}\right)\right)=\cos \left(x-\frac{\pi }{2}\right)$$

Recalling Section, we see from this formula that the graph of $y=\sin (x)$ is the result of shifting the graph of $y=\cos (x)$ to the right $\frac{\pi }{2}$ units. A visual inspection confirms this.

Now that we know the basic shapes of the graphs of $y=\cos (x)$ and $y=\sin (x)$, we can use Theorem in Section to graph more complicated curves. To do so, we need to keep track of the movement of some key points on the original graphs. We choose to track the values $x=0$, $\frac{\pi }{2}$, $\pi$, $\frac{3\pi }{2}$ and $2\pi$. These `quarter marks' correspond to quadrantal angles, and as such, mark the location of the zeros and the local extrema of these functions over exactly one period. Before we begin our next example, we need to review the concept of the `argument' of a function as first introduced in Section. For the function $f(x)=1-5\cos (2x-\pi )$, the argument of $f$ is $x$. We shall have occasion, however, to refer to the argument of the cosine, which in this case is $2x-\pi$. Loosely stated, the argument of a trigonometric function is the expression `inside' the function.

The functions in Example Example 1 are examples of sinusoids. Roughly speaking, a sinusoid is the result of taking the basic graph of $f(x)=\cos (x)$ or $g(x)=\sin (x)$ and performing any of the transformations⁵ mentioned in Section. Sinusoids can be characterized by four properties: period, amplitude, phase shift and vertical shift. We have already discussed period, that is, how long it takes for the sinusoid to complete one cycle. The standard period of both $f(x)=\cos (x)$ and $g(x)=\sin (x)$ is $2\pi$, but horizontal scalings will change the period of the resulting sinusoid. The amplitude of the sinusoid is a measure of how `tall' the wave is, as indicated in the figure below. The amplitude of the standard cosine and sine functions is $1$, but vertical scalings can alter this.

The phase shift of the sinusoid is the horizontal shift experienced by the fundamental cycle. We have seen that a phase (horizontal) shift of $\frac{\pi }{2}$ to the right takes $f(x)=\cos (x)$ to $g(x)=\sin (x)$ since $\cos \left(x-\frac{\pi }{2}\right)=\sin (x)$. As the reader can verify, a phase shift of $\frac{\pi }{2}$ to the left takes $g(x)=\sin (x)$ to $f(x)=\cos (x)$. The vertical shift of a sinusoid is exactly the same as the vertical shifts in Section. In most contexts, the vertical shift of a sinusoid is assumed to be $0$, but we state the more general case below. The following theorem, which is reminiscent of Theorem in Section, shows how to find these four fundamental quantities from the formula of the given sinusoid.

We note that in some scientific and engineering circles, the quantity $\varphi$ mentioned in Theorem is called the phase of the sinusoid. Since our interest in this book is primarily with graphing sinusoids, we focus our attention on the horizontal shift $-\frac{\varphi }{\omega }$ induced by $\varphi$.

The proof of Theorem is a direct application of Theorem in Section and is left to the reader. The parameter $\omega$, which is stipulated to be positive, is called the (angular) frequency of the sinusoid and is the number of cycles the sinusoid completes over a $2\pi$ interval. We can always ensure $\omega >0$ using the Even/Odd Identities.⁶ We now test out Theorem using the functions $f$ and $g$ featured in Example Example 1. First, we write $f(x)$ in the form prescribed in Theorem,

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

so that $A=3$, $\omega =\frac{\pi }{2}$, $\varphi =-\frac{\pi }{2}$ and $B=1$. According to Theorem, the period of $f$ is $\frac{2\pi }{\omega }=\frac{2\pi }{\pi /2}=4$, the amplitude is $|A|=|3|=3$, the phase shift is $-\frac{\varphi }{\omega }=-\frac{-\pi /2}{\pi /2}=1$ (indicating a shift to the right $1$ unit) and the vertical shift is $B=1$ (indicating a shift up $1$ unit.) All of these match with our graph of $y=f(x)$. Moreover, if we start with the basic shape of the cosine graph, shift it $1$ unit to the right, $1$ unit up, stretch the amplitude to $3$ and shrink the period to $4$, we will have reconstructed one period of the graph of $y=f(x)$. In other words, instead of tracking the five `quarter marks' through the transformations to plot $y=f(x)$, we can use five other pieces of information: the phase shift, vertical shift, amplitude, period and basic shape of the cosine curve. Turning our attention now to the function $g$ in Example Example 1, we first need to use the odd property of the sine function to write it in the form required by Theorem

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

We find $A=-\frac{1}{2}$, $\omega =2$, $\varphi =-\pi$ and $B=\frac{3}{2}$. The period is then $\frac{2\pi }{2}=\pi$, the amplitude is $\left|-\frac{1}{2}\right|=\frac{1}{2}$, the phase shift is $-\frac{-\pi }{2}=\frac{\pi }{2}$ (indicating a shift right $\frac{\pi }{2}$ units) and the vertical shift is up $\frac{3}{2}$. Note that, in this case, all of the data match our graph of $y=g(x)$ with the exception of the phase shift. Instead of the graph starting at $x=\frac{\pi }{2}$, it ends there. Remember, however, that the graph presented in Example Example 1 is only one portion of the graph of $y=g(x)$. Indeed, another complete cycle begins at $x=\frac{\pi }{2}$, and this is the cycle Theorem is detecting. The reason for the discrepancy is that, in order to apply Theorem, we had to rewrite the formula for $g(x)$ using the odd property of the sine function. Note that whether we graph $y=g(x)$ using the `quarter marks' approach or using the Theorem, we get one complete cycle of the graph, which means we have completely determined the sinusoid.

Note that each of the answers given in Example Example 2 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 $(\frac{1}{2},\frac{1}{2})$ taking $A=-2$. In this case, the phase shift is $\frac{1}{2}$ so $\varphi =-\frac{\pi }{6}$ for an answer of $S(x)=-2\sin \left(\frac{\pi }{3}x-\frac{\pi }{6}\right)+\frac{1}{2}$. Alternatively, we could have extended the graph of $y=f(x)$ to the left and considered a sine function starting at $(-\frac{5}{2},\frac{1}{2})$, and so on. Each of these formulas determine the same sinusoid curve and their formulas are all equivalent using identities. Speaking of identities, if we use the sum identity for cosine, we can expand the formula to yield

$$C(x)=A\cos (\omega x+\varphi )+B=A\cos (\omega x)\cos (\varphi )-A\sin (\omega x)\sin (\varphi )+B.$$

Similarly, using the sum identity for sine, we get

$$S(x)=A\sin (\omega x+\varphi )+B=A\sin (\omega x)\cos (\varphi )+A\cos (\omega x)\sin (\varphi )+B.$$

Making these observations allows us to recognize (and graph) functions as sinusoids which, at first glance, don't appear to fit the forms of either $C(x)$ or $S(x)$.

It is important to note that in order for the technique presented in Example Example 3 to fit a function into one of the forms in Theorem, the arguments of the cosine and sine function much match. That is, while $$f(x)=\cos (2x)-\sqrt{3}\sin (2x)$$ is a sinusoid, $$g(x)=\cos (2x)-\sqrt{3}\sin (3x)$$ is not.⁹ It is also worth mentioning that, had we chosen $A=-2$ instead of $A=2$ as we worked through Example Example 3, our final answers would have looked different. The reader is encouraged to rework Example Example 3 using $A=-2$ to see what these differences are, and then for a challenging exercise, use identities to show that the formulas are all equivalent. The general equations to fit a function of the form $$f(x)=a\cos (\omega x)+b\sin (\omega x)+B$$ into one of the forms in Theorem are explored in Exercise.

Graphs of the Secant and Cosecant Functions

We now turn our attention to graphing $y=\sec (x)$. Since $\sec (x)=\frac{1}{\cos (x)}$, we can use our table of values for the graph of $y=\cos (x)$ and take reciprocals. We know from Section that the domain of $F(x)=\sec (x)$ excludes all odd multiples of $\frac{\pi }{2}$, and sure enough, we run into trouble at $x=\frac{\pi }{2}$ and $x=\frac{3\pi }{2}$ since $\cos (x)=0$ at these values. Using the notation introduced in Section, we have that as $x\to {\frac{\pi }{2}}^{-}$, $\cos (x)\to 0^{+}$, so $\sec (x)\to \mathrm{\infty }$. (See Section for a more detailed analysis.) Similarly, we find that as $x\to {\frac{\pi }{2}}^{+}$, $\sec (x)\to -\mathrm{\infty }$; as $x\to {\frac{3\pi }{2}}^{-}$, $\sec (x)\to -\mathrm{\infty }$; and as $x\to {\frac{3\pi }{2}}^{+}$, $\sec (x)\to \mathrm{\infty }$. This means we have a pair of vertical asymptotes to the graph of $y=\sec (x)$, $x=\frac{\pi }{2}$ and $x=\frac{3\pi }{2}$. Since $\cos (x)$ is periodic with period $2\pi$, it follows that $\sec (x)$ is also.¹⁰ Below we graph a fundamental cycle of $y=\sec (x)$ along with a more complete graph obtained by the usual `copying and pasting.'¹¹

$$\begin{array}{cccc}& & & \\ \hfill x& \hfill \cos (x)& \hfill \sec (x)& \hfill (x,\sec (x))\\ & & & \\ \hfill 0& \hfill 1& \hfill 1& \hfill (0,1)\\ & & & \\ \hfill \frac{\pi }{4}& \hfill \frac{\sqrt{2}}{2}& \hfill \sqrt{2}& \hfill (\frac{\pi }{4},\sqrt{2})\\ & & & \\ \hfill \frac{\pi }{2}& \hfill 0& \hfill \text{undefined}& \\ & & & \\ \hfill \frac{3\pi }{4}& \hfill -\frac{\sqrt{2}}{2}& \hfill -\sqrt{2}& \hfill (\frac{3\pi }{4},-\sqrt{2})\\ & & & \\ \hfill \pi & \hfill -1& \hfill -1& \hfill (\pi ,-1)\\ & & & \\ \hfill \frac{5\pi }{4}& \hfill -\frac{\sqrt{2}}{2}& \hfill -\sqrt{2}& \hfill (\frac{5\pi }{4},-\sqrt{2})\\ & & & \\ \hfill \frac{3\pi }{2}& \hfill 0& \hfill \text{undefined}& \\ & & & \\ \hfill \frac{7\pi }{4}& \hfill \frac{\sqrt{2}}{2}& \hfill \sqrt{2}& \hfill (\frac{7\pi }{4},\sqrt{2})\\ & & & \\ \hfill 2\pi & \hfill 1& \hfill 1& \hfill (2\pi ,1)\end{array}$$

As one would expect, to graph $y=\csc (x)$ we begin with $y=\sin (x)$ and take reciprocals of the corresponding $y$-values. Here, we encounter issues at $x=0$, $x=\pi$ and $x=2\pi$. Proceeding with the usual analysis, we graph the fundamental cycle of $y=\csc (x)$ below along with the dotted graph of $y=\sin (x)$ for reference. Since $y=\sin (x)$ and $y=\cos (x)$ are merely phase shifts of each other, so too are $y=\csc (x)$ and $y=\sec (x)$.

$$\begin{array}{cccc}& & & \\ \hfill x& \hfill \sin (x)& \hfill \csc (x)& \hfill (x,\csc (x))\\ & & & \\ \hfill 0& \hfill 0& \hfill \text{undefined}& \\ & & & \\ \hfill \frac{\pi }{4}& \hfill \frac{\sqrt{2}}{2}& \hfill \sqrt{2}& \hfill (\frac{\pi }{4},\sqrt{2})\\ & & & \\ \hfill \frac{\pi }{2}& \hfill 1& \hfill 1& \hfill (\frac{\pi }{2},1)\\ & & & \\ \hfill \frac{3\pi }{4}& \hfill \frac{\sqrt{2}}{2}& \hfill \sqrt{2}& \hfill (\frac{3\pi }{4},\sqrt{2})\\ & & & \\ \hfill \pi & \hfill 0& \hfill \text{undefined}& \\ & & & \\ \hfill \frac{5\pi }{4}& \hfill -\frac{\sqrt{2}}{2}& \hfill -\sqrt{2}& \hfill (\frac{5\pi }{4},-\sqrt{2})\\ & & & \\ \hfill \frac{3\pi }{2}& \hfill -1& \hfill -1& \hfill (\frac{3\pi }{2},-1)\\ & & & \\ \hfill \frac{7\pi }{4}& \hfill -\frac{\sqrt{2}}{2}& \hfill -\sqrt{2}& \hfill (\frac{7\pi }{4},-\sqrt{2})\\ & & & \\ \hfill 2\pi & \hfill 0& \hfill \text{undefined}& \end{array}$$

Once again, our domain and range work in Section is verified geometrically in the graph of $y=G(x)=\csc (x)$.

Note that, on the intervals between the vertical asymptotes, both $F(x)=\sec (x)$ and $G(x)=\csc (x)$ are continuous and smooth. In other words, they are continuous and smooth on their domains.¹² The following theorem summarizes the properties of the secant and cosecant functions. Note that all of these properties are direct results of them being reciprocals of the cosine and sine functions, respectively.

In the next example, we discuss graphing more general secant and cosecant curves.

Before moving on, we note that it is possible to speak of the period, phase shift and vertical shift of secant and cosecant graphs and use even/odd identities to put them in a form similar to the sinusoid forms mentioned in Theorem. Since these quantities match those of the corresponding cosine and sine curves, we do not spell this out explicitly. Finally, since the ranges of secant and cosecant are unbounded, there is no amplitude associated with these curves.

Graphs of the Tangent and Cotangent Functions

Finally, we turn our attention to the graphs of the tangent and cotangent functions. When constructing a table of values for the tangent function, we see that $J(x)=\tan (x)$ is undefined at $x=\frac{\pi }{2}$ and $x=\frac{3\pi }{2}$, in accordance with our findings in Section. As $x\to {\frac{\pi }{2}}^{-}$, $\sin (x)\to 1^{-}$ and $\cos (x)\to 0^{+}$, so that $\tan (x)=\frac{\sin (x)}{\cos (x)}\to \mathrm{\infty }$ producing a vertical asymptote at $x=\frac{\pi }{2}$. Using a similar analysis, we get that as $x\to {\frac{\pi }{2}}^{+}$, $\tan (x)\to -\mathrm{\infty }$; as $x\to {\frac{3\pi }{2}}^{-}$, $\tan (x)\to \mathrm{\infty }$; and as $x\to {\frac{3\pi }{2}}^{+}$, $\tan (x)\to -\mathrm{\infty }$. Plotting this information and performing the usual `copy and paste' produces:

$$\begin{array}{ccc}& & \\ \hfill x& \hfill \tan (x)& \hfill (x,\tan (x))\\ & & \\ \hfill 0& \hfill 0& \hfill (0,0)\\ & & \\ \hfill \frac{\pi }{4}& \hfill 1& \hfill (\frac{\pi }{4},1)\\ & & \\ \hfill \frac{\pi }{2}& \hfill \text{undefined}& \\ & & \\ \hfill \frac{3\pi }{4}& \hfill -1& \hfill (\frac{3\pi }{4},-1)\\ & & \\ \hfill \pi & \hfill 0& \hfill (\pi ,0)\\ & & \\ \hfill \frac{5\pi }{4}& \hfill 1& \hfill (\frac{5\pi }{4},1)\\ & & \\ \hfill \frac{3\pi }{2}& \hfill \text{undefined}& \\ & & \\ \hfill \frac{7\pi }{4}& \hfill -1& \hfill (\frac{7\pi }{4},-1)\\ & & \\ \hfill 2\pi & \hfill 0& \hfill (2\pi ,0)\end{array}$$

From the graph, it appears as if the tangent function is periodic with period $\pi$. To prove that this is the case, we appeal to the sum formula for tangents. We have:

$$\tan (x+\pi )=\frac{\tan (x)+\tan (\pi )}{1-\tan (x)\tan (\pi )}=\frac{\tan (x)+0}{1-(\tan (x))(0)}=\tan (x),$$

which tells us the period of $\tan (x)$ is at most $\pi$. To show that it is exactly $\pi$, suppose $p$ is a positive real number so that $\tan (x+p)=\tan (x)$ for all real numbers $x$. For $x=0$, we have $$\tan (p)=\tan (0+p)=\tan (0)=0$$, which means $p$ is a multiple of $\pi$. The smallest positive multiple of $\pi$ is $\pi$ itself, so we have established the result. We take as our fundamental cycle for $y=\tan (x)$ the interval $(-\frac{\pi }{2},\frac{\pi }{2})$, and use as our `quarter marks' $x=-\frac{\pi }{2}$, $-\frac{\pi }{4}$, $0$, $\frac{\pi }{4}$ and $\frac{\pi }{2}$. From the graph, we see confirmation of our domain and range work in Section.

It should be no surprise that $K(x)=\cot (x)$ behaves similarly to $J(x)=\tan (x)$. Plotting $\cot (x)$ over the interval $[0,2\pi ]$ results in the graph below.

$$\begin{array}{ccc}& & \\ \hfill x& \hfill \cot (x)& \hfill (x,\cot (x))\\ & & \\ \hfill 0& \hfill \text{undefined}& \\ & & \\ \hfill \frac{\pi }{4}& \hfill 1& \hfill (\frac{\pi }{4},1)\\ & & \\ \hfill \frac{\pi }{2}& \hfill 0& \hfill (\frac{\pi }{2},0)\\ & & \\ \hfill \frac{3\pi }{4}& \hfill -1& \hfill (\frac{3\pi }{4},-1)\\ & & \\ \hfill \pi & \hfill \text{undefined}& \\ & & \\ \hfill \frac{5\pi }{4}& \hfill 1& \hfill (\frac{5\pi }{4},1)\\ & & \\ \hfill \frac{3\pi }{2}& \hfill 0& \hfill (\frac{3\pi }{2},0)\\ & & \\ \hfill \frac{7\pi }{4}& \hfill -1& \hfill (\frac{7\pi }{4},-1)\\ & & \\ \hfill 2\pi & \hfill \text{undefined}& \end{array}$$

From these data, it clearly appears as if the period of $\cot (x)$ is $\pi$, and we leave it to the reader to prove this.¹³ We take as one fundamental cycle the interval $(0,\pi )$ with quarter marks: $x=0$, $\frac{\pi }{4}$, $\frac{\pi }{2}$, $\frac{3\pi }{4}$ and $\pi$. A more complete graph of $y=\cot (x)$ is below, along with the fundamental cycle highlighted as usual. Once again, we see the domain and range of $K(x)=\cot (x)$ as read from the graph matches with what we found analytically in Section.

The properties of the tangent and cotangent functions are summarized below. As with Theorem, each of the results below can be traced back to properties of the cosine and sine functions and the definition of the tangent and cotangent functions as quotients thereof.

As with the secant and cosecant functions, it is possible to extend the notion of period, phase shift and vertical shift to the tangent and cotangent functions as we did for the cosine and sine functions in Theorem. Since the number of classical applications involving sinusoids far outnumber those involving tangent and cotangent functions, we omit this. The ambitious reader is invited to formulate such a theorem, however.

Exercises

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

1. $y=3\sin (x)$
2. $y=\sin (3x)$
3. $y=-2\cos (x)$
4. $y=\cos \left(x-\frac{\pi }{2}\right)$
5. $y=-\sin \left(x+\frac{\pi }{3}\right)$
6. $y=\sin (2x-\pi )$
7. $y=-\frac{1}{3}\cos \left(\frac{1}{2}x+\frac{\pi }{3}\right)$
8. $y=\cos (3x-2\pi )+4$
9. $y=\sin \left(-x-\frac{\pi }{4}\right)-2$
10. $y=\frac{2}{3}\cos \left(\frac{\pi }{2}-4x\right)+1$
11. $y=-\frac{3}{2}\cos \left(2x+\frac{\pi }{3}\right)-\frac{1}{2}$
12. $y=4\sin (-2\pi x+\pi )$
13. $y=\tan \left(x-\frac{\pi }{3}\right)$
14. $y=2\tan \left(\frac{1}{4}x\right)-3$
15. $y=\frac{1}{3}\tan (-2x-\pi )+1$
16. $y=\sec \left(x-\frac{\pi }{2}\right)$
17. $y=-\csc \left(x+\frac{\pi }{3}\right)$
18. $y=-\frac{1}{3}\sec \left(\frac{1}{2}x+\frac{\pi }{3}\right)$
19. $y=\csc (2x-\pi )$
20. $y=\sec (3x-2\pi )+4$
21. $y=\csc \left(-x-\frac{\pi }{4}\right)-2$
22. $y=\cot \left(x+\frac{\pi }{6}\right)$
23. $y=-11\cot \left(\frac{1}{5}x\right)$
24. $y=\frac{1}{3}\cot \left(2x+\frac{3\pi }{2}\right)+1$
25. $$f(x)=\sqrt{2}\sin (x)+\sqrt{2}\cos (x)+1$$
26. $$f(x)=3\sqrt{3}\sin (3x)-3\cos (3x)$$
27. $$f(x)=-\sin (x)+\cos (x)-2$$
28. $$f(x)=-\frac{1}{2}\sin (2x)-\frac{\sqrt{3}}{2}\cos (2x)$$
29. $$f(x)=2\sqrt{3}\cos (x)-2\sin (x)$$
30. $$f(x)=\frac{3}{2}\cos (2x)-\frac{3\sqrt{3}}{2}\sin (2x)+6$$
31. $$f(x)=-\frac{1}{2}\cos (5x)-\frac{\sqrt{3}}{2}\sin (5x)$$
32. $$f(x)=-6\sqrt{3}\cos (3x)-6\sin (3x)-3$$
33. $$f(x)=\frac{5\sqrt{2}}{2}\sin (x)-\frac{5\sqrt{2}}{2}\cos (x)$$
34. $$f(x)=3\sin \left(\frac{x}{6}\right)-3\sqrt{3}\cos \left(\frac{x}{6}\right)$$
35. In Exercises -, you should have noticed a relationship between the phases $\varphi$ for the $S(x)$ and $C(x)$. Show that if $$f(x)=A\sin (\omega x+\alpha )+B$$, then $$f(x)=A\cos (\omega x+\beta )+B$$ where $\beta =\alpha -\frac{\pi }{2}$.
36. Let $\varphi$ be an angle measured in radians and let $P(a,b)$ be a point on the terminal side of $\varphi$ when it is drawn in standard position. Use Theorem and the sum identity for sine in Theorem to show that $$f(x)=a\sin (\omega x)+b\cos (\omega x)+B$$ (with $\omega >0$) can be rewritten as $$f(x)=\sqrt{a^2+b^2}\sin (\omega x+\varphi )+B$$.
37. With the help of your classmates, express the domains of the functions in Examples Example 4 and Example 5 using extended interval notation. (We will revisit this in Section.)
38. ${\sin }^2(x)+{\cos }^2(x)=1$
39. ${\sec }^2(x)-{\tan }^2(x)=1$
40. $\cos (x)=\sin \left(\frac{\pi }{2}-x\right)$
41. $\tan (x+\pi )=\tan (x)$
42. $$\sin (2x)=2\sin (x)\cos (x)$$
43. $\tan \left(\frac{x}{2}\right)=\frac{\sin (x)}{1+\cos (x)}$
44. $$f(x)=\cos (3x)+\sin (x)$$. Is this function periodic? If so, what is the period?
45. $f(x)=\frac{\sin (x)}{x}$. What appears to be the horizontal asymptote of the graph?
46. $f(x)=x\sin (x)$. Graph $y=\pm x$ on the same set of axes and describe the behavior of $f$.
47. $f(x)=\sin \left(\frac{1}{x}\right)$. What's happening as $x\to 0$?
48. $f(x)=x-\tan (x)$. Graph $y=x$ on the same set of axes and describe the behavior of $f$.
49. $$f(x)=e^{-0.1x}\left(\cos (2x)+\sin (2x)\right)$$. Graph $y=\pm e^{-0.1x}$ on the same set of axes and describe the behavior of $f$.
50. $$f(x)=e^{-0.1x}\left(\cos (2x)+2\sin (x)\right)$$. Graph $y=\pm e^{-0.1x}$ on the same set of axes and describe the behavior of $f$.
51. Show that a constant function $f$ is periodic by showing that $f(x+117)=f(x)$ 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 another way, 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'.

In Exercises -, graph one cycle of the given function. State the period of the function.

In Exercises -, use Example Example 3 as a guide to show that the function is a sinusoid by rewriting it in the forms $C(x)=A\cos (\omega x+\varphi )+B$ and $S(x)=A\sin (\omega x+\varphi )+B$ for $\omega >0$ and $0\le \varphi <2\pi$.

In Exercises -, verify the identity by graphing the right and left hand sides on a calculator.

In Exercises -, graph the function with the help of your calculator and discuss the given questions with your classmates.

Answers

1. $y=3\sin (x)$ Period: $2\pi$ Amplitude: $3$ Phase Shift: $0$ Vertical Shift: $0$

   

2. $y=\sin (3x)$ Period: $\frac{2\pi }{3}$ Amplitude: $1$ Phase Shift: $0$ Vertical Shift: $0$

   

3. $y=-2\cos (x)$ Period: $2\pi$ Amplitude: $2$ Phase Shift: $0$ Vertical Shift: $0$

   

4. $y=\cos \left(x-\frac{\pi }{2}\right)$ Period: $2\pi$ Amplitude: $1$ Phase Shift: $\frac{\pi }{2}$ Vertical Shift: $0$

   

5. $y=-\sin \left(x+\frac{\pi }{3}\right)$ Period: $2\pi$ Amplitude: $1$ Phase Shift: $-\frac{\pi }{3}$ Vertical Shift: $0$

   

6. $y=\sin (2x-\pi )$ Period: $\pi$ Amplitude: $1$ Phase Shift: $\frac{\pi }{2}$ Vertical Shift: $0$

   

7. $y=-\frac{1}{3}\cos \left(\frac{1}{2}x+\frac{\pi }{3}\right)$ Period: $4\pi$ Amplitude: $\frac{1}{3}$ Phase Shift: $-\frac{2\pi }{3}$ Vertical Shift: $0$

   

8. $y=\cos (3x-2\pi )+4$ Period: $\frac{2\pi }{3}$ Amplitude: $1$ Phase Shift: $\frac{2\pi }{3}$ Vertical Shift: 4

   

9. $y=\sin \left(-x-\frac{\pi }{4}\right)-2$ Period: $2\pi$ Amplitude: $1$ Phase Shift: $-\frac{\pi }{4}$ (You need to use $y=-\sin \left(x+\frac{\pi }{4}\right)-2$ to find this.)¹⁴ Vertical Shift: $-2$

   

10. $y=\frac{2}{3}\cos \left(\frac{\pi }{2}-4x\right)+1$ Period: $\frac{\pi }{2}$ Amplitude: $\frac{2}{3}$ Phase Shift: $\frac{\pi }{8}$ (You need to use $y=\frac{2}{3}\cos \left(4x-\frac{\pi }{2}\right)+1$ to find this.)¹⁵ Vertical Shift: $1$

    

11. $y=-\frac{3}{2}\cos \left(2x+\frac{\pi }{3}\right)-\frac{1}{2}$ Period: $\pi$ Amplitude: $\frac{3}{2}$ Phase Shift: $-\frac{\pi }{6}$ Vertical Shift: $-\frac{1}{2}$

    

12. $y=4\sin (-2\pi x+\pi )$ Period: $1$ Amplitude: $4$ Phase Shift: $\frac{1}{2}$ (You need to use $y=-4\sin (2\pi x-\pi )$ to find this.)¹⁶ Vertical Shift: $0$

    

13. $y=\tan \left(x-\frac{\pi }{3}\right)$ Period: $\pi$

    

14. $y=2\tan \left(\frac{1}{4}x\right)-3$ Period: $4\pi$

    

15. $y=\frac{1}{3}\tan (-2x-\pi )+1$ is equivalent to $y=-\frac{1}{3}\tan (2x+\pi )+1$ via the Even / Odd identity for tangent. Period: $\frac{\pi }{2}$

    

16. $y=\sec \left(x-\frac{\pi }{2}\right)$ Start with $y=\cos \left(x-\frac{\pi }{2}\right)$ Period: $2\pi$

    

17. $y=-\csc \left(x+\frac{\pi }{3}\right)$ Start with $y=-\sin \left(x+\frac{\pi }{3}\right)$ Period: $2\pi$

    

18. $y=-\frac{1}{3}\sec \left(\frac{1}{2}x+\frac{\pi }{3}\right)$ Start with $y=-\frac{1}{3}\cos \left(\frac{1}{2}x+\frac{\pi }{3}\right)$ Period: $4\pi$

    

19. $y=\csc (2x-\pi )$ Start with $y=\sin (2x-\pi )$ Period: $\pi$

    

20. $y=\sec (3x-2\pi )+4$ Start with $y=\cos (3x-2\pi )+4$ Period: $\frac{2\pi }{3}$

    

21. $y=\csc \left(-x-\frac{\pi }{4}\right)-2$ Start with $y=\sin \left(-x-\frac{\pi }{4}\right)-2$ Period: $2\pi$

    

22. $y=\cot \left(x+\frac{\pi }{6}\right)$ Period: $\pi$

    

23. $y=-11\cot \left(\frac{1}{5}x\right)$ Period: $5\pi$

    

24. $y=\frac{1}{3}\cot \left(2x+\frac{3\pi }{2}\right)+1$ Period: $\frac{\pi }{2}$

    

25. $$f(x)=\sqrt{2}\sin (x)+\sqrt{2}\cos (x)+1=2\sin \left(x+\frac{\pi }{4}\right)+1=2\cos \left(x+\frac{7\pi }{4}\right)+1$$
26. $$f(x)=3\sqrt{3}\sin (3x)-3\cos (3x)=6\sin \left(3x+\frac{11\pi }{6}\right)=6\cos \left(3x+\frac{4\pi }{3}\right)$$
27. $$f(x)=-\sin (x)+\cos (x)-2=\sqrt{2}\sin \left(x+\frac{3\pi }{4}\right)-2=\sqrt{2}\cos \left(x+\frac{\pi }{4}\right)-2$$
28. $$f(x)=-\frac{1}{2}\sin (2x)-\frac{\sqrt{3}}{2}\cos (2x)=\sin \left(2x+\frac{4\pi }{3}\right)=\cos \left(2x+\frac{5\pi }{6}\right)$$
29. $$f(x)=2\sqrt{3}\cos (x)-2\sin (x)=4\sin \left(x+\frac{2\pi }{3}\right)=4\cos \left(x+\frac{\pi }{6}\right)$$
30. $$f(x)=\frac{3}{2}\cos (2x)-\frac{3\sqrt{3}}{2}\sin (2x)+6=3\sin \left(2x+\frac{5\pi }{6}\right)+6=3\cos \left(2x+\frac{\pi }{3}\right)+6$$
31. $$f(x)=-\frac{1}{2}\cos (5x)-\frac{\sqrt{3}}{2}\sin (5x)=\sin \left(5x+\frac{7\pi }{6}\right)=\cos \left(5x+\frac{2\pi }{3}\right)$$
32. $$f(x)=-6\sqrt{3}\cos (3x)-6\sin (3x)-3=12\sin \left(3x+\frac{4\pi }{3}\right)-3=12\cos \left(3x+\frac{5\pi }{6}\right)-3$$
33. $$f(x)=\frac{5\sqrt{2}}{2}\sin (x)-\frac{5\sqrt{2}}{2}\cos (x)=5\sin \left(x+\frac{7\pi }{4}\right)=5\cos \left(x+\frac{5\pi }{4}\right)$$
34. $$f(x)=3\sin \left(\frac{x}{6}\right)-3\sqrt{3}\cos \left(\frac{x}{6}\right)=6\sin \left(\frac{x}{6}+\frac{5\pi }{3}\right)=6\cos \left(\frac{x}{6}+\frac{7\pi }{6}\right)$$