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Graphs of the Other Trigonometric Functions

Analyze the graph of y=tan x.

Analyzing the Graph of

We will begin with the graph of the tangent function, plotting points as we did for the sine and cosine functions. Recall that

\[\tan \ x=\frac{\sin \ x}{\cos \ x}\]

The period of the tangent function is \(\pi\) because the graph repeats itself on intervals of \(k\pi\) where \(k\) is a constant. If we graph the tangent function on \(-\frac{\pi }{2}\) to \(\frac{\pi }{2},\) we can see the behavior of the graph on one complete cycle. If we look at any larger interval, we will see that the characteristics of the graph repeat.

We can determine whether tangent is an odd or even function by using the definition of tangent.

\[\begin{array}{llll}\tan (-x)=\frac{\sin (-x)}{\cos (-x)} & \begin{array}{lll} & & \end{array}\text{Definition of tangent}. \\ =\frac{-\sin \ x}{\cos \ x} & \begin{array}{lll} & & \end{array}\text{Sine is an odd function, cosine is even}. \\ =-\frac{\sin \ x}{\cos \ x} & \begin{array}{lll} & & \end{array}\text{The quotient of an odd and an even function is odd}. \\ =-\tan \ x & \begin{array}{lll} & & \end{array}\text{Definition of tangent}.\end{array}\]

Therefore, tangent is an odd function. We can further analyze the graphical behavior of the tangent function by looking at values for some of the special angles, as listed in .

\(x\)	\(-\frac{\pi }{2}\)	\(-\frac{\pi }{3}\)	\(-\frac{\pi }{4}\)	\(-\frac{\pi }{6}\)	0	\(\frac{\pi }{6}\)	\(\frac{\pi }{4}\)	\(\frac{\pi }{3}\)	\(\frac{\pi }{2}\)
\(\tan (x)\)	undefined	\(-\sqrt{3}\)	–1	\(-\frac{\sqrt{3}}{3}\)	0	\(\frac{\sqrt{3}}{3}\)	1	\(\sqrt{3}\)	undefined

These points will help us draw our graph, but we need to determine how the graph behaves where it is undefined. If we look more closely at values when \(\frac{\pi }{3}

\(x\)	1.3	1.5	1.55	1.56
\(\tan \ x\)	3.6	14.1	48.1	92.6

As \(x\) approaches \(\frac{\pi }{2},\) the outputs of the function get larger and larger. Because \(y=\tan \ x\) is an odd function, we see the corresponding table of negative values in .

\(x\)	−1.3	−1.5	−1.55	−1.56
\(\tan \ x\)	−3.6	−14.1	−48.1	−92.6

We can see that, as \(x\) approaches \(-\frac{\pi }{2},\) the outputs get smaller and smaller. Remember that there are some values of \(x\) for which \(\cos \ x=0.\) For example, \(\cos (\frac{\pi }{2})=0\) and \(\cos (\frac{3\pi }{2})=0.\) At these values, the tangent function is undefined, so the graph of \(y=\tan \ x\) has discontinuities at \(x=\frac{\pi }{2}\text{ and }\frac{3\pi }{2}.\) At these values, the graph of the tangent has vertical asymptotes. represents the graph of \(y=\tan \ x.\) The tangent is positive from 0 to \(\frac{\pi }{2}\) and from \(\pi\) to \(\frac{3\pi }{2},\) corresponding to quadrants I and III of the unit circle.

Graphing Variations of

As with the sine and cosine functions, the tangent function can be described by a general equation.

\[y=A\tan (Bx)\]

We can identify horizontal and vertical stretches and compressions using values of \(A\) and \(B.\) The horizontal stretch can typically be determined from the period of the graph. With tangent graphs, it is often necessary to determine a vertical stretch using a point on the graph.

Because there are no maximum or minimum values of a tangent function, the term amplitude cannot be interpreted as it is for the sine and cosine functions. Instead, we will use the phrase stretching/compressing factor when referring to the constant \(A.\)

Condensed — the full section is in OpenStax Precalculus 2e.

Analyzing the Graphs of

The secant was defined by the reciprocal identity \(\text{sec}\ x=\frac{1}{\cos \ x}.\) Notice that the function is undefined when the cosine is 0, leading to vertical asymptotes at \(\frac{\pi }{2},\) \(\frac{3\pi }{2},\) etc. Because the cosine is never more than 1 in absolute value, the secant, being the reciprocal, will never be less than 1 in absolute value.

We can graph \(y=\text{sec}\ x\) by observing the graph of the cosine function because these two functions are reciprocals of one another. See . The graph of the cosine is shown as a dashed orange wave so we can see the relationship. Where the graph of the cosine function decreases, the graph of the secant function increases. Where the graph of the cosine function increases, the graph of the secant function decreases. When the cosine function is zero, the secant is undefined.

The secant graph has vertical asymptotes at each value of \(x\) where the cosine graph crosses the x-axis; we show these in the graph below with dashed vertical lines, but will not show all the asymptotes explicitly on all later graphs involving the secant and cosecant.

Note that, because cosine is an even function, secant is also an even function. That is, \(\text{sec}(-x)=\text{sec}\ x.\)

As we did for the tangent function, we will again refer to the constant \(|A|\) as the stretching factor, not the amplitude.

Similar to the secant, the cosecant is defined by the reciprocal identity \(\text{csc}\ x=\frac{1}{\sin \ x}.\) Notice that the function is undefined when the sine is 0, leading to a vertical asymptote in the graph at \(0,\) \(\pi ,\) etc. Since the sine is never more than 1 in absolute value, the cosecant, being the reciprocal, will never be less than 1 in absolute value.

We can graph \(y=\text{csc}\ x\) by observing the graph of the sine function because these two functions are reciprocals of one another. See . The graph of sine is shown as a dashed orange wave so we can see the relationship. Where the graph of the sine function decreases, the graph of the cosecant function increases. Where the graph of the sine function increases, the graph of the cosecant function decreases.

Condensed — the full section is in OpenStax Precalculus 2e.

Graphing Variations of

For shifted, compressed, and/or stretched versions of the secant and cosecant functions, we can follow similar methods to those we used for tangent and cotangent. That is, we locate the vertical asymptotes and also evaluate the functions for a few points (specifically the local extrema). If we want to graph only a single period, we can choose the interval for the period in more than one way. The procedure for secant is very similar, because the cofunction identity means that the secant graph is the same as the cosecant graph shifted half a period to the left. Vertical and phase shifts may be applied to the cosecant function in the same way as for the secant and other functions.The equations become the following.

\[y=A\text{sec}(Bx-C)+D\]\[y=A\text{csc}(Bx-C)+D\]

Condensed — the full section is in OpenStax Precalculus 2e.

Analyzing the Graph of

The last trigonometric function we need to explore is cotangent. The cotangent is defined by the reciprocal identity \(\text{cot}\ x=\frac{1}{\tan \ x}.\) Notice that the function is undefined when the tangent function is 0, leading to a vertical asymptote in the graph at \(0,\pi ,\) etc. Since the output of the tangent function is all real numbers, the output of the cotangent function is also all real numbers.

We can graph \(y=\text{cot}\ x\) by observing the graph of the tangent function because these two functions are reciprocals of one another. See . Where the graph of the tangent function decreases, the graph of the cotangent function increases. Where the graph of the tangent function increases, the graph of the cotangent function decreases.

The cotangent graph has vertical asymptotes at each value of \(x\) where \(\tan \ x=0;\) we show these in the graph below with dashed lines. Since the cotangent is the reciprocal of the tangent, \(\text{cot}\ x\) has vertical asymptotes at all values of \(x\) where \(\tan \ x=0,\) and \(\text{cot}\ x=0\) at all values of \(x\) where \(\tan \ x\) has its vertical asymptotes.

Graphing Variations of

We can transform the graph of the cotangent in much the same way as we did for the tangent. The equation becomes the following.

\[y=A\text{cot}(Bx-C)+D\]

Example

Try it.

Determine the stretching factor, period, and phase shift of \(y=3\text{cot}(4x),\) and then sketch a graph.

Solution

Step 1. Expressing the function in the form \(f(x)=A\text{cot}(Bx)\) gives \(f(x)=3\text{cot}(4x).\)
Step 2. The stretching factor is \(|A|=3.\)
Step 3. The period is \(P=\frac{\pi }{4}.\)
Step 4. Sketch the graph of \(y=3\tan (4x).\)
Step 5. Plot two reference points. Two such points are \((\frac{\pi }{16},3)\) and \((\frac{3\pi }{16},-3).\)
Step 6. Use the reciprocal relationship to draw \(y=3\text{cot}(4x).\)
Step 7. Sketch the asymptotes, \(x=0,\ x=\frac{\pi }{4}.\)

The blue graph in shows \(y=3\tan (4x)\) and the green graph shows \(y=3\text{cot}(4x).\)

Condensed — the full section is in OpenStax Precalculus 2e.

Using the Graphs of Trigonometric Functions to Solve Real-World Problems

Many real-world scenarios represent periodic functions and may be modeled by trigonometric functions. As an example, let’s return to the scenario from the section opener. Have you ever observed the beam formed by the rotating light on a fire truck and wondered about the movement of the light beam itself across the wall? The periodic behavior of the distance the light shines as a function of time is obvious, but how do we determine the distance? We can use the tangent function.

Example

Try it.

Suppose the function \(y=5\tan (\frac{\pi }{4}t)\) marks the distance in the movement of a light beam from the top of a police car across a wall where \(t\) is the time in seconds and \(y\) is the distance in feet from a point on the wall directly across from the police car.

ⓐ Find and interpret the stretching factor and period.
ⓑ Graph on the interval \([0,5].\)
ⓒ Evaluate \(f(1)\) and discuss the function’s value at that input.

Solution

ⓐ We know from the general form of \(y=A\tan (Bt)\) that \(|A|\) is the stretching factor and \(\frac{\pi }{B}\) is the period.
The vertical stretch factor of 5 means that the beam will have moved 5 feet in the one-quarter period before or after the half-period mark. This corresponds to the y-value of the standard tangent function being 1 at one-quarter of the period away from the center of the period, multiplied by the stretching factor of 5.

The period is \(\frac{\pi }{\frac{\pi }{4}}=\frac{\pi }{1}⋅\frac{4}{\pi }=4.\) This means that every 4 seconds, the beam of light sweeps the wall. The distance from the spot across from the police car grows larger as the police car approaches.
ⓑ To graph the function, we draw an asymptote at \(t=2\) and use the stretching factor and period. See
ⓒ period: \(f(1)=5\tan (\frac{\pi }{4}(1))=5(1)=5;\) after 1 second, the beam of has moved 5 ft from the spot across from the police car.

Key Equations

Shifted, compressed, and/or stretched tangent function	\(y=A\ \tan (Bx-C)+D\)
Shifted, compressed, and/or stretched secant function	\(y=A\ \text{sec}(Bx-C)+D\)
Shifted, compressed, and/or stretched cosecant function	\(y=A\ \text{csc}(Bx-C)+D\)
Shifted, compressed, and/or stretched cotangent function	\(y=A\ \text{cot}(Bx-C)+D\)

Key Concepts

The tangent function has period \(\pi .\)
\(f(x)=A\tan (Bx-C)+D\) is a tangent with vertical and/or horizontal stretch/compression and shift. See , , and .
The secant and cosecant are both periodic functions with a period of \(2\pi .\) \(f(x)=A\text{sec}(Bx-C)+D\) gives a shifted, compressed, and/or stretched secant function graph. See and .
\(f(x)=A\text{csc}(Bx-C)+D\) gives a shifted, compressed, and/or stretched cosecant function graph. See and .
The cotangent function has period \(\pi\) and vertical asymptotes at \(0,\pm \pi ,\pm 2\pi ,...\)
The range of cotangent is \((-\infty ,\infty ),\) and the function is decreasing at each point in its range.
The cotangent is zero at \(\pm \frac{\pi }{2},\pm \frac{3\pi }{2},...\)
\(f(x)=A\text{cot}(Bx-C)+D\) is a cotangent with vertical and/or horizontal stretch/compression and shift. See and .
Real-world scenarios can be solved using graphs of trigonometric functions. See .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Sketch a graph of one period of the function \(y=0.5\tan (\frac{\pi }{2}x).\)

Révèle la réponse

First, we identify \(A\) and \(B.\)
Because \(A=0.5\) and \(B=\frac{\pi }{2},\) we can find the stretching/compressing factor and period. The period is \(\frac{\pi }{\frac{\pi }{2}}=2,\) so the asymptotes are at \(x=\pm 1.\) At a quarter period from the origin, we have
\[\begin{array}{l}f(0.5)=0.5\tan (\frac{0.5\pi }{2}) \\ =0.5\tan (\frac{\pi }{4}) \\ =0.5\end{array}\]
This means the curve must pass through the points \((0.5,0.5),\) \((0,0),\) and \((-0.5,-0.5).\) The only inflection point is at the origin. shows the graph of one period of the function.
Sketch a graph of \(f(x)=3\tan (\frac{\pi }{6}x).\)
Graph one period of the function \(y=-2\tan (\pi x+\pi )-1.\)
Révèle la réponse
- Step 1. The function is already written in the form \(y=A\tan (Bx-C)+D.\)
- Step 2. \(A=-2,\) so the stretching factor is \(|A|=2.\)
- Step 3. \(B=\pi ,\) so the period is \(P=\frac{\pi }{|B|}=\frac{\pi }{\pi }=1.\)
- Step 4. \(C=-\pi ,\) so the phase shift is \(\frac{C}{B}=\frac{-\pi }{\pi }=-1.\)
- Step 5-7. The asymptotes are at \(x=-\frac{3}{2}\) and \(x=-\frac{1}{2}\) and the three recommended reference points are \((-1.25,1),\) \((-1,-1),\) and \((-0.75,-3).\) The graph is shown in .
How would the graph in look different if we made \(A=2\) instead of \(-2?\)

Révèle la réponse

It would be reflected across the line \(y=-1,\) becoming an increasing function.
Find a formula for the function graphed in .
Révèle la réponse
The graph has the shape of a tangent function.
- Step 1. One cycle extends from –4 to 4, so the period is \(P=8.\) Since \(P=\frac{\pi }{|B|},\) we have \(B=\frac{\pi }{P}=\frac{\pi }{8}.\)
- Step 2. The equation must have the form \(f(x)=A\tan (\frac{\pi }{8}x).\)
- Step 3. To find the vertical stretch \(A,\) we can use the point \((2,2).\) \[2=A\tan (\frac{\pi }{8}⋅2)=A\tan (\frac{\pi }{4})\]
Because \(\tan (\frac{\pi }{4})=1,\) \(A=2.\)
This function would have a formula \(f(x)=2\tan (\frac{\pi }{8}x).\)
Find a formula for the function in .

Révèle la réponse

\(g(x)=4\tan (2x)\)
Graph one period of \(f(x)=2.5\text{sec}(0.4x).\)
Révèle la réponse
- Step 1. The given function is already written in the general form, \(y=A\text{sec}(Bx).\)
- Step 2. \(A=2.5\) so the stretching factor is \(\text{2}\text{.5}\text{.}\)
- Step 3. \(B=0.4\) so \(P=\frac{2\pi }{0.4}=5\pi .\) The period is \(5\pi\) units.
- Step 4. Sketch the graph of the function \(g(x)=2.5\cos (0.4x).\)
- Step 5. Use the reciprocal relationship of the cosine and secant functions to draw the cosecant function.
- Steps 6–7. Sketch two asymptotes at \(x=1.25\pi\) and \(x=3.75\pi .\) We can use two reference points, the local minimum at \((0,2.5)\) and the local maximum at \((2.5\pi ,-2.5).\) shows the graph.
Graph one period of \(f(x)=-2.5\text{sec}(0.4x).\)

Révèle la réponse

This is a vertical reflection of the preceding graph because \(A\) is negative.
Graph one period of \(y=4\text{sec}(\frac{\pi }{3}x-\frac{\pi }{2})+1.\)
Révèle la réponse
- Step 1. Express the function given in the form \(y=4\text{sec}(\frac{\pi }{3}x-\frac{\pi }{2})+1.\)
- Step 2. The stretching/compressing factor is \(|A|=4.\)
- Step 3. The period is \[\begin{array}{l}\frac{2\pi }{|B|}=\frac{2\pi }{\frac{\pi }{3}} \\ =\frac{2\pi }{1}⋅\frac{3}{\pi } \\ =6\end{array}\]
- Step 4. The phase shift is \[\begin{array}{l}\frac{C}{B}=\frac{\frac{\pi }{2}}{\frac{\pi }{3}} \\ =\frac{\pi }{2}⋅\frac{3}{\pi } \\ =1.5\end{array}\]
- Step 5. Draw the graph of \(y=A\text{sec}(Bx),\) but shift it to the right by \(\frac{C}{B}=1.5\) and up by \(D=1.\)
- Step 6. Sketch the vertical asymptotes, which occur at \(x=0,x=3,\) and \(x=6.\) There is a local minimum at \((1.5,5)\) and a local maximum at \((4.5,-3).\) shows the graph.
Graph one period of \(f(x)=-6\text{sec}(4x+2)-8.\)
Graph one period of \(f(x)=-3\text{csc}(4x).\)
Révèle la réponse
- Step 1. The given function is already written in the general form, \(y=A\text{csc}(Bx).\)
- Step 2. \(|A|=|-3|=3,\) so the stretching factor is 3.
- Step 3. \(B=4,\) so \(P=\frac{2\pi }{4}=\frac{\pi }{2}.\) The period is \(\frac{\pi }{2}\) units.
- Step 4. Sketch the graph of the function \(g(x)=-3\sin (4x).\)
- Step 5. Use the reciprocal relationship of the sine and cosecant functions to draw the cosecant function.
- Steps 6–7. Sketch three asymptotes at \(x=0,\ x=\frac{\pi }{4},\) and \(x=\frac{\pi }{2}.\) We can use two reference points, the local maximum at \((\frac{\pi }{8},-3)\) and the local minimum at \((\frac{3\pi }{8},3).\) shows the graph.
Graph one period of \(f(x)=0.5\text{csc}(2x).\)
Sketch a graph of \(y=2\text{csc}(\frac{\pi }{2}x)+1.\) What are the domain and range of this function?
Révèle la réponse
- Step 1. Express the function given in the form \(y=2\text{csc}(\frac{\pi }{2}x)+1.\)
- Step 2. Identify the stretching/compressing factor, \(|A|=2.\)
- Step 3. The period is \(\frac{2\pi }{|B|}=\frac{2\pi }{\frac{\pi }{2}}=\frac{2\pi }{1}⋅\frac{2}{\pi }=4.\)
- Step 4. The phase shift is \(\frac{0}{\frac{\pi }{2}}=0.\)
- Step 5. Draw the graph of \(y=A\text{csc}(Bx)\) but shift it up \(D=1.\)
- Step 6. Sketch the vertical asymptotes, which occur at \(x=0,x=2,x=4.\)
The graph for this function is shown in .
Given the graph of \(f(x)=2\cos (\frac{\pi }{2}x)+1\) shown in , sketch the graph of \(g(x)=2\text{sec}(\frac{\pi }{2}x)+1\) on the same axes.
Determine the stretching factor, period, and phase shift of \(y=3\text{cot}(4x),\) and then sketch a graph.
Révèle la réponse
- Step 1. Expressing the function in the form \(f(x)=A\text{cot}(Bx)\) gives \(f(x)=3\text{cot}(4x).\)
- Step 2. The stretching factor is \(|A|=3.\)
- Step 3. The period is \(P=\frac{\pi }{4}.\)
- Step 4. Sketch the graph of \(y=3\tan (4x).\)
- Step 5. Plot two reference points. Two such points are \((\frac{\pi }{16},3)\) and \((\frac{3\pi }{16},-3).\)
- Step 6. Use the reciprocal relationship to draw \(y=3\text{cot}(4x).\)
- Step 7. Sketch the asymptotes, \(x=0,\ x=\frac{\pi }{4}.\)
The blue graph in shows \(y=3\tan (4x)\) and the green graph shows \(y=3\text{cot}(4x).\)
Sketch a graph of one period of the function \(f(x)=4\text{cot}(\frac{\pi }{8}x-\frac{\pi }{2})-2.\)
Révèle la réponse
- Step 1. The function is already written in the general form \(f(x)=A\text{cot}(Bx-C)+D.\)
- Step 2. \(A=4,\) so the stretching factor is 4.
- Step 3. \(B=\frac{\pi }{8},\) so the period is \(P=\frac{\pi }{|B|}=\frac{\pi }{\frac{\pi }{8}}=8.\)
- Step 4. \(C=\frac{\pi }{2},\) so the phase shift is \(\frac{C}{B}=\frac{\frac{\pi }{2}}{\frac{\pi }{8}}=4.\)
- Step 5. We draw \(f(x)=4\tan (\frac{\pi }{8}x-\frac{\pi }{2})-2.\)
- Step 6-7. Three points we can use to guide the graph are \((6,2),(8,-2),\) and \((10,-6).\) We use the reciprocal relationship of tangent and cotangent to draw \(f(x)=4\text{cot}(\frac{\pi }{8}x-\frac{\pi }{2})-2.\)
- Step 8. The vertical asymptotes are \(x=4\) and \(x=12.\)
The graph is shown in .
Suppose the function \(y=5\tan (\frac{\pi }{4}t)\) marks the distance in the movement of a light beam from the top of a police car across a wall where \(t\) is the time in seconds and \(y\) is the distance in feet from a point on the wall directly across from the police car.
1. ⓐ Find and interpret the stretching factor and period.
2. ⓑ Graph on the interval \([0,5].\)
3. ⓒ Evaluate \(f(1)\) and discuss the function’s value at that input.
Révèle la réponse
1. ⓐ We know from the general form of \(y=A\tan (Bt)\) that \(|A|\) is the stretching factor and \(\frac{\pi }{B}\) is the period.
  The vertical stretch factor of 5 means that the beam will have moved 5 feet in the one-quarter period before or after the half-period mark. This corresponds to the y-value of the standard tangent function being 1 at one-quarter of the period away from the center of the period, multiplied by the stretching factor of 5.
  
  The period is \(\frac{\pi }{\frac{\pi }{4}}=\frac{\pi }{1}⋅\frac{4}{\pi }=4.\) This means that every 4 seconds, the beam of light sweeps the wall. The distance from the spot across from the police car grows larger as the police car approaches.
2. ⓑ To graph the function, we draw an asymptote at \(t=2\) and use the stretching factor and period. See
3. ⓒ period: \(f(1)=5\tan (\frac{\pi }{4}(1))=5(1)=5;\) after 1 second, the beam of has moved 5 ft from the spot across from the police car.
Explain how the graph of the sine function can be used to graph \(y=\text{csc}\ x.\)

Révèle la réponse

Since \(y=\text{csc}\ x\) is the reciprocal function of \(y=\sin \ x,\) you can plot the reciprocal of the coordinates on the graph of \(y=\sin \ x\) to obtain the y-coordinates of \(y=\text{csc}\ x.\) The x-intercepts of the graph \(y=\sin \ x\) are the vertical asymptotes for the graph of \(y=\text{csc}\ x.\)
How can the graph of \(y=\cos \ x\) be used to construct the graph of \(y=\text{sec}\ x?\)
Explain why the period of \(\tan \ x\) is equal to \(\pi .\)

Révèle la réponse

Answers will vary. Using the unit circle, one can show that \(\tan (x+\pi )=\tan \ x.\)
Why are there no intercepts on the graph of \(y=\text{csc}\ x?\)
How does the period of \(y=\text{csc}\ x\) compare with the period of \(y=\sin \ x?\)

Révèle la réponse

The period is the same: \(2\pi .\)
\(f(x)=\tan \ x\)
\(f(x)=\text{sec}\ x\)

Révèle la réponse

IV
\(f(x)=\text{csc}\ x\)
\(f(x)=\text{cot}\ x\)

Révèle la réponse

III
\(f(x)=2\tan (4x-32)\)
\(h(x)=2\text{sec}(\frac{\pi }{4}(x+1))\)

Révèle la réponse

period: 8; horizontal shift: 1 unit to left
\(m(x)=6\text{csc}(\frac{\pi }{3}x+\pi )\)
If \(\tan \ x=-1.5,\) find \(\tan (-x).\)

Révèle la réponse

1.5
If \(\text{sec}\ x=2,\) find \(\text{sec}(-x).\)
If \(\text{csc}\ x=-5,\) find \(\text{csc}(-x).\)

Révèle la réponse

5
If \(x\sin \ x=2,\) find \((-x)\sin (-x).\)
\(\text{cot}(-x)\cos (-x)+\sin (-x)\)

Révèle la réponse

\(-\text{cot}x\cos x-\sin x\)
\(\cos (-x)+\tan (-x)\sin (-x)\)
\(f(x)=2\tan (4x-32)\)

Révèle la réponse

stretching factor: 2; period: \(\frac{\pi }{4};\) asymptotes: \(x=\frac{1}{4}(\frac{\pi }{2}+\pi k)+8,\text{ where }k\text{ is an integer}\)
\(h(x)=2\text{sec}(\frac{\pi }{4}(x+1))\)
\(m(x)=6\text{csc}(\frac{\pi }{3}x+\pi )\)

Révèle la réponse

stretching factor: 6; period: 6; asymptotes: \(x=3k,\text{ where }k\text{ is an integer}\)
\(j(x)=\tan (\frac{\pi }{2}x)\)
\(p(x)=\tan (x-\frac{\pi }{2})\)

Révèle la réponse

stretching factor: 1; period: \(\pi ;\) asymptotes: \(x=\pi k,\text{ where }k\text{ is an integer}\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

A \cup B,\ A \cap B,\ A \setminus B

union, intersection, difference
In either; in both; in A but not B.

\approx

approximately equal
Equal to the precision shown, not exactly.

\neq

not equal
The two sides are different.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

\binom{n}{k}

binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).

Euler's number
2.71828…, the base whose exponential is its own derivative.

imaginary unit
i² = −1.

\theta

theta
The usual name for an angle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of numbers.

How to: Graphs of the Other Trigonometric Functions

Analyze the graph of y=tan x.
Graph variations of y=tan x.
Analyze the graphs of y=sec x and y=csc x.
Graph variations of y=sec x and y=csc x.
Analyze the graph of y=cot x.
Graph variations of y=cot x.
The stretching factor is
The period is

Questions people ask

What is a function, really?

A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.

Why do we need complex numbers?

Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.

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Parts of this page are adapted from OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Plus en Precalculus

Complex numbers Polynomial functions Rational functions Sequences and series The binomial theorem Conic sections Vectors Exponential and logarithmic functions Polynomial division and the remainder theorem Parametric equations and polar coordinates