In part 1 of this blog series, I shared a foolproof method for graphing transformed sine and cosine functions using the Eight Blanks method.  I am going to show you how to extend this method to cover the other four trigonometric functions you typically encounter in Trig class: tangent, cotangent, secant, and cosecant.

The base tangent and cotangent graphs

As before, I want to start by first examining the base graphs for tangent and cotangent. If you have never seen these in class, there are several websites where you can learn the basics of them, including MathHints and Purplemath.

Here is the base tangent curve:

Base tangent curve. Generated by Desmos.

Like the sine and cosine curves, the tangent curve can be characterized by five landmark positions, marked in the graph above with green arrows. However, there are several differences between the landmarks here and the ones for sine and cosine:

• At Landmarks 1 and 5, the curve crosses the centerline, just like sine, but Landmark 3 is a vertical asymptote, and the actual curve does not touch this asymptote.
• The tangent curve does not have an amplitude, as such, since it goes all the way to positive and negative infinity in the vertical direction. However, at Landmark 2, the curve is exactly one unit above the centerline, and at Landmark 4 the curve is exactly one unit below the centerline. So in the Eight Blanks method, these locations are going to fill in for our “amplitude”.
• Important! The period is [math]\pi[/math], not [math]2\pi[/math] as it is for sine and cosine

Some clients find this view of tangent a little confusing, as they are used to seeing the tangent curve as one continuous curve that goes from [math]x = -\frac{\pi}{2}[/math] to [math]x = \frac{\pi}{2}[/math]. But the Eight Blanks method requires that the base curve start from [math]x = 0[/math], so I will always display it in “two pieces” as shown above.

Here is the curve for cotangent:

Base cotangent curve. Generated by Desmos.

For the cotangent curve,

• Landmarks 1 and 5 are vertical asymptotes
• Landmark 3 is the centerline crossing.
• As with tangent, Landmark 2 is exactly one unit above the centerline, and Landmark 4 is exactly one unit below the centerline.  These will serve as our “amplitude” points.
• As with tangent, the period is [math]\pi[/math], not [math]2\pi[/math].

Transformations of tangent and cotangent graphs

Just as with sine and cosine, we can write transformed versions of the tangent and cotangent graphs in the form

[math]f(x) = A\; tan [b(x\, -\, h)]\; +\; k[/math]       OR       [math]f(x) = A\; cot [b(x\, -\, h)]\; +\; k[/math]

where

• [math]A[/math] is the “amplitude” / vertical stretch. Remember that the tangent and cotangent curves don’t really have an amplitude, but we are still going to use this value to set the height of the curve at Landmarks 2 and 4.
• [math]b[/math] is the horizontal stretch. This is not the period, but the period can be calculated from [math]b[/math] using the formula: [math](Period) = \frac{\pi}{b}[/math]. Note that this is not the same as the formula for sine and cosine! We write it as [math]\frac{\pi}{b}[/math] and not [math]\frac{2\pi}{b}[/math] because the base period for tangent and cotangent is [math]\pi[/math], not [math]2\pi[/math].
• [math]h[/math] is the horizontal shift.
• [math]k[/math] is the vertical shift.

The Eight Blanks method for tangent

Let’s see how we can implement the Eight Blanks method for a transformed tangent curve. Suppose I am asked to graph the function

$$y = 4\; tan [\frac{3}{2}(x\; – \frac{\pi}{3})]\; -\; 1$$

As before, I will start by figuring out what all the parameters are.

• The “amplitude” is [math]A = 4[/math].
• The horizontal stretch is [math]b = \frac{3}{2}[/math], so the period is [math]\pi / (\frac{3}{2}) = \frac{2\pi}{3}[/math].
• The horizontal shift is [math]h = \frac{\pi}{3}[/math] (to the right).
• The vertical shift is [math]k = -1[/math] (down).

As before, I will draw my Eight Blanks:

I’ll start with the three vertical blanks. This part is just the same as it was for sine and cosine:

• The middle blank (centerline) is the value of [math]k[/math], so I write [math]-1[/math] here.
• The top blank takes [math]k + A[/math], so I write [math](-1) + (4) = 3[/math].
• The bottom blank takes [math]k\; – A[/math], so I write [math](-1)\: – (4) = -5[/math].

Now I will set the five horizontal blanks. This also repeats the process I used for sine and cosine:

• The first blank is the value of [math]h[/math], so I write [math]\frac{\pi}{3}[/math] here.
• The last blank is the value in the first blank plus one whole period, so in this space I will place [math](\frac{\pi}{3}) + (\frac{2\pi}{3}) = \pi[/math].
• The third blank takes the average between the first and last blanks, so I compute [math]((\frac{\pi}{3}) + (\pi)) / 2 = \frac{2\pi}{3}[/math] and put that value in the third blank space.
• The second blank is the average between the first and third, so in this space will go [math]((\frac{\pi}{3}) + (\frac{2\pi}{3})) / 2 = \frac{\pi}{2}[/math].
• Finally, the fourth blank is the average between the third and fifth blanks, so I will write [math]((\frac{2\pi}{3}) + (\pi)) / 2 = \frac{5\pi}{6}[/math] here.

Here is the complete set of Eight Blanks:

Now I am ready to plot points, using the landmark pattern for tangent:

• Landmark 1 crosses the centerline, so the coordinates are (horizontal position 1, centerline) = [math](\frac{\pi}{3}, -1)[/math].
• Landmark 2 is one “amplitude” above the centerline, so its coordinates are (horizontal position 2, top) = [math](\frac{\pi}{2}, 3)[/math].
• Landmark 3 is a vertical asymptote, so I place the asymptote at [math]x =[/math] (horizontal position 3) [math]= \frac{2\pi}{3}[/math] and don’t plot a point here.
• Landmark 4 is one “amplitude” below the centerline, so the coordinates should be (horizontal position 4, bottom) = [math](\frac{5\pi}{6}, -5)[/math]
• Finally, Landmark 5 crosses the centerline again, so I pick coordinates (horizontal position 5, centerline) = [math](\pi, -1)[/math].

And here is the graph:

Transformed tangent curve. Generated by Desmos.

The Eight Blanks method for cotangent

The process for graphing cotangent is practically identical to the process for graphing tangent, except that the vertical asymptotes and the centerline points are at different landmarks. Suppose I want to graph

$$y = 3\; cot [\frac{1}{2}(x + \frac{\pi}{4})]\; +\; 2$$

Again I will start with the parameters:

• The “amplitude” is [math]A = 3[/math].
• The horizontal stretch is [math]b = \frac{1}{2}[/math], so the period is [math]\pi / (\frac{1}{2}) = 2\pi[/math].
• The horizontal shift is [math]h = -\frac{\pi}{4}[/math] (to the left).
• The vertical shift is [math]k = 2[/math] (up).

I will draw my Eight Blanks as usual and start with the three vertical blanks:

• The middle blank (centerline) is the value of [math]k[/math], so I place a [math]2[/math] here.
• The top blank is [math]k + A[/math], so I write [math](2) + (3) = 5[/math].
• The bottom blank is [math]k\; – A[/math], so I write [math](2)\: – (3) = -1[/math].

Now for the five horizontal blanks:

• I put the value of [math]h[/math] into the first blank, so I write [math]-\frac{\pi}{4}[/math] here.
• In the last blank I put the value in the first blank plus one whole period, so in this space I will place [math](-\frac{\pi}{4}) + (2\pi) = \frac{7\pi}{4}[/math].
• In the third blank, I place the average between the first and last blanks, so I compute [math]((-\frac{\pi}{4}) + (\frac{7\pi}{4})) / 2 = \frac{3\pi}{4}[/math] and put that value in the third blank space.
• In the second blank, I am supposed to put the average between the first and third, so in this space will go [math]((-\frac{\pi}{4}) + (\frac{3\pi}{4})) / 2 = \frac{\pi}{4}[/math].
• Finally, the fourth blank is to be filled with the average between the third and fifth blanks, so I will write [math]((\frac{3\pi}{4}) + (\frac{7\pi}{4})) / 2 = \frac{5\pi}{4}[/math] here.

Here is the complete set of Eight Blanks:

As before, I will use the information from the Eight Blanks to plot points:

• The first and fifth landmarks are asymptotes, so I place an asymptote at both [math]x =[/math] (horizontal position 1) [math]= -\frac{\pi}{4}[/math], and [math]x =[/math] (horizontal position 5) [math]= \frac{7\pi}{4}[/math]
• The second landmark is one “amplitude” above centerline, so I place it at (horizontal position 2, top) = [math](\frac{\pi}{4}, 5)[/math]
• The third landmark goes back to the centerline, so the next point goes to (horizontal position 3, center) = [math](\frac{3\pi}{4}, 2)[/math]
• Finally, the fourth landmark is placed one “amplitude” below the centerline, so it goes to (horizontal position 4, bottom) = [math](\frac{5\pi}{4}, -1)[/math]

And here is the graph:

Transformed cotangent curve. Generated by Desmos.

Secant and cosecant graphs

Finally, let’s look at transformations of secant and cosecant curves. I am actually going to start with cosecant because it is a little easier to visualize over one period.

Here is the base graph for cosecant:

Base cosecant curve. Generated by Desmos.

As sites such as Purplemath or MathHints will tell you, the cosecant curve (in blue) looks like a set of alternating right-side up and upside down U’s. I intentionally placed the sine curve (in green) on the same graph as the cosecant curve. Since the cosecant is the reciprocal of sine, these two curves have a special relationship:

• At each horizontal position where the sine curve has an x-intercept, the cosecant curve has a vertical asymptote, which the curve itself does not touch. This occurs in the graph above at Landmarks 1, 3, and 5.
• If we look at both the peak (Landmark 2) and the trough (Landmark 4) of the sine curve, the secant curve just barely “kisses” the sine curve in those two locations.

A similar relationship exists between the secant curve and the cosine curve:

Base secant curve. Generated by Desmos.

This suggests a convenient strategy for drawing these curves: when we want to create a secant or cosecant graph, we can first draw the corresponding cosine or sine graph, and then fit the secant or cosecant curve on top of these, placing both the asymptotes and the U’s according to the shape of the sine or cosine.

Let’s try a couple of examples.

Transformations of secant graphs

Suppose I am asked to draw the graph for

$$y = \frac{3}{2}\; sec [\frac{\pi}{3} (x\: -\: 2)]\: +\: \frac{5}{2}$$

I am going to start by creating the graph for the corresponding cosine equation:

$$y = \frac{3}{2}\; cos [\frac{\pi}{3} (x\: -\: 2)]\: +\: \frac{5}{2}$$

Fortunately, I already worked out this curve in Part 1 of this blog, so I can immediately plot out the correct coordinate points for the five landmarks: [math](2, 4)[/math], [math](\frac{7}{2}, \frac{5}{2})[/math], [math](5, 1)[/math], [math](\frac{13}{2}, \frac{5}{2})[/math], and [math](8, 4)[/math], after which I will draw out the curve.

Transformed cosine curve. Generated by Desmos.

Now I can draw the actual transformed secant curve. At every point at which the cosine curve crosses the centerline, I place a vertical asymptote, and then draw my U’s so that they just barely touch the peaks and troughs of the cosine:

Transformed secant curve. Generated by Desmos.

If I want my answer to look cleaner, I can choose to erase my cosine curve before turning in my work.

Transformations of cosecant graphs

I can follow the same process for drawing cosecant graphs. Suppose I want to graph

$$y = 3\; csc [2(x\: +\: \frac{\pi}{6})]\; -\; 1$$

Just as before, I will first graph the corresponding reciprocal trig function, in this case sine:

$$y = 3\; sin [2(\:x +\: \frac{\pi}{6})]\; -\; 1$$

which again I already graphed in Part 1 of this blog. I found at that time that the landmark points were [math](-\frac{\pi}{6}, -1)[/math], [math](\frac{\pi}{12}, 2)[/math], [math](\frac{\pi}{3}, -1)[/math], [math](\frac{7\pi}{12}, -4)[/math], and [math](\frac{5\pi}{6}, -1)[/math]. I can use these points to plot out the sine curve:

Transformed sine curve. Generated by Desmos.

To generate the transformed cosecant curve, I will place vertical asymptotes on the centerline crossings and draw U’s to touch the peaks and troughs:

Transformed cosecant curve. Generated by Desmos.

I hope you find my post helpful! If you would like more help for any level of Math, I would love to work with you on 1:1 tutoring. Please see my website at https://andrewjeungtutoring.com .