8.4 Trigonometric Functions

Learning Objectives

By the end of this section, you will be able to:

• view sine, cosine, tangent, cosecant, secant, and cotangent as functions of an input angle
• use the unit circle to find values for trigonometric functions at common angles
• recognize the graphs of sine, cosine, and tangent on sight, and sketch them yourself

In the previous section, we learned the trigonometric "ratios" defined using the sides of a right triangle, relative to a chosen acute angle. We also saw that trig ratios still make sense even when considering an angle that isn't acute, by drawing a two-dimensional plane and sketching reference triangles. At this point, it'll hopefully be one short step for man to transition to thinking of sine, cosine, and tangent (etc.) as functions. Because side ratios are proportional for similar triangles, it didn't actually matter what exactly the side lengths were in the triangles we were looking at; everything really boiled down to depending on the angle \(\theta\). If we consider \(\theta\) to be the independent input variable, feed it into sine, for example, and see what "\(\sin \theta\)" spits out as an output, we're seeing sine as a function of \(\theta\).

The Unit Circle

An excellent tool to have on instant recall is what's called "the" unit circle. The name "unit" indicates that the radius of the circle is \(1\) (thanks Latin! still undefeated) and the circle is drawn on a 2D plane, centered at the origin. If you draw a ray at a particular angle \(\theta\) on that circle just like we did in the last section, you will pick out a particular point on the circle. The coordinates of that point give you the values of cosine and sine for that angle! Just repeat to yourself "cosine, sine" until you remember the order, because it matters! Here's the angle \( \pi/6\) as an example: \(\cos(\pi/6) = \frac{\sqrt{3}}{2}\) and \(\sin(\pi/6) = \frac{1}{2}\), and as long as the radius is \(1\), those are the exact coordinates of that point on the circle.

If you know your unit circle, you are going to have a MUCH easier time in calculus courses (and beyond). Now, here's the whole thing, but I'm going to give you a trick so you don't have to like rote memorize every tiny detail separately.

When I myself took Calc I, approximately 187 years ago, my professor handed out a blank version of this to us every single day (yes we had class five days a week at 7:45am (uphill both ways my guy)) and made us fill it out from memory. Now, the trick is that all you really need to know is the first quadrant, and you can figure everything else out from there! Observe:

If I know the values at \(\pi/3\), then I can extrapolate that information to find the coordinates for \(2\pi/3\). The two points are the exact same height, so they have the same \(y\)-value. They're also symmetric across the \(y\)-axis, so their \(x\)-values are the same number but opposite sign! If you study the full circle carefully, you will see that using symmetry and common sense, as well as paying attention to which coordinates should be negative or positive according to the \(x\)- and \(y\)-axes, it's a piece of cake to figure out any location with logic! Work smarter, not harder. You only need to find the values that you need in the moment on a problem.

Example \(\PageIndex{1}\)

Use the unit circle to evaluate the trig functions:

1. \( \sin\left( \frac{5\pi}{4}\right)\)
2. \( \cos(\pi)\)
3. \( \tan \left( \frac{11\pi}{6} \right) \)
4. \( \csc\left( \frac{5\pi}{4}\right) \)

Solution

1. Look at the unit circle (or eventually, sketch it yourself) to see that the coordinates at \(5\pi/4\) are \( \left( -\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)\). This makes sense because if I sketch in a reference triangle, it will be 45-45-90, and being in Quadrant III, I have both \(x\)- and \(y\)-coordinates negative. The \(y\)-coordinate is who tells us about sine, so my answer is \(-\frac{1}{\sqrt{2}}\).
2. Looking at \(\pi\), the coordinates are \( (-1,0)\), and the \(x\)-coordinate is who tells us about cosine, so the answer is \(-1\).
3. "Wait a second," you say, "we've been scammed! There's nothing on the unit circle mentioning tangent!" It's okay, we just remember the fact that \( \tan \theta = \dfrac{\sin \theta}{\cos \theta}\) and use the sine and cosine values to compute! Looking at \(11\pi/6\), the sine value is \(-1/2\) and the cosine value is \(\sqrt{3}/2\). Dividing those (order matters!), we get the answer \( -1/\sqrt{3}\).
4. Recall that \(\csc \theta = \frac{1}{\sin \theta}\), so we can just take our answer from part 1 and flip it over: \( - \sqrt{2} \).

Exercise \(\PageIndex{1}\)

Use the unit circle to evaluate the trig functions:

1. \( \sin\left(0\right)\)
2. \( \cos\left(\frac{\pi}{4}\right)\)
3. \( \tan \left( \frac{5\pi}{4} \right)\)
4. \( \sec \left(\frac{\pi}{4}\right)\)

Answer

1. \(0\)
2. \( \frac{1}{\sqrt{2}}\)
3. \( 1\)
4. \( \sqrt{2}\)

The Graphs of Trig Functions

I want to emphasize that the unit circle is a picture tool that we use to figure out the values of sine and cosine, but it is NOT somehow a graph of sine or cosine! Trig functions have graphs like any other function, where the independent variable (on the horizontal axis) is the input \(\theta\), and the output of the function gives you a height or \(y\)-value on the vertical axis. Here are the graphs of the Big Three.

Graph	Observations
	Periodic behavior: The function starts at \((0,0)\), does a hill, does a valley, and returns to the height of 0 by \(x = 2\pi\)...and then it repeats the same pattern. This is called being periodic with period \(2\pi\), because it takes from 0 to \(2\pi\) to do one full cycle. The points called out on the graph are the easiest ones to find on the unit circle, the cardinal points. Sine passes through \( (0,0)\). That is, \(\sin(0) = 0\). He is not symmetric about the \(y\)-axis. He is symmetric about the origin. The function values are bounded between \(-1\) and \(1\). Aka, the function never goes higher than \(1\) or lower than \(-1\). Sine is nice and smooth and continuous, meaning you can draw him without ever having to pick up your pencil. The function is defined for negative values of \(x\): these are just negative angles, which were discussed in the last section and its exercises. In fact, his domain is all real numbers.
	Periodic behavior: Cosine also repeats his behavior, starting the cycle over every time you travel a distance of \(2\pi\). You can see he starts at a height of \(1\) at \(x = 0\), performs a full down-and-up cycle, and comes back to a height of \(1\) at \(x = 2\pi\). The points labeled are again the easiest ones to find from the unit circle. You can compare them to confirm. Cosine passes through \( (0,1)\). That is, \(\cos(0) = 1\). He is symmetric about the \(y\)-axis. His \(y\)-value heights are also trapped between \(-1\) and \(1\). Cosine is also nice and smooth and continuous, with domain all real numbers again.
	Periodic behavior: Tangent repeats himself, but his period is only \(\pi\). Vertical asymptotes: Tangent is undefined at any \(x\)-value that is an odd multiple of \(\pi/2\). This is because at those locations, \(\cos x\) is zero, and you can't compute \(\tan x = \frac{ \sin x}{\cos x}\)! You see vertical asymptotes at those locations, and tangent hugs them but never touches them. This means tangent is not continuous because you have to keep picking up your pencil to hop over those asymptotes! The locations where tangent is \(1\) are \(\pi/4\) and \(5\pi/4\). These are the "quarters" locations in Quadrants I and III, and these are the locations where sine and cosine have the same value. Thus when you divide them to find tangent, you get \(1\). Tangent passes through \( (0,0)\), and passes through the \(x\)-axis at every multiple of \(\pi\). In math symbols, you can write "\(\tan x = 0\) whenever \(x = nx\) for some integer \(n\)." Due to those undefined values, the domain of tangent is all real numbers except odd multiples of \( \pi/2\). You can write this in math as "\(x \neq (2k-1)\frac{\pi}{2}\) for \(k\) any integer." Try some options for \(k\) to see that this expression will give you odd multiples like \( -\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}\), etc.

To sketch these graphs yourself, you really only need a few factoids and a sprinkling of artistic talent, as long as you stare at those pictures until they're burned into your retinas so you have an idea of what trig functions should look like...

To sketch the graph of sine or cosine,

1. Sketch a baby unit circle in the margin and label the cardinal points \( (1,0), (0,1), (-1,0),\) and \( (0,-1)\). These give you the values of sine and cosine at the inputs \(0, \frac{\pi}{2}, \pi, \frac{3\pi}{2},\) respectively.
2. Draw the axes for your function's graph and mark those same angles on the horizontal axis. Then use your unit circle to plot the function's points at those values.
3. Use your knowledge of symmetry and the hilly shape of sine/cosine to smoothly connect the dots and follow the pattern to repeat periodically.

To sketch the graph of tangent,

1. Sketch the \(x\)- and \(y\)-axes and mark the \(x\)-values \(\frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi, \frac{5\pi}{4}, \frac{3\pi}{2}, ...\) etc.
2. Recall the locations where tangent is \(0\) (anywhere that sine is 0): at \(..., -\pi, 0, \pi, 2\pi, 3\pi, ...\) etc. At those locations, the function crosses the \(x\)-axis, so plot those roots.
3. Recall the locations where tangent is \(1\): at \(\frac{\pi}{4}\) and \(\frac{5\pi}{4}\), as discussed above. Plot those points, \( (\pi/4, 1)\) and \(5\pi/4\).
4. Recall the locations where tangent is \(-1\): at \(-\frac{\pi}{4}\) and \(\frac{3\pi}{4}\), etc. Plot those points too.
5. Sketch in the vertical asymptotes: dotted vertical lines passing through \(\frac{\pi}{2}, \frac{3\pi}{2}, \) etc. on the \(x\)-axis.
6. Use your knowledge of the shape of tangent to smoothly connect the dots and follow the asymptotes up and down. Repeat the behavior periodically.

Example \(\PageIndex{2}\)

Sketch the graphs of \(\sin(x)\) and \(\tan(x)\).

Solution

To graph \(\sin x\), I use the unit circle to pick out the points \( (x, \sin x)\) that are easiest: \( (0,0), (\pi/2, 1), (\pi,0), (3\pi/2, -1), (2\pi,0) \). I drop those in, and also note to myself that I'm not going to be going above \(y = 1\) or below \(y = -1\):

Now that I can see how my repeating wave pattern is going to look, I smoothly sketch it in:

To graph \(\tan x\), I plot some signpost points \( (0,0) \), \( (-\pi/4, -1)\) and \( (\pi/4, 1)\), and maybe one more round of that pattern. I also drop in the vertical asymptotes between which I will draw my repeating lines.

Then, knowing that \(\tan\) will go smoothly through these points and hug the vertical asymptotes, and repeat itself periodically, I sketch in the graph.

Exercise \(\PageIndex{2}\)

Sketch the graph of \(\cos x\).

Answer

You can check your answer by looking back at the graph of \(\cos x\) given in the table earlier.

There are of course graphs for the reciprocal trig functions cosecant, secant, and cotangent as well. They are...less intuitive and in my experience don't seem to come up all that much later on? Ask your professor if you need to have these memorized. For now, here's what they look like, just so you're aware. I'll guide you through some exploration of these in the exercises section!