\( \definecolor{flatred}{RGB}{192, 57, 43} \definecolor{flatblue}{RGB}{41, 128, 185} \definecolor{flatgreen}{RGB}{39, 174, 96} \definecolor{flatpurple}{RGB}{142, 68, 173} \definecolor{flatlightblue}{RGB}{52, 152, 219} \definecolor{flatlightred}{RGB}{231, 76, 60} \definecolor{flatlightgreen}{RGB}{46, 204, 113} \definecolor{flatteal}{RGB}{22, 160, 133} \definecolor{flatpink}{RGB}{243, 104, 224}\)

February 13, 2021

Fundamental Identities

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Textbook Reference: Section 5.3 The Other Trigonometric Functions

So far you've been introduced to the two most famous circular functions, the sine and cosine, which respectively represent the height and overness of an angle projected onto the unit circle. These two functions are so useful in math as a whole that, if we really wanted, we could do everything for the semester in terms of these two functions. However, there are four useful ratios we can build out of these two functions that are so commonly encountered that we'll give them their own names: the tangent, the cotangent, the secant, and the cosecant.

The tangent function

By far the most useful of these new functions is the tangent function, which is written \(\color{flatgreen}\tan(\theta)\). The most useful intuition to motivate the tangent is this:

The tangent of an angle
is the slope of the terminal side.

Drag the point to change the angle.

Yes, this is the exact same "rise-over-run" slope you've encountered in your previous algebra classes! Remeber that the slope of a line is the ratio between the vertical change (the "rise") and the horizontal change (the "run"): \[{\color{flatgreen}\text{slope}}=\dfrac{\color{flatblue}\Delta y}{\color{flatred}\Delta x}=\dfrac{\color{flatblue}y_2-y_1}{\color{flatred}x_1-x_1}\] Thinking about this in terms of our unit circle, you might realize that this is just the height divided by the overness! This gives us a really nice way to calculate the tangent of any* angle: \[{\color{flatgreen}\tan(\theta)}=\dfrac{\color{flatblue}\sin(\theta)}{\color{flatred}\cos(\theta)}\]

*...well, almost any angle. You'll see what I mean later.

The reciprocal functions

The sine, cosine, and tangent functions are the most well-known circular functions. However, in certain applications, it's sometimes easier to work with their reciprocals, which we'll define as follows:

  • The cosecant function is the reciprocal of the sine function. \[{\color{flatlightblue}\csc(\theta)}=\dfrac1{\color{flatblue}\sin(\theta)}\]
  • The secant function is the reciprocal of the cosine function. \[{\color{flatlightred}\sec(\theta)}=\dfrac1{\color{flatred}\cos(\theta)}\]
  • The cotangent function is the reciprocal of the tangent function. \[{\color{flatlightgreen}\cot(\theta)}=\dfrac1{\color{flatgreen}\tan(\theta)}\]


Be careful not to get these mixed up! It's a common mistake to think that \({\color{flatlightred}\sec(\theta)}=\dfrac1{\color{flatblue}\sin(\theta)}\) and \({\color{flatlightblue}\csc(\theta)}=\dfrac1{\color{flatred}\cos(\theta)}\). A mnemonic that may help is to look at the third letter of each reciprocal function's name:

  • SECANT is the reciprocal of COSINE.
  • COSECANT is the reciprocal of SINE.
  • COTANGENT is the reciprocal of TANGENT.

(By the way, if you want to know where the names "tangent" and "secant" actually come from, you might want to look at Extension Problem 2!)


Suppose \(\color{flatpurple}\phi=\dfrac\pi6\). We can see from the unit circle that \(\color{flatblue}\sin(\phi)=\dfrac12\) and \(\color{flatred}\cos(\phi)=\dfrac{\sqrt{3}}2\).

  • The tangent can be found by dividing the two: \[{\color{flatgreen}\tan(\phi)}=\dfrac{\color{flatblue}\sin(\phi)}{\color{flatred}\cos(\phi)}=\dfrac{\color{flatblue}1/2}{\color{flatred}\sqrt{3}/2}=\dfrac12\cdot\dfrac2{\sqrt{3}}=\color{flatgreen}\dfrac1{\sqrt{3}}\]
  • The other functions are just reciprocals of our other three: \[ \begin{align*} {\color{flatlightblue}\csc(\phi)}&=\dfrac{1}{\color{flatblue}\sin(\phi)}=\dfrac{1}{\color{flatblue}1/2}={\color{flatlightblue}2}\\ {\color{flatlightred}\sec(\phi)}&=\dfrac{1}{\color{flatred}\cos(\phi)}=\dfrac{1}{\color{flatred}\sqrt{3}/2}={\color{flatlightred}\dfrac2{\sqrt{3}}}\\ {\color{flatlightgreen}\cot(\phi)}&=\dfrac{1}{\color{flatgreen}\tan(\phi)}=\dfrac{1}{\color{flatgreen}1/\sqrt{3}}=\color{flatlightgreen}\sqrt{3}\\ \end{align*} \]


Once you get a feel for the circular functions, you start noticing a whole bunch of relationships between them — like the same numbers getting passed around, or certain ways that they complement each other. The equation \[{\color{flatgreen}\tan(\theta)}=\dfrac{\color{flatblue}\sin(\theta)}{\color{flatred}\cos(\theta)}\] is an example of such a relationship — given the values of any two of these basic functions, you can find the third.

This is an example of what's called an identity in mathematics: an equation that's always true for any valid numbers that get plugged into it. Pick any value of \(\theta\) you want, and as long as \(\color{flatblue}\sin(\theta)\), \(\color{flatred}\cos(\theta)\), and \(\color{flatgreen}\tan(\theta)\) are all defined, you can be sure that the above equation will hold true.

We'll see many examples of identities as you learn more about the circular functions. It's important not to think of them as just a bunch of things to memorize — after all, in the real world, you can always just look them up if you need them. What's important is understanding them and seeing where they come from.

A caveat

Remember how I said we can find the tangent of "almost" any angle, and put a note saying "*...well, almost any angle. You'll see what I mean later."?

It's later.

What happens if we try to find the tangent of a right angle (\(\pi/2\) radians)?

Look what happens if we try to use our formula from earlier: \[ {\color{flatgreen}\tan(\pi/2)}=\dfrac{\color{flatblue}\sin(\pi/2)}{\color{flatred}\cos(\pi/2)}=\dfrac{\color{flatblue}1}{\color{flatred}0} \]

Uh-oh. We just divided by zero.

That's a bad thing, right? That's why most people say that the tangent of a right angle is "undefined." They also say that the slope of a vertical line is undefined for the same reason, so that makes sense, since we just said earlier that the tangent and the slope are related. From a function point of view, we would say that \(\pi/2\) isn't in the domain of the tangent function — it's an invalid input.

So when you're calculating these functions, if you run into an angle that makes you divide by zero, you know you've found an angle for which that function is undefined. And you just have to accept that sometimes these functions just don't work.

But I'm going to let you in on a little secret.

If you want, instead of thinking of \(1/0\) as "undefined", you can actually think about it as "infinity", if you're really careful about what that means. Going into how to do that is a bit of a rabbit hole, and you can regard it as optional, but if you want to know how to break the rules and actually divide by zero, you might want to look into it sometime. 😉

Preview Activity 3

Answer these questions and submit your answers as a document on Moodle. (Please submit as .docx or .pdf if possible.)

  1. Find the value of all six circular functions of \(\theta=\dfrac\pi4\).
  2. If \({\color{flatlightblue}\csc(\theta)}=\dfrac1{\color{flatblue}\sin(\theta)}\), what do you think \(\dfrac1{\color{flatlightblue}\csc(\theta)}\) should be, and why?
    What about \(\dfrac1{\color{flatlightred}\sec(\theta)}\) and \(\dfrac1{\color{flatlightgreen}\cot(\theta)}\)?
  3. We saw earlier that \({\color{flatgreen}\tan(\theta)}=\dfrac{\color{flatblue}\sin(\theta)}{\color{flatred}\cos(\theta)}\). What do you think \(\dfrac{\color{flatred}\cos(\theta)}{\color{flatblue}\sin(\theta)}\) should be, and why?
  4. What do you think should be the value of \(\color{flatlightgreen}\cot(\pi/2)\)? Why?
  5. Answer AT LEAST one of the following questions:
    1. What was something you found interesting about this reading?
    2. What was an "a-ha" moment you had while doing this reading?
    3. What was the muddiest point of this reading for you?
    4. What question(s) do you have about anything you've read?


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