In 2013, The Onion thrilled the nation’s math teachers by, at long last, mentioning them:

It’s a headline that bites right to the center of the Tootsie Pop that is our mathematical curriculum. If math is a deafening barrage of arbitrary things to memorize, then trig cranks the volume of gibberish up to 11. I mean, cot^{2}(x) + 1 = csc^{2}(x)? Why do I care again?

Of course, all of trig really just boils down to two functions:

This pair of functions has many offspring, several of which we ask students to learn. Evelyn Lamb lays out some more obscure cousins in a wonderful post:

As Lamb explains, every function in this motley crew once served a useful purpose, back in the days before automated computing. Now, they’re mostly obsolete. So that means we can narrow our attention to the essential few, like sine, cosine, and tangent, right?

Wrong, I say! In the spirit of life imitating The Onion, I propose that all humans memorize the following utterly essential trig functions:

25 thoughts on “13 Trig Functions You Need to Memorize Right Now”

As a German, I am now of course a big fan of deutschetangent. All of these functions should be considered as possible activation functions for neural networks. This would give us beautiful new members of that family like the frankensine networks. There should be ways to combine each of them with the word “logistic” and I am already looking for a PhD student to start work on logistic frankensine networks.

I vote we rename “sine” to “rise”, “cosine” to “run”, and “tangent” to “slope”. There, trig made simple!

Also, all the important trig identities can be derived directly from either Pythagoras, or by drawing a right triangle inside a rectangle like the image in the Angle Sum section on Wikipedia.

True. For high school students with strong skills and investment in math, I think the development of those identities from first principles makes for a nice little case study in proof and the growth of mathematical theory. Sadly, I’ve never seen a standardized test (even IB or A-Levels here in the UK) that pushes the theory in this way; exam writers are much more concerned with getting kids to deploy the formulas than with pushing them to understand their origins.

Ha, that’s fun! I glanced at it long enough to see that every chunk is a full period, but you get your first discontinuity where sin(3x) and sin(4x) are stitched together, and after that I suspect the periods start to begin and end in very peculiar places. I’ll have to explore it further!

Okay, thought about it! For other readers, spoilers follow (unless my reasoning is screwy, in which case, gibberish follows):

The function partitions its domain into many pieces.

Over any given piece, the function completes a full cycle of sin(nx), so the integral over that piece is 0.

The pieces get shorter and shorter, with piece length descending towards 0.

Integrating the function from 0 to n, we include some number of complete pieces (each with integral 0), and one partial piece at the end.

The integral of this partial piece is unknown. But it certainly fits inside a rectangle with height 2 (from the function’s minimum to its maximum) and width equal to the length of the whole piece. Let’s call this our “rectangular bound.”

As n –> infinity, the length of that final partial piece goes to 0. Thus, the area of our “rectangular bound” goes to 0. And so the whole integral goes to 0.

Hence, our final answer is 0.

(Now, if we took the floor function BEFORE taking the limit, we’d be in trouble, because if the integral is slightly negative, it’d go to -1, and if the integral is slightly positive, it’d go to 0. I suspect we’d vacillate unpredictably between these two possibilities, and the limit would not ultimately exist.)

Also, you may notice that my argument never uses the fact that n is an integer. So this improper integral is actually, against all appearances, convergent!

Thanks. You’re the right track. But your explanation needs to be more rigor. Plus, you actually need to have n to be an integer else the limit does not exist.

You can review the solution that was posted by Julian Poon, it’s the rigorous proof needed to solve this.

I’ve enjoyed your blog. Been lurking on this site for years without commenting. This site deserved way more recognition.

Really? I’m not convinced that you need n to be an integer. I haven’t tried writing out a full analytic proof, but give me epsilon > 0, and I can obviously find N such that the final wavelength (of that last little piece) is less than epsilon. If the wavelength is less than epsilon, then the integral (bounded by 1 x that final wavelength) is also less than epsilon. i.e., integral converges to 0. I really can’t see where you need n to be an integer!

Anyway, thanks for reading, the problem is lots of fun, and I’ll check out Julian’s proof.

typo: on the right track *

Do post a solution there while you’re at it! HAHA

I’ll try to write up my solution in a more readable form!

As it is, I think Julian’s solution runs into a problem, because a basically identical argument would tell us that the integral of sin(x) from 0 to infinity is convergent (to 0) as well.

Anyway, I think your frankensine is a super ingenious construction. I’ll be showing it off around my department tomorrow!

I guess it would be fitting to define the ambigusecant using the improvisine?
ambigusecant(Θ) = 0.5 * ( sec(Θ) + csc(Θ) + improvisine(Θ) * (sec(Θ) – csc(Θ)) )

There’s only one trig function! cos(theta) = sin(theta+pi/2). For some reason the intro physics book I teach from switches from one to another while talking about waves, even though there’s a phase already in the definition of a sinusoidal wave. Students who think they’re different functions are perplexed. Sigh.

As a German, I am now of course a big fan of deutschetangent. All of these functions should be considered as possible activation functions for neural networks. This would give us beautiful new members of that family like the frankensine networks. There should be ways to combine each of them with the word “logistic” and I am already looking for a PhD student to start work on logistic frankensine networks.

I’m a quick learner: Cocoasin(pi) = Choco-pi !

http://hanswisbrun.nl/winkel/choco-pi-3/

Although: Cocoacos(pi) fits even better !

http://hanswisbrun.nl/winkel/choco-pi-3/

cococococococosine(θ) = 1 if θ is acting cowardly AND user has never seen an actual chicken; 0 otherwise

Ah yes – abbreviated in some textbooks as gob(θ).

I vote we rename “sine” to “rise”, “cosine” to “run”, and “tangent” to “slope”. There, trig made simple!

Also, all the important trig identities can be derived directly from either Pythagoras, or by drawing a right triangle inside a rectangle like the image in the Angle Sum section on Wikipedia.

True. For high school students with strong skills and investment in math, I think the development of those identities from first principles makes for a nice little case study in proof and the growth of mathematical theory. Sadly, I’ve never seen a standardized test (even IB or A-Levels here in the UK) that pushes the theory in this way; exam writers are much more concerned with getting kids to deploy the formulas than with pushing them to understand their origins.

AWESOME!! LOL

Your post inspired me to post this question: https://brilliant.org/problems/grandiose-gibberish/

Ha, that’s fun! I glanced at it long enough to see that every chunk is a full period, but you get your first discontinuity where sin(3x) and sin(4x) are stitched together, and after that I suspect the periods start to begin and end in very peculiar places. I’ll have to explore it further!

Okay, thought about it! For other readers, spoilers follow (unless my reasoning is screwy, in which case, gibberish follows):

The function partitions its domain into many pieces.

Over any given piece, the function completes a full cycle of sin(nx), so the integral over that piece is 0.

The pieces get shorter and shorter, with piece length descending towards 0.

Integrating the function from 0 to n, we include some number of complete pieces (each with integral 0), and one partial piece at the end.

The integral of this partial piece is unknown. But it certainly fits inside a rectangle with height 2 (from the function’s minimum to its maximum) and width equal to the length of the whole piece. Let’s call this our “rectangular bound.”

As n –> infinity, the length of that final partial piece goes to 0. Thus, the area of our “rectangular bound” goes to 0. And so the whole integral goes to 0.

Hence, our final answer is 0.

(Now, if we took the floor function BEFORE taking the limit, we’d be in trouble, because if the integral is slightly negative, it’d go to -1, and if the integral is slightly positive, it’d go to 0. I suspect we’d vacillate unpredictably between these two possibilities, and the limit would not ultimately exist.)

Also, you may notice that my argument never uses the fact that n is an integer. So this improper integral is actually, against all appearances, convergent!

Thanks. You’re the right track. But your explanation needs to be more rigor. Plus, you actually need to have n to be an integer else the limit does not exist.

You can review the solution that was posted by Julian Poon, it’s the rigorous proof needed to solve this.

I’ve enjoyed your blog. Been lurking on this site for years without commenting. This site deserved way more recognition.

Really? I’m not convinced that you need n to be an integer. I haven’t tried writing out a full analytic proof, but give me epsilon > 0, and I can obviously find N such that the final wavelength (of that last little piece) is less than epsilon. If the wavelength is less than epsilon, then the integral (bounded by 1 x that final wavelength) is also less than epsilon. i.e., integral converges to 0. I really can’t see where you need n to be an integer!

Anyway, thanks for reading, the problem is lots of fun, and I’ll check out Julian’s proof.

typo: on the right track *

Do post a solution there while you’re at it! HAHA

I’ll try to write up my solution in a more readable form!

As it is, I think Julian’s solution runs into a problem, because a basically identical argument would tell us that the integral of sin(x) from 0 to infinity is convergent (to 0) as well.

Anyway, I think your frankensine is a super ingenious construction. I’ll be showing it off around my department tomorrow!

I guess it would be fitting to define the ambigusecant using the improvisine?

ambigusecant(Θ) = 0.5 * ( sec(Θ) + csc(Θ) + improvisine(Θ) * (sec(Θ) – csc(Θ)) )

And don’t forget all their hyperbolic analogues.

It is actually possible for improvisine(x)=improvisine(x).

It’s a set of measure zero, so it’s not possible if you define probability as a ratio of the measures of the success set to the total set of outcomes…

Also, one you must never use:

Cocasine

which is like cocoasine, but faster. It gets its math homework done faster, but then drops out to go be a drug dealer faster.

This was disappointing 😦

Completely ignoring the humorous aspects… 🙂

There’s only one trig function! cos(theta) = sin(theta+pi/2). For some reason the intro physics book I teach from switches from one to another while talking about waves, even though there’s a phase already in the definition of a sinusoidal wave. Students who think they’re different functions are perplexed. Sigh.