My New Favourite Trig Function

[@blackpenredpen]

cas(x) is such a great idea!

[@FreshBeatles]

Do a video on it :)

[@DrBarker]

@@FreshBeatles I'd like to see that! I'm sure there are lots more fun properties to explore.

[@MyOneFiftiethOfADollar]

I liked Dr Barker’s usage of := as an assignment statement for new function . Have seen some strange mistakes by students with the f(x)=expression form such as x=expression/f , where student was treating it as conditional equation by solving for x😀

[@kristianthulin]

Better than sac(x)

[@orangegaming9562]

@@kristianthulin way better

[@mychessnotebook8653]

Very fun video. Now define cash(x) = cosh(x) + sinh(x), and discuss its properties (smile).

[@pedrosso0]

Cash, money.

[@akssumusic]

Wouldn’t that just be e^x

[@pedrosso0]

@@akssumusic cosh(x)+sinh(x)=e^x/2+e^(-x)+e^x-e^(-x)/2=e^x seems to check out

[@CodingDragon04]

I mean if you invest your cash the amount of cash you have grows exponentially so this makes sense.

[@好吧-h6k]

@@CodingDragon04 If you invest no money, you get a dollar

[@alxjones]

From the derivative, you can see immediately that cas is the negative of it's own second derivative. With a bit of chain rule, we get that cas(wx) are all eigenfunctions of the second derivative operator with eigenvalue -w^2. Of course, this is wholly unsurprising, as cos and sin are also such eigenfunctions, and the eigenspaces are vector spaces, so they are closed under addition.

[@elijahbuck6499]

Actually the second derivative is -sin(x)-cos(x), which is the negative of cas(x), the issue is that you have to remember the rule of derivatives for function concatenations, and we’re combining cas(x) with -x

[@alxjones]

@@elijahbuck6499 Missed a minus sign, as always. Thanks for pointing that out! I've corrected it.

[@evanlewis2349]

This is so interesting because cis(x) is also its negative second derivative. e^ix = y ie^ix = dy/dx -e^ix = d^2y/dx^2

[@alxjones]

@@evanlewis2349 Yep! The sin and cos functions form a basis for the eigenspace over C of the second derivative operator, so any A*sin(wx) + B*cos(wx) will also be an eigenfunction (and conversely, all eigenfunctions have this form) for some complex A, B, w. When we fix w to be real, we force the eigenvalue (-w^2) to be negative. Since cis(x) has A = 1, B = i, and w = 1, it inherits this property as well.

[@Walczyk]

This was too obvious for op to put it in the video

[@zhelyo_physics]

I really like how your videos start with something really simple to digest and over time you are exposed to some brilliant and beautiful maths. Thanks!

[@DrBarker]

Thank you - I'm glad you enjoy them!

[@ConManAU]

What a neat function! If I did everything correctly, I think it’s fairly easy to prove that Acas(x) is actually the unique family of solutions to the DE f’(x) = f(-x). The trick is to differentiate the DE to get f’’(x) = -f’(-x) = -f(x) (being very careful with the chain rule and substitution involved). Then f(x) = Acos(x) + Bsin(x), and putting that into the original DE gives the constraint A = B, so f(x) = Acos(x) + Asin(x) = Acas(x).

[@Jack_Callcott_AU]

This is very useful to me, because many times I have wondered what sin (a) +cos(a) was in terms of other trig functions.

[@FadkinsDiet]

Put it in a graphing calculator like Desmos and play around with it. You'll quickly see it looks like a sinusoidal just scaled and shifted. Then use the sum or difference formula for cos or sin to line it up

[@Jack_Callcott_AU]

@@FadkinsDiet Thanks! From what I gather any linear combination of sin and cos functions is also sinusoidal, but what about sin^2(x) and sqrt(sin(x)) ? 💎

[@Beanybag2]

@@Jack_Callcott_AU sin²x + cos²x = 1, and that's a constant. But ANY smooth, differentiable function can be decomposed into an expanded combination of sinusoidal functions with different phases, that's what the Fourier transform does. So every analytic function could be said to be sinusoidal.

[@Jack_Callcott_AU]

@@Beanybag2 Thanks for the reply! I don't know much really about Fourier series. You know how vast the field of mathematics is, so it's impossible to cover everything adequately. So, a function has to be smooth and differentiable to be expressed as a Fourier series ?

[@PaperQuestion]

@@FadkinsDiet you can also write it as sqrt(2) cos(a-pi/4), or sqrt(2) sin(a+pi/4)

[@gilhmm]

I dont know why i enjoy watching this video. I really appreciate your passion for just playing with math, thanks!

[@DrBarker]

Thank you!

[@orangegaming9562]

Teachers are now gonna force us to remember Cas^2(x) -2sinx*cosx=1

[@DrBarker]

Haha, I bet there are even more trig identities we could find involving cas!

[@phwaedih]

or cas²(x) + cas²(-x) = 2

[@jessstuart7495]

The Hartley transform is very useful for digital signal processing, because you can do everything with plain old real numbers (fast). Unfortunately this gets a lot less attention than it deserves at most universities.

[@DrBarker]

I found the comparison with Fourier transforms interesting when I was reading up on "cas". Any idea why the Hartley transform isn't more well known?

[@jessstuart7495]

@@DrBarker, The Hartley Transform was only developed in 1942. Fourier Transform methods using complex numbers had been established for at-least a century by this time. I think the reason the Hartley Transform isn't well known is a hold-over from pre-computer days. Electrical engineers are taught to analyze circuits and systems using fourier methods (complex frequency domain). When you're solving circuits or calculating Fourier Series coefficients by hand there isn't really much of a penalty for using complex numbers. When you're processing digital signals using a CPU however, being able to use hardware supported data types (real numbers) for calculations instead of a complex data type, can give you a real speed advantage.

[@Beanybag2]

@@jessstuart7495 i found this hard to believe given that a real, discrete Fourier transform can use a real valued matrix and has an exceptionally fast runtime in O(nlogn). Nothing about the unitary group requires working with complex numbers, perse. And so yes, in fact, most of the time the Discrete Hartley Transform is equivalent to the Discrete Fourier Transform. But indeed, it seems in some very special examples, the Hartley Transform does have advantages! Which surprised me a lot.

[@DrBarker]

@@jessstuart7495 That's really interesting - I suppose this change could be possible if universities were to teach Fourier analysis with a much heavier emphasis on practical/computational applications, rather than just the theory and by-hand calculations, which is how I was introduced to the topic.

[@philipoakley5498]

Folks don't like too many choices nor changing the status quo ;-) The Mellin / Wavelet transforms are also of interest and could also be covered.

[@angelmendez-rivera351]

One trigonometric function I like is the function tec(x) = [1 + sin(x)]/cos(x) = sec(x) + tan(x). What I like about this function is that it is to the tangent function what cas is to the sine function. Also, tec'(x) = sec(x)·tec(x), and the Maclaurin expansion of tec has coefficients that are fundamental: they actually provide a deep connection between the Euler numbers and Bernoulli numbers. Also, notice that, since sec(x)^2 - tan(x)^2 = 1, we have that tec(-x) = sec(-x) + tan(-x) = sec(x) - tan(x), so tec(x)·tec(-x) = 1, hence tec(-x) = 1/tan(x).

[@DrBarker]

Really interesting! I didn't know tan(x) + sec(x) had a name. I guess "tan + sec" turns into "tec". Since the Bernoulli numbers appear in the expansion of tan, and the Euler numbers appear in the expansion of sec, I guess it makes sense that they should be related, because tan and sec are closely related. It would be interesting to know how the expansion of tec helps to reveal this connection.

[@That_One_Guy...]

set also works !

[@cycklist]

That's great. How rare to see something completely unfamiliar. Thank you for sharing.

[@petrparizek9945]

You haven't talked about one specific property which I think is super-duper-mega-useful. If you replace cos(x) with cos(-x), the values are exactly the same. If you replace sin(x) with sin(-x), you get the same thing as if you multiplied them all by -1 or as if you phase-shifted the argument by a half-period. However, if you replace cas(x) with cas(-x), you get the same thing as if you phase-shifted the argument by a quarter-period. This means that, unlike ordinary sines and cosines (where you always need to have both of them available), here the one single funcion of cas(x) actually allows you to phase-shift your sinusoid by any possible amount that you desire. The amount of phase-shift depends on the ratio of the amplitudes of the positive-frequency part vs. the negative-frequency part. And this, exactly this is the reason why the so-called "Hartley transform" expresses everything in terms of cas, not sin or cos.

[@DrBarker]

This is really interesting - I hadn't actually seen an explanation before for why cas is used in the Hartley transform, but that makes a lot of sense! Thanks for sharing!

[@petrparizek9945]

When I'm now thinking about it further, I'm realizing that you could even take it from the opposite viewpoint and ask: How exactly should I adjust my sinusoidal function so that f(-x) is equal to phase-shifting f(+x) specifically by a quarter-period? So, you see, I'm deliberately going about finding such a function rather than just picking a function and finding that it meets the condition. And I'll soon realize that cos(x) and sin(x) do indeed differ in phase by a quarter-period. So I might definitely use this fact somewhere. But I also know that sin(-x) is equal to -sin(+x), which means that sin(x) + sin(-x) = 0 for any x. These two equalities can help me a lot. Now suppose that f(x) is equal to sin(x) + something that we don't know what it is, let's call it blabla(x). So we have f(x) + f(-x) = 2×blabla(x), while we also have f(x) - f(-x) = 2×sin(x). So the final part of my question then is: "How do I make the sum equal to 2×cos(x)?" That's easy. Our blabla(x) is actually cos(x), which means that f(x) = cos(x) + sin(x) while f(-x) = cos(-x) + sin(-x) = cos(+x) - sin(+x). And there's our cas(x).

[@RozarSmacco]

Cas(x) is used extensively in the Hartley Transform (Ronald Bracewell)

[@anon6514]

A lovely invention. Really nice how all of the +|- combinations for A and B are represented in Cas(A+B)

[@MyOneFiftiethOfADollar]

Love your usage of := or assignment to create new function. Equals(=) gets used in conditional sense, identity sense and function assignment sense in US! Some educators believe students “should be aware of context” I side with students and use triple horizontal bars for identities to remove possible ambiguity with conditional equality. So what kind of domain restriction do you need for CAS(x) to have an inverse? Sine and cosine are 1-1 on different intervals.

[@DrBarker]

I suppose, ideally, we ought to specify the range of values of x when defining a function, or saying when an equation holds, e.g. cas(x) = cos(x) + sin(x) for all x ∈ ℝ. For the inverse, it's easiest to see from the form cas(x) = sqrt{2} sin(x+pi/2) that we need to restrict to [-sqrt{2}, sqrt{2}].

[@MyOneFiftiethOfADollar]

@@DrBarker so our latest favorite function has a range of [_sqrt(2) , sqrt(2)] , a period of PI, and we can restrict the domain to [PI/4 , 5PI/4] so inverse cas(x) will have a legal residence 😀 this is fun!

[@Muhammed_English314]

@@DrBarker Just a correction: it's cas(x) = √2 sin(x+π/4)

[@leftleg4024]

Triple bars are used for congurences though

[@MyOneFiftiethOfADollar]

@@leftleg4024 you are right. This is quite the crisis! (a,b) can be an interval of real numbers or a point in the plane. What to do about this symbol overload problem that afflicts mathematics?! Triple bar for congruences is accompanied by modulo n to remove any confusion that might arise using same symbol for identities

[@divyansharora6788]

This is Simply Brilliant!! Cas is my favorite trig. function from now on!!

[@philipoakley5498]

The +/- 45 degree (+/- pi/4) part is reminiscent to the Dual Tree Complex Wavelet transform which has a symmetry about the pi/4 point of the regular sin/cosine forms.

[@elliotb1137]

Honestly I like this quite a bit and it would look interesting as a solution to some of those second order ODEs

[@danthewalsh]

The last proof for cas(A+B) seemed rather circuitous. Observe that 1/2(cas(x)+cas(-x))= cos(x). Similarly, 1/2(cas(x)-cas(-x))=sin(x). This allows us to immediately express any results written entirely in terms of standard trig functions by using only cas, which you already have after your very first step when you apply the angle-addition formulae.

[@davidchuong714]

This is a really cool function! Surprised I’ve never heard about it before

[@ilankaboom7009]

Very clear video and at a really nice pace

[@stapler942]

I love the function Computer Algebra System of x!

[@rosskemp2444]

I've seen cis(x) = cos(x) + i*sin(x) used as shorthand.

[@shreysrivastavcivil9667]

Am gonna use it

[@sotirisgkiouleas9521]

cas rules everything around me

[@drdca8263]

I think the derivation at the end would have been simpler/shorter if you first expressed cosine and sine in terms of cas (as the even and odd parts of it), and then plugged in those expressions into the result of the first step.

[@DrBarker]

This is a great suggestion! You're right - it would definitely make the derivation much cleaner. I think this also gives more of an insight into where the factor of 1/2 comes from.

[@saperoi]

That's it, you've convinced me it's the best trig function. I used to prefer cohavercosine

[@vinicus508]

It’s funny that 17 yr old me hated trigonometry cus its so fucking hard and it has so much stuff to it. But now 23 yr old me loves trigonometry cus of how powerful it is and all that stuff I hated about it are actually what makes it so good though hard to comprehend.

[@Luke-mr4ew]

If cas is hard to distinguish from cos, why not rename it to 'sac' for sine and cosine ?

[@DrBarker]

I guess then it might look like "sec", seems like we can't win!

[@frzferdinand72]

sac(x) can make for excellent innuendos and possibly saxophone jokes, I like that better

[@HDitzzDH]

Differentiating and integrating cas(x) is just the conjugate of cas(x) which in itself is pretty interesting.

[@sethdon1100]

Isn’t that sqrt(2)cos(x-pi/4)?

[@odysseasv7734]

Nice stuff really liked the calculation at the end

[@DrBarker]

Thank you!

[@HugoHabicht12]

Love it too, thx for sharing, Dr Barker 🤩

[@FuzzyFirechu]

If you square it it's equal to 1 + sin(2x)

[@That_One_Guy...]

Also equal to 1+(cas(2x)-cas(-2x))/2 or 1 + cas(3pi/4 - 2x)/sqrt(2)

[@curtiswfranks]

It is (up to rescaling) halfway between sin and cos. I have used this notion independently.

[@bigbadbith8422]

Bloody brilliant! Just sent this to my students and they think so too.

[@bigbadbith8422]

Just five months on - and I still think this is brilliant 😀

[@parsecgilly1495]

... the discovery of the century

[@epherum7838]

thats cool and all but what font is that in the thumbnail ? looks really nice

[@Linkedblade]

In my precalc class, grade 11, we used cis(x). Not sure how popular that is and that's cos(x)+i*sin(x)

[@deoradh]

You'll find the cis(x) all over the place in modern physics.

[@marianaldenhoevel7240]

There is a lot of cassing in this video. You might want to add a language warning :-). Thank you very much, I enjoyed it.

[@mesory]

There is also cis(x) which is cos(x) + isin(x) or just eulers formula

[@joaobaptista4610]

Is there any usefull reason why to use cas(x) instead of using the form R*sin(x + theta)?

[@easymathematik]

Its an abbrevation. A couple of hundred years ago trig functions like haversin(x) or versin(x) or coversin(x) and so on were trendy. They are abbrevations for example versin(x) = 1 - cos(x). Is there a special reason for that? Not in the first moment. But in the second. It is sometimes useful to get rid of the constant term in the taylor series expansion. Or similar reasons according to taylor serieses. Some of them were used in sea navigation. To write sin+cos as cas is not a bad idea. No sqrt(2), no shift. Why not. :)

[@CppExpedition]

WAIT A MINUTE, THAT'S PRETTY INTERESTING!

[@novemtrigintillionaire7684]

If cas stands for cos and sin, that will make me extremely happy.

[@joefarrow1599]

I guess it would be natual to define cms(x) = cos(x) - sin(x), and then ask if any trig identities look natural in the new cas/cms basis. I guess cms(a+b) might depend only on cms and not cas, and you have cms^2 + cas^2 = 2, cms'(x) = -cms(-x) and cms(-x) = cas(x), cas(-x) = cms(x). It's nice that the derivatives and the addition identity decouple, so maybe some more complex identities would decouple or have some interesting properties too Edit: I see that it _looks_ like cas(a+b) decouples, but that's because cas(-x) = cms(x), so I would rather say that it hasn't decoupled, it's just written in terms of both functions and the net result is to turn a two term identity for cos(a+b) into a four term one in cas and cms, so maybe not so useful

[@QuiescentPilot]

cas(x), meaning cosine and sine? I think I prefer “sine and cosine.” sac(x)

[@bernhardbauer5301]

Don't we have already enough trig-functions? Do we really need sek and cosek, cas and casek and all these stuff?

[@mohammedbasharat9584]

You refer to a paper on fourier analysis where cas(x) first appeared? Any more on this?link to the paper?

[@DrBarker]

Here you go: gwern.net/doc/math/1942-hartley.pdf

[@ultimatedude5686]

It seems like cas is it's own cofunction!

[@DrBarker]

This is a fun fact! I guess it was inevitable since cos are sin are each others' cofunctions.

[@DatBoi_TheGudBIAS]

doesn't Dat mean dat cash (muny) =e^x?

[@dr.mikelitoris]

Yessir Dr goatker uploaded

[@jorgepedreirapedreira678]

It's really really cool idea...

[@Tydox]

cas(x) reminds me of cis(x), The question is can cas(x) be useful for anything? Shorten some proofs or it’s properties be any useful? Maybe some math majors can share their opinion/experience.

[@DrBarker]

The only application I'm aware of is the Hartley transform (related to Fourier transforms), which involves an integral of f(t)cas(wt)dt. There must be lots of papers out there which use "cos(x) + sin(x)" at some point though.

[@resonatedvirtue8746]

Great video sir! I'd love to discover where this would be of use tho Also, if this makes it into Indian textbooks Engineering entrances are gonna be hell lot tough.

[@mr.d8747]

*But how did cas(A)cos(B) + cas(-A)sin(-B) turn into -cas(A+B)?*

[@SimchaWaldman]

I shall now cry from joy! 😭😭😭

[@pauljackson3491]

1 problem is, as you said, cas(x) is similar to cos(x). Now do exsecant and versin. and halversin.

[@DrBarker]

That's a fun idea! There's something very intriguing about those old "forgotten" trig functions.

[@Tom-sp3gy]

Beautiful !

[@abdullahahmed3843]

It’d make differentiation so much easier

[@michatarnowski5905]

Thats The first time I see someone writing “x” this way.

[@FreeGroup22]

duh, your function is just sqrt(2)cos(x-pi/4)

[@rogerluo7753]

It's sad that my cas do not have CAS(X)

[@kepler4192]

Now we need arcas(x), cash(x), and arcash(x)

[@DrBarker]

cash(x) is a very interesting point! It's actually just e^x.

[@kepler4192]

@@DrBarker that’s actually really cool!

[@That_One_Guy...]

Here it is : - arccas(x) = pi/4 -+ arccos(x/sqrt(2)) i'm not really sure how to convert arccos into arccas though - cash(x) just like Dr Barker said - arccash(x)= ln(x)

[@kepler4192]

@@That_One_Guy... interesting

[@lox7182]

by the way you can use (cas(A) +cas(-A))/2 = cos(A) and (cas(A) - cas(-A))/2 = sin(A) to your advantage

[@j.r.8176]

isn't it just sqrt(2)sin(x- pi/4)?? why do we need a new function? it's just sin

[@bartolhrg7609]

Taylor expansion should be good?

[@bluelive23]

shifted by 1/4th pi and amplified by sqrt(2)

[@rosiefay7283]

1/4th? Is that 1/2nd of 1/2nd?

[@ThomasDeLello]

Wouldn't cas(x) always be = 0...? Wouldn't the resultant of a sine wave and a cosine wave built on the same variable be a flat line...?

[@nachiketagrawal5154]

No. Take x=45° as an example. cas(x) = cos (x) + sin(x) If cas was always 0, it would imply for all x, cos(x) = -sin(x) ie tan(x) = -1 which is clearly not always the case

[@Qsdd0]

Ah, but normalizing cas to sin(x+pi/4) gives us an even prettier addition rule getting rid of the coefficient.

[@rosiefay7283]

Exactly, proving that, with sin and cos, cas is needless.

[@Bruh-bk6yo]

cas(x)=cosx+sinx sas(x)=cosx-sinx ?

[@DrBarker]

I don't think a standard function "sas(x)" exists. I guess it can be whatever you want it to be!

[@stardust-r8z]

Getting the addition formula for cas would be much faster if we first notice: casx = sinx + cosx cas(-x) = -sinx + cosx cosx = 1/2 (casx + cas(-x)) sinx = 1/2 (casx - cas(-x)) They look like the exponential forms for sine and cosine :) cas(a+b) = sin(a+b) + cos(a+b) = sinacosb + sinbcosa + cosacosb - sinasinb = cosb(sina + cosa) + sinb(cosa - sina) = cosbcasa + sinbcas(-a) = 1/2 (casb + cas(-b))casa + 1/2 (casb - cas(-b))cas(-a) = 1/2 (casacasb + casacas(-b) + cas(-a)casb - cas(-a)cas(-b)) Or, we can also notice (motivated by addition formulas for sine and cosine): (The terms with * are those with signs opposite to that in the cas(a+b) expansion) cas(a+b) = sinacosb + sinbcosa + cosacosb - sinasinb casacasb = (sina + cosa)(sinb + cosb) = *sinasinb + sinacosb + cosasinb + cosacosb casacas(-b) = (sina + cosa)(-sinb + cosb) = -sinasinb + sinacosb - *sinbcosa + cosacosb cas(-a)casb = (-sina + cosa)(sinb + cosb) = -sinasinb - *sinacosb + cosasinb + cosacosb cas(-a)cas(-b) = (-sina + cosa)(-sinb + cosb) = *sinasinb - *sinacosb - *sinbcosa + cosacosb Negate cas(-a)cas(-b) to reduce the number of opposite-sign terms: -cas(-a)cas(-b) = -sinasinb + sinacosb + sinbcosa - *cosacosb We notice that the * terms add up to cas(a+b), so we get: cas(a+b) = casacasb + casacas(-b) + cas(-a)casb - cas(-a)cas(-b) - cas(a+b) cas(a+b) = 1/2 (casacasb + casacas(-b) + cas(-a)casb - cas(-a)cas(-b))

[@whatelseison8970]

I don't know how you managed to say and write cos(B) so many times without so much as a hint that you even wanted to say "Hey hey hey" in a Bill Cosby voice.

[@bernardlemaitre4701]

very clever ! and sinx + cosy = 2 sinx+y /2 + ....

[@DepFromDiscord]

Why not write (sqrt(2))/2 as 1/(sqrt(2))?

[@samueldevulder]

This is a cos(B) show !

[@alexsoft55]

Maybe you can define casi(x):=cos(x) + isin(x)

[@gudmundurjonsson4357]

Already exists, called cis(x) :)

[@77Chester77]

Very interesting. Good content thou!

[@ddiq47]

The final result reminded me of truth tables

[@Nicolobos77]

is it casin?

[@quantummekanix5132]

what about the integral? I got -cas(x)

[@quantummekanix5132]

+c

[@attilatakacs8673]

R is also could be 2/root2

[@pl4t1n00b]

My personal favorite is tan(x) + sec(x)

[@MartinMllerSkarbiniksPedersen]

Does all americans write x as two curves? In Europe I think everyone is using two straigth lines.

[@rodge4411]

Dr Barker is English. In England x is written as two curves (in mathematics)

[@MartinMllerSkarbiniksPedersen]

@@rodge4411 OK. I haven't seen that in Denmark but it was also 30+ years since I went there.

[@MartinMllerSkarbiniksPedersen]

@@rodge4411 Does pupils in schools also learn that or only in college/university in UK?

[@dmitripogosian5084]

It is just that modern students tend write everything in block letters and have problems with cursive writing

[@MartinMllerSkarbiniksPedersen]

@@MaruSurfs ? Really ? Multiply is a small dot or nothing. Not x ?!

[@mirkodobrota4861]

where is cosA*cosB term

[@gudmundurjonsson4357]

You could probably make a horrible looking differential equation by substituting the derivative into the a+b identity. Fun function

[@DrBarker]

This is a fun idea! I wonder if cas is the only function which satisfies the functional equation f(a+b) = f(a)f(b) + f(a)f(-b) + f(-a)f(b) - f(-a)f(-b)?

[@PubicGore]

@@DrBarker Future video, perhaps?

[@minimath5882]

cas(2x)=?

[@TheFerdi265]

nice, so cas(A+B) is actually 1/2 * (cas(A) + cas(-A)) * (cas(B) + cas(-B))

[@tatecolores]

Spanish subtitles?

[@akssumusic]

cas^2 (x) = sin(2x) + 1

[@easymathematik]

It´s get interesting, if you consider cas( x_1 + x_2 + ... + x_n) = f( cas(x_1), cas(x_2), ..., cas(x_n) ) Funny task.

[@sphakamisozondi]

Mmmh, I kind of like the idea of the cas(x) function.

[@whatitmeans]

First time a saw someone writing the letter "x" differently from doing just two crossed straight lines hahaha

[@borekking4769]

Nice function:)

[@Tomaplen]

@drpeyam fook it doesnt work

[@syedmdabid7191]

Versx=1--sinx, covers(x) =1--cosx, Cas(x) = sinx+cosx

[@emanuellandeholm5657]

Then we have the Euler function caz(x) = cos x + i sin x

[@TooManyEditsProductions]

I've never heard my name so many times before

[@benamiel6180]

In israel we use cis(x) as cosx +isinx. My fav

[@hgfuhgvg]

I got lost in the final moments

[@ciemathhomework405]

This is interesting: if f(x) = cas(x), f''(x) = f(-x) = -f(x)

[@grinreaperoftrolls7528]

There’s a bill cos(B) joke just waiting to be made