USA here

Assume that
[math]a = adjacent\\
h = hypotenuse\\[/math]
Therefore, we can say that:
[math]\cos(\theta) = \frac{a}{h}\\
\sec(\theta) = \frac{1}{\cos(\theta)}\\
\sec(\theta) = \frac{1}{\frac{a}{h}}\\
\sec(\theta) = \frac{h}{a}[/math]

Which makes more sense because it doesn't assign an arbitrary function to an easily manipulatable fraction (which the secant function is). Sure, secant makes things look cleaner, but overall, more annoying to understand.

My teacher taught us the definition of sec and csc and called them "irrelevant trig functions that are going the way of the versine and vercosine".

>not using cos^-1
Bunch of brainlets in this thread

We use sec

t. Britbong

lul

czech student here, confirming we never use sec

Yes, the only notation that we use is cos, sin, tan and their inverses; nothing else.
I get mad when I see sec and the other stuff Americans use. They are pointless and I always have to look up what they mean.

oh the only exception (in Greece at least) is in schools where they use 1/tan=cot (1/εφ=σφ)

It's completely useless. In any question/problem that involves sec, cosec, cot, the first thing anyone ever does is change it to 1/cos, etc, so there is literally no point other than to look """neater""".

Who gives a fuck? Personally I appreciate secant having its own name because of the geometric motivation. It's a bit of notation that references the meaning of the function in a non-cumbersome way.

because in europe people are expected to solve things by hand at least to undergraduate level. and it allows to transform equations more easily

Because Eurofags are complete idiots. If they used 1/cos instead of sec, they should also be using sin/cos instead of tan. Also if you take the derivative of tan, you get sec^2. Taking the derivative of sec, you get sec*tan. Eurofags don't like making patterns to remember these handy dandy derivatives and integrals, so they insist that Americans are stupid for remembering one more notation.

Are you really this stupid?

Is that the smartest thing you could come up with? It's true and you don't want to admit it

arccosine

THIS

sec is just another way of saying 1/cos, now tan is a way of solving sin and cos.

can confirm
t.spaghetti

I've always wondered why some people use this dumb sec/cot notation instead of calling things what they are. Thanks to OP, I now know that the culprits are once again retarded burgers and their autistic notation and units.

I did some reading and it seems that the reason for labelling these functions inependently is because back in the day when you had to read from tables if you computed it from cos tables you'd get an error. Best to have it's own dedicated table.

isn't sec useful for trig identities?

no, it just makes them more complicated and harder to memorize

you don't need to memorize them if you understand how they work

Because Yuros are brainlets who need everything spelled out for them. All the Sec formulas are easier to remember than theid 1/cos equivalent. It's cleaner and more convenient to study and use. Since Sec, Csc and Cot have behave similary in many cases. If you consider these meaningless why don't you just use sqrt of -1 instead of i?

brainlet spotted
i != sqrt(-1)

Hihi, Ballsec...
Wait a sec, that means dry in French.
Join the sec-t.
I like this sec-tion.
No, this information is sec-ret.
Don't post this on Veeky Forums, we need /sec/.
Secc, the retarded version of succ.
This guy never had secs.

you can say that sec is another way of solving 1/cos as well. Sec is a good notation, especially when remembering simple derivatives. If you use 1/cos each time, first it's not a beautiful way of describing it, and its harder to do simple calculations, since you always have to account for 1/cos. This is even more complicated when doing integrals with multiple powers of trig identities. Its just much more simpler using things like sec and csc since they have a general simplification of their derivatives and integrals

Why to waste precious brain power to remember them all? Sin and Cos is all you need.

Is it 1994 already?

>not using [math]\sin x, \sin \left(\frac{\pi}{2}-x \right), \frac{\sin x}{\sin \left(\frac{\pi}{2}-x \right)}, \frac{\sin \left(\frac{\pi}{2}-x \right)}{\sin x}, \frac{1}{\sin \left(\frac{\pi}{2}-x \right)},
\frac{1}{\sin x}[/math]

Brainlets please

>not using [math]\sin x, \sin \left(\frac{\pi}{2}-x \right), \frac{\sin x}{\sin \left(\frac{\pi}{2}-x \right)}, \frac{\sin \left(\frac{\pi}{2}-x \right)}{\sin x}, \frac{1}{\sin \left(\frac{\pi}{2}-x \right)}, \frac{1}{\sin x}[/math]

I've just learnt what's a sec thx op. So far this board has taught me :
[x] tau is another way to say 2 pi
[x] sec is the inverse of a cosine
[] genuine new knowledge and not just facetious other way to depict common concepts

Aussie here. What is cos? We only use sin/tan. Makes more cents.

My teacher (asian) said cos was goin the way of the white man.

why would you need to remember any trigonometric formulas you brainlet? just fucking learn how to derive them

>not using glorious symmetric function [math]\cos x[/math]

Veeky Forums taught me Latex. If you've been here for a month and you haven't learned a thing then you may be an absolute brainlet. Go lurk the /sqt/.

>>not using sinx,sin(π2−x),sinxsin(π2−x),sin(π2−x)sinx,1sin(π2−x),1sinx
not using [math] \sin x, \sqrt{1-\sin^2 x}, \frac{\sin x}{\sqrt{1-\sin^2 x}} [/math]

>
> >not using glorious symmetric function cosx
Because it's better a function as it is 0 at x=0.

>USA here.
>Today my Calculus II professor mentioned that in Europe (He's European) they do not use sec, but rather just 1/cos.
>Just looking for conformation on this, and if so why? Is there some advantage I just don't see?
I'm Spanish and I use the sec function.... But I've never used the versed functions or the excess functions.

>h-hey guys wait up, im still deriving the formula to integrate this simple problem i could just apply a formula too and finish in a second

>h-hey guys wait up, this problem isn't one I've memorised the formula for, what do I do?! Halp!

It's defined as sqrt(-1) in many books

Europeans don't need cookie cutter recipes for indentities and derivates, so we don't use them.
I understand the quality of US education is at an all time low and it's easier for students to remember some formulas than to understand what is actually happening.

back in the day every mathematician had a book of trig tables

they used different named functions because then you would then have them listed in the table.

that's why we have all these uniquely named trig functions

really all trig functions are derived from sin though

>[math]\pi[/math]

Holland and Germany
We use 1/cos
Why would you want to memorise more functions

>not using [eqn]\frac{e^{ix}-e^{-ix}}{2i},\ \frac{e^{ix}+e^{-ix}}{2},\ i\frac{1-e^{2ix}}{1+e^{2ix}},\ i\frac{1+e^{2ix}}{1-e^{2ix}},\ \frac{2}{e^{ix}+e^{-ix}},\ \frac{2i}{e^{ix}-e^{-ix}}[/eqn]

>not writing [math]z=\operatorname{m}(x)=e^{2\pi i x}[/math]
>bothering to convert imaginary to real

First time hearing about sec. Why would you use it when it's literally cos^-1 (aka 1/cos)? Do you have something for 1/sin too? It seems completely pointless.
We only mention cot as 1/tan, but mostly just use tan.

Fractions are so ugly though. I much prefer the way secant looks on paper to 1/cosine

We don't do that in Britain.

Well, sec(x) is just easier to write. I mean, for Americans and Brits the relationship between sec(x) and cos(x), and likewise cosec(x)->sin(x) and cot(x)->tan(x) is instantaneous.

It's kind of like the English language, when you think about it. Our language is sprawling with useless words, but to us we can instantaneously connect words together, and words with their abstract relations. Euros can't do that because their words aren't as geared to that sort of behaviour; their words convey feeling and expression, rather than relation and thought.

Ultimately, Euros have small brains.

Yes we do its clearly in the A-level curriculum for every exam board.

Your professor is right, and for us it's the other way: why do Americans insist on giving things like 1/cos a special name? Why not just call it what it is?

>all the sec formulas

Like for example...?

You might be joking, but this really is the superior way.

Such an irrelevant function doesn't need a special name.

Its a historical term from when they used trig for navigation. It was easier to list secant as such. It's not a specifically American invention.

I bet he doesn't know a versine from a haversine

oh, so its just simplifying it

cosΘ is easier to graph than secΘ

>I bet
there is no wagering at Veeky Forums, Grandpa

1/sin = csc

1/tan= cot

they teach this to you in middle school geometry and use it throughout schooling in US

[math]\cos^{-1}=\arccos\neq\frac{1}{\cos}[/math]

It's easier to see known identities when expressions contain just three basic trigonometric function. Whilst amerifats learn by heart tons of identities with all these sec's and god knows what Europeans know the basic ones and the necessary rules, so they can do the same things with much less effort and rote memorization

>not using [math]\sum_{n=0}^\infty (-1)^n\dfrac{x^{2n+1}}{\left(2n+1\right)!}[/math], [math]\sum_{n=0}^\infty (-1)^n\dfrac{x^{2n}}{\left(2n\right)!}[/math]

i'm an american and i use 1/cos because all standard trig identities reduce to cos/sin identities

this, using power series representation of sin and cos, Cauchy product for products and termwise differentiation and integration is superior to all other notations for brainlets

We have wonderful Latin so fuck you.

>sec is the inverse of a cosine
careful how you say that, the inverse of cosine is arccosine.

B8

>inverse
this is why I always say reciprocal,
to specify multiplicative inverse

found the engineering student

po-tay-toe, po-tah-toe

greek here, first time i hear of this

kekd

tan is also redundant. Frankly we can also get rid of one of sin or cos, since cos(x) = sin(pi/2 -x)

Whateverrr you say, Fred Astaire.

>not using the reciprocal of the derivative of the series expansion of sin x with an approximation of n=100

sin(x)/cos(x) * 1/cos(x) = sin(x) / cos^2(x) = -sin(x) /(sin^2(x) - 1)