What is the deal with (hyperbolic) trig functions?
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- Опубліковано 20 сер 2024
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This is the first time in my life that I see a graphical representation of what hyperbolic sine and cosine stand for. Would have been useful when I took calculus in college.
Same here! Thank you Professor Penn!
The graphic for the hyperbolic functions is the motivation to call the inverse functions "AREA hyperbolic sine/cosine".
It's literally the area parametrized after x enclosed with the x-ordinate with the hyperbola.
Which means you can derive an integral representation.
I learnt that cosh was the shape of a hanging chain, but i never saw this before.
@@easymathematik Thank you. I never have thought about that. Always thought it was some Greek or Latin stuff.
Cosh and sinh are just cos and sin moved around a bit in the complex plane.
E.g. cos(iz) = cosh(z), meaning cosh and cos are just 90 degree rotations of each other.
This explains why the identities are so similar.
cosh(iz) = cos(z)
sin(iz) = i sinh(z)
sinh(iz) = i sin(z)
So the Taylor Series expansions, instead of alternating addition and subtraction, exclusively adds terms?
@@DanGRV GOOD, thanks... Easily proved from the definitions
I feel like it's easier to understand the motivation for all of this in the context of the definitions of the unit circle (i.e. x^2 + y^2 = 1) and the unit hyperbola (i.e. x^2 - y^2 = 1)
That's the first thing he shows after listing the formulas on the left side of the board.
Your comment tell me more than the whole video, tanks!
@@angelmendez-rivera351 so am I correct in assuming that hyperbolic trig functions work only under the assumption of a perfect hyperbole x^2-y^2=1 ?
@@twt2718 Yes, kind of, like the trig functions work only for the unit circle.
@@friedrichschumann740ok. That makes perfect sense. Thanx
I like this kind of video: where you talk about the intuitions about mathematical relationships that you find valuable, rather than just solving a problem or giving a lecture-type talk that presents material for a viewer who doesn't know it.
I know what you mean... I've came across my own discoveries that wasn't directly taught. What I came to understand is that there is a direct relationship between linear equations and the trigonometric functions. The slope-intercept form of any arbitrary line is y = mx+b where b is the y-intercept and m is the slope defined as rise over run (y2-y1) / (x2-x1) or dy/dx. We will set and keep b equal to 0 so that it is fixed at the origin. To see the direct relationship between linear equation we can set the slope m equal to 1. This gives us the equation y = x. This is an identity equation and it is still linear. We will use two points from this line P0 and P1. P0 will remain fixed at (0,0) and P1 can be any other point on that line but for simplicity we will state that P0x < P1x && P0y < P1y to keep the point within the first quadrant.
We know that the linear equation y = x bisects both the x - (horizontal) and y - (vertical) axes. We also know that the two axes x && y are perpendicular or orthogonal to each creating a 90 degree right angle which is also equivalent to PI/2. Since the line y=x where m = (y2-y1)/(x2-x1) = dy/dx = 1 and b = 0 bisects the coordinate axes, this gives us theta/2 which now gives us a 45 degree angle or PI/4 radians value.
We can choose any point on the line y=x for P1 leaving P0 fixed at the origin (0,0). We will then draw a vertical line starting on the x-axis at (x2,0) up to the point (x2,y2) that is on y=x. This vertical line will be perpendicular to the x-axis and parallel to the y-axis. This then creates a right triangle ABC from the points P0 = (0,0), P1 = (x,0), P2 = (x,y). The orientation and winding order CCW (Counter Clockwise) puts this triangle and its interior angle from the origin into standard form. Its interior angle is above the x-axis and below the line y=x and we will call this angle theta and label it (t).
If we look at the line y=x and use (0,0) and any other point on the line the slope m=dy/dx is also equal to sin(t)/cos(t). We also know that sin(t)/cos(t) = tan(t) This means we can rewrite the original slope-intercept form of the line y=mx+b within any of the following contexts: {y = (dy/dx)x + b); y = (sin(t))/(cos(t)))x + b; y=tan(t)x + b}. They all equate and evaluate to the same thing, they only differ in the perspective of inspection. The original form of the slope-intercept y=mx+b is generalized from its slope m where it is the ratio between the change in y and the change in x. This is its linearity perspective. The dy/dx form is the simplification of (y2-y1)/(x2-x1) into dy/dx where dy and dx in this case is the change in x and y respectively as opposed to dy/dx where d is the derivative. This is more of a geometrical or vector approach. The sin/cos variant does relate to linearity but this perspective comes from the trigonometric view based on their definitions in regards to right triangles.
To confirm this we can take the points (3,3) and (5,5) as they both fall on the line y=x. We know that b = 0 and that m = 1. Since (5-3)/(5-3) = 2/2 = 1. We can also show that sin(45)/cos(45) = 1 where 45 is in degrees and similarly sin(PI/4)/cos(pi/4) = 1. Also tan(45) or tan(PI/4) = 1.
We know that m is also a/b which is also a fraction. However a/b in this case is not strictly limited to integers since both a and b can be any real or complex number. So the connection that I am seeing between Linear Equations and the Trigonometric functions from the property that y=mx+b can also be written as y = tan(t)x + b is that m = tan(t). This connection tells me that all fractions, all divisions, all ratios and all percentages are in fact results from the tangent function. This is the relationship that allows both geometrical triangles, the unit circle and linear equations in regards to their slopes to all have a direct relationship with the Pythagorean Theorem.
This is all possible simply because 1+1 = 2 and the equation 1+1 = 2 is linear. Even the equation 1 = 1 or 0 = 0 satisfies these relationships. And this isn't limited to just linear equations. The trigonometric functions are also directly related to all of the polynomials since y = x is a polynomial, it's just that the order of the exponent of x in this case is simply 1 as it is omitted because it is understood. Now, once you can generate your polynomials x^1, x^2, x^3, x^4, .... We can separate them into two sets S0 = {x^1, x^3, x^5, ... } and S1= {x^2, x^4, x^6, ... } respectively and when we consider these to be functions of x as in f(x) instead of equations of y, these sets gives us our Odd and Even Functions.
And what makes this even more interesting is that since sin(t) is an Odd function and cos(t) is an Even function there is also a direct relationship sin(t) and S0 as well as cos(t) and S1. This is why the power series are able to approximate the sine and cosine functions. Most people only see 1+1 = 2 and simple arithmetic yet what many don't know or understand at first sight is that 1+1 = 2 sets the foundations for all mathematics from Algebra to Geometry to Trigonometry to Calculus to Linear Algebra and Vector Calculus. Why? Because the equation 1+1 = 2 generates a Unit Circle where its center point is fixed at the point (1,0).
More than just that, if we take the general equation of the Circle: (x-h)^2 + (y-k)^2 = r^2 where the point (h,k) is the center of the circle the point (x,y) lies on the circle and r is the magnitude of its radius, we can place its center at the origin (0,0) and give it a unit length of 1. This will give us (x-0)^2 + (y-0)^2 = 1 which simplifies to x^2 + y^2 = 1. From the unit circle the trigonometric functions can be defined once again but this time it isn't based on the linearity of the slope m in regards to the angle theta (t). This time they are defined with respect to both quadratics and the Pythagorean Theorem. This is due to the fact that x^2 + y^2 = r^2 for all tense and purposes is the Pythagorean Theorem for we can replace x,y,r with A,B,C giving us A^2 + B^2 = C^2.
Finally to sum it all up (pun intended)... with regards to the vector notation of points and lines and their geometric properties with respect to the slope m, the angle theta (t), the unit circle itself being the Pythagorean Theorem is also how and why the Cosine of an angle is related to the dot product of two vectors. Another curious observation is that when we look at normal fractions such 1/2, 2/3, 7/16, etc... We don't immediate associate them with either being a slope or gradient, nor the result from the tangent function, yet what is curious is that the tangent function is the ratio between an Odd function over an Even function, and it is known that an Odd Function divided by ann Odd Function returns an Even Function and And an Odd Function over an Even Function Or an Even Function over an Odd Function will return an Odd Function and this is true because Tan(t) is an Odd Function which is proven by Sin(t) / Cos(t) which is Odd / Even resulting in an Odd Function which also makes the slope m = (y2-y1)/(x2-x1) = dy/dx an odd function. And this makes sense because slope is linearity and linear equations are Odd Functions. And this property of the slope of a line in relation to its generated angle being the tangent of that angle is what also enables us to calculate derivatives and integrals.
Without knowing it, numbers such as sqrt(2), PI, e, and even the complex numbers are embedded within 1+1 = 2. Even geometry, trigonometry, the Unit Circle, the Pythagorean Theorem, etc. are embedded within 1+1 = 2.
@@skilz8098 present more of that, it is useful
I’ve always found the hyperbolic functions very obscure.
Thank you, professor.
8:40 Dr. Penn notices he left the 'h' off the sinh in the hyperbolic pythagorean identity. Good lookin' out, doc.
So satisfying
Why did you spoil it bro
It's notable that while x can be considered the angle elapsed by the parametrization for the first area, the same is not true for the second. For both, we are considering x to be the arclength of the curve from the point (1,0).
Very interesting. In fact, in the second area the angle would be arctan(tanh(x)), which is not equal to x, isn't it?
yes, but only if the "arclength" in the second case is taken with respect to a non-Euclidean metric (specifically, the metric which defines the distance between (x_1, y_1) and (x_2, y_2) as sqrt{(x_1 - x_2)^2 - (y_1 - y_2)^2}, i.e. as the square root of the *difference* of squares, as opposed to the usual square root of the sum of squares) - if you compute the arclength in the second case with respect to the standard (Euclidean) metric then you get the integral from 0 to x of sqrt{cosh^2(t) + sinh^2(t)} dt, but this is not equal to x because cosh^2(t) + sinh^2(t) is not equal to 1; it is cosh^2(t) *minus* sinh^2(t) that equals 1 (so we have to use the other metric to change the plus to a minus so that we can use the "pythagorean hyperbolic trig identity")
At school (70s in the UK) we called them 'cosh', 'shine', 'than' (unvoiced 'th', as in 'thick'), 'shec' and 'coshec'.
"Singe" and "tange" are more common, I think.
We called them cosh, shine and tansh.
Regional differences!
@@iankr hyperbolic trig used to be in alevel math? Now it is only taught in alevel further math
@@mao4859 Yep
you forgot to mention the similarities between their argument-sum identities. for example:
cosh(x+y) = cosh(x)cosh(y) + sinh(x)sinh(y)
Same one. Just keep in mind there are two sinh's so there are two 'i's.
cosh(x+y) = cosh(x)cosh(y) - i*sinh(x)*i*sinh(y)
cosh(x+y) = cosh(x)cosh(y) + sinh(x)sinh(y)
This is why the velocity addition function in special relativity looks like the tangent addition, because it is.
This is the first time I see anyone relate the 2. It was always so random that the proportions of some triangle and the half-sum of exponentials shared the same name. None of my professors ever explained why, it was always like "here's the hyperbolic formulas btw ok next topic".
When you start talking about sinh and cosh, you've generally just been talking about sin and cos as solutions to differential equations or being related to complex exponentials. For example, cos is the half-sum of exp(ix) and exp(-ix), so it's not too surprising that without the "i", it's named similarly. On the other hand, they never explained why "hyperbolic".
That's a nice video. Fun fact, there's a generalization of the cosine and sine functions for any convex set, that contains the origin. And it's really useful for solving some problems of optimal control.
In this generalization, the basic properties of trigonometric functions hold for the 2 pairs of functions: coordinates on the convex set, and coordinates on the polar set. Actualy, I work on the same genralization for hyperbolic functions.
@@user-pj3le2cx6f that’s really interesting. Are you a grad student?
@@itszeen7855Currently, I am in the third year of my math studies at Moscow State University.
Interesting! I would like to look into this more. Can you point me in the right direction?
.
You really have made your videos much easier to understand, via minor adjustments to the presentation style. Also, I really like how you provided *so much* explanation of the hyperbolic trig functions here.
The double integral stuff at 2:48 seems a bit off topic to me, but that's just because I am not used to taking using double integrals to find areas of non-functional areas. To clarify, the reason we need to use a double integral instead of a single integral to find the area of a circle is that the circle is not just the area under a functional curve.
The many links between regular and hyperbolic trig functions will include similar relationships for 3 times the argument and the argument itself. For instance, where 4(cos t)^3 - 3cos t = cos 3t, you have 4 (cosh t)^3 - 3cosh t = cosh 3t.
Finding the 3 real solutions of the generic cubic equation, valid for p and q forming a negative discriminant in X^3 + px + q = 0, where the discriminant is 4p^3+27q^2, can also then be applied with a positive discriminant by using cosh t as substitute variable instead of cos t.
In both cases, it only takes re-writing the canonic equation WLOG to impose the coefficient of the term in x (« p ») as 3/4 ( Viete, French mathematician, used a similar trick in a particular equation of the 45th degree offered as a challenge by the Dutch ambassador to the King of France ).
The solution by Viete only covered the case with 3 real solutions obtained from finding 3 adequate angles whose cosine met a value based on p and q . Thanks to the hyperbolic trig , the single real value when there is only one can be found in the way. The other, complex, solutions are conjugates and therefore known from their sum or product.
5:37 I really dislike doing this integral by trig sub, but coincidentally, its much easier to do with a hyperbolic trig sub t=cosh(v)
Yeah but like, I'm watching this video to motivate cosh and sinh, doesn't seem very logical to use them in their own motivation xD
Let's not forget the Pink Floyd song, "Sinh on You Crazy Diamond." Okay, now that that's out of my system...
Something which I only became aware of relatively recently is that inverse hyperbolic functions take an 'ar-' prefix rather than an 'arc-' prefix, e.g., inverse sinh is written arsinh rather than arcsinh in analogy with inverse sine being arcsin(e).
A function known as the gudermannian associates the circular and hyperbolic functions in a different manner, explicitly, sin with tanh, cos with sech, tan with sinh, cot with csch, sec with cosh, and csc with coth. Graphically, the hyperbolic functions appear to be "stretched out" versions of their circular counterparts. Furthermore, algebraic relations are preserved, e.g. sin²x + cos²x = tanh²x + sech²x = 1.
Here is an observation that blows my mind: start with the Taylorseries of cosh, split it into 2 parts, the exponents divisible by 4 and the others. You get 2 nice even functions, strictly increasing on the positive side , their sum being the cosh, and you may ask which one is occasionally bigger. So you take the difference and you get: cos x. Can you see why this behaviour is somehow beyond my imagination?
Wow that was a satisfying integral, even though I haven't learned about hyperbolic trig
0:03 sin(x)/x = 1, sinh(x)/x = 1 so sin(x) = sinh(x)
7:35 Homework?
10:19 Good Place To Stop
First line:
lim(x->0) sin(x)/x = 1
lim(x->0) sinh(x)/x = 1
you forgot the limit
ah so h = 1 then
@@haziqthebiohazard3661 "sometimes my genius is.... it's almost frightening"
Pretty cool stuff. I've known about the similarity of the two sets of functions with their derivatives, identities, etc. But I never really understood the whole "hyperbolic angle" concept until you showed that the areas were equal.
I -just realised- that cosh and sinh are just the even and odd parts of the exponential function
@Lakshya Gadhwal Every function f(x) can be broken down as the sum of an even function plus an odd function:
f(x) = ( f(x) + f(-x) )/2 + ( f(x) - f(-x) )/2
For the exponential function, those two functions are cosh (even) and sinh (odd)
@Lakshya Gadhwal cosh(x) + sinh(x) = e^x, cosh is an even fn, and sinh is an odd fn, that's essentially the definition of even and odd part of a fn. Basically any function (which is defined on a suitable domain, preferably throughout the real number line, but at least on a symmetric domain) can be expresses as a sum of an even and an odd fn exactly, and those functions are called the even and odd parts of the original function respectively
Yup. You could also look at the Taylor series for e^x. Cosh(x) is just all the even terms, Sinh(x) is all the odd terms.
@Number_8 u should have thanked him bro
What correlates nicely with the fact cos and sin are [almost] just even and odd parts of imaginary exponential function
And what about "eliptical sine and cossine"?
They are known as "Jacobian elliptic functions".
They generalize sin/cos and sinh/cosh.
They are called:
sinus amplitudinis
cosinus amplitudinis
@@easymathematik why Jacobian? Related to jacpboan in calculus?
@@leif1075 Yes, its named after him.
I've always wanted to make up a math problem where you had to use cos(t) and sin(k) to calculate the cost of a sink.
factor in the cost of wooden log(s)
"Ha!" That was the noise I made at the end. Nice.
U should have used a hypebolic subsiton u=coshx u get that the integral simplfies to sinh^2 which is super easy to calculate cause its a bunch if exponentiols
Where’d he learn his blackboard penmanship? None of my math professors come close to his level of legibility 💙 Great Job
An equation on the blackboard is wrong :: it should be "cosh² x - sinh²x = 1", but the second h is missing.
Why is it that trigonometry [ basically circular functions and hyperbolic functions ] have retained so many sub functions - sin, cos, tan, cosec, sec, cosh, sinh, tanh cosh, coth ....... ? In the old days calculations were very tedious, and extensive use was made of tables of functions, and logarithms, natural and base 10, and logarithms of trig functions. Nowadays with a cheap calculator much of that is necessary. With access to math software, the tables are even more obsolete.
The whole subject needs streamlining and a lot of house cleaning, and most of the material, identities etc reworked. Particularly glaring is an apparent reluctance to mix classical trigonometry with complex numbers, except for a nod to Euler and de Moivre. This is a pity because it simplifies a lot of otherwise difficult proofs - for example deriving sin(3x) in terms of sin(x).
This sounds extremely interesting. Do you have any suggestions for where to learn more about this? (I'm not sure what search terms to use to get started.)
I don't see why there being six trig functions and six hyperbolic trig functions is really a problem - you can of course use your own "housecleaning" in your personal work though. For example an easy way to cut down from six functions to three in both cases is to fully embrace the notation sin^2(x), cos^3(x), etc., and extend this to negative integer exponents, allowing one to write sec(x) = cos^{-1}(x), cosec(x) = sin^{-1}(x), cot(x) = tan^{-1}(x), and similarly for the hyperbolic functions. However if you are going to do this then you must use a different notation for inverse functions (the notation I personally use is to just write f^{-} instead of f^{-1} - in fact I use this more generally in abstract algebra; e.g. I will write the inverse of an element x in a group as x^{-} instead of x^{-1})
@@nordicexile7378you can read about trig tables and logarithm tables on wikipedia, if that's what you're interested in.
@@nordicexile7378 I'm glad I was able to spark your interest. I regret I don't have any suggestions as to how or. where to search the topic.
The ideas I was expressing are entirely my own, and have developed over a period of about 70 years, starting in the age when multiple calculations were done with 10 inch slide rules, and before that by laborious calculations done using tables contained in fat books.
With the introduction of computer software which allowed calculations, original BASIC had, as far as I remember only SIN( ) COS( ) and ATAN( ). This greatly simplified things, and frankly many of the trig identities became unnecessary. More modern software and also calculators, these days built into phones and packing a lot of functions in addition to circular and hyperbolic functions, make calculation almost a trivial exercise.
Bearing all this in mind it's a little sad to see that the method of teaching mathematics has not progressed very much, and nothing has been thrown away. Teaching methods also tend to keep subjects in watertight compartments : I mentioned that complex numbers could be used to simplify the interface between traditional Euclidean Geometry and Algebra as traditionally taught.
As an illustration of the latter try comparing the expansion of (1+I*t)^3 with the expression of tan(3x) in terms of tan(x) then try to write a proof of the identity based only on Euclidean Geometry.
There is another neat trick which can be very useful, but as far as I know is never taught. All the circular functions relate back to right triangles. A Pythagorean triplet can be used to express right triangles with integer values. Consider the group { 2*u ,u^2-1, u^2+1 }. Using integer values for u, any Pythagorean triplet can be found. However if real values of u are used any circular function can be expressed in terms of the parameter u. Thus t(u) = 2*u/(u^2-1), s(u)=2*u/(u^2+1) and we can also relate these values to tan(x) and sin(x) by realizing that x = arctan( t(u) ) ...
I close with the thought that " experiment is the essence of science ". This was recognized when " SCIENTIA " meant KNOWLEDGE. The same people would also have said SCIENTIA POTENTIA EST.
@@schweinmachtbree1013 I agree that the way mathematics is practiced is often notation driven. Recall the classic row between Newton and Leibnitz, and their use of very different notations to express essentially the same concept. To misuse a popular phrase, I am beneath all that. We are all free to use, discard, invent whatever we want. All that matters in the end is that truth is maintained. As my grannie used to say " there's more than one way to skin a cat ". I'm never likely to attempt such an undertaking, but I believe she was probably right.
Are there corresponding parabolic trig functions?
You can define them. But the relations between them are not so "nice".
Yes, but they aren't very 'pretty' in or in Michael's words they aren't
nice' comparison with the regular and hyperbolic sin and cos functions.
People sometimes define them as sinp(x) = x & cosp(x) = 1. If we allow sinp(x) as the variable x, we then can work in terms of cosp(x). I'm not sure if this is the 'general' definition of sinp(x) and cosp(x), but that's how I learned them. For what's it's worth I studied them in "advance math" in high school, knowing that I'd be a math major in college, which is exactly what i am.
Also you will see, that parabolic sine and parabolic cosine are solutions of polynomials. In other words:
cosp(x), sinp(x) are expressible with radicals.
"Classical" sin,cos and hyperbolic sin/cos are not expressible with radicals.
Won't add to good answers above, but want to note that you asked a very good question.
How about elliptic trig functions?
So it's not standard to call sinh "Sinch" (like winch, but with an s) and cosh "Kosh" (like the start of Kosher)?
Its popular, but not 100% accepted, a lot of people call sinh "shine" , rather than "sinch"
In Russian we sometimes call them _shinus_ and _chosinus_ , but this jargon isn't quite common either.
@@rubetz528 Is that хосинус or чосинус?
Say what they are: hyperbolic sine and hyperbolic cosine.
@@bethhentges I feel tempted to pronounce the "nh" as in Portuguese, and "cosh" as the blunt instrument.
The way the title of the video is phrased makes it sound like the opening line of a Seinfeld episode.
Now i understand why my engineering teachers didnt explain hyperbolic trig calculus
In the split complex numbers, j^2=+1 and exp(xj)=cosh(x)+sinh(x)j.
This is the analogue of exp(xi)=cos(x)+sin(x)i.
Hyperbolic functions can be really cool. There are lots of really cool properties of hyperbolas and related functions and they share a lot of similarities with their cousins, sin, cos, tan and inverses. I think of them as all two sides of the same functions except with the domain turned by 90 degrees on the complex plane. There's almost always a mirror hyperbolic property for regular trigonometric functions.
(1-x^2)^(1/2) solves for y in the implicit function x^2+y^2 = 1. Notice that this function is an involution (its own inverse) for the domain (0,1), and likewise, you can plot the two types of hyperbolas solving for x^2-y^2 =1 with (x^2-1)^(1/2) and (x^2+1)^(1/2), and those two functions are inverses of each other, and you can see that they are an extension of the (1-x^2)^(1/2) such that their are rotated by a quarter turn on the imaginary axis.
1/2 * (x + (x^2-1)^(1/2))^n + (x - (x^2-1)^(1/2))^n) yields chebyshev type 1 polynomials and
1/2 * (x + (x^2+1)^(1/2))^n + (x - (x^2+1)^(1/2))^n) yields the same coefficients except that they're all positive
and you can decompose the hyperbolic version into their e^x and e^-x equivalents with each half of that function.
(x/n + ((x/n)^2+1)^(1/2))^n yields a very good approximation for e^x for relatively small n and it always stays positive unlike the taylor series or
(1 + x/n)^n or other polynomial approximations. You can construct good approximations for all the exponential and hyperbolic functions with that form and they remain well behaved to +- infinity
The relationship between trig and hyperbolic trig functions always fascinated me. Especially when you go into complex numbers.
Nice t-shirt! BTW, it's true what's written on it.
In Germany, we say cosinus hyperboliculs and sinus hyperbolicus. The inverse functions are area cosinus hyperboliculs and area sinus hyperbolicus. Explaining the latter might be worth another video.
Am I imagining it, or are there also such things as elliptical trig functions?
They are known as "Jacobian elliptic functions".
They generalize sin/cos and sinh/cosh.
They are also known as
sine amplitudinis
cosine amplitudinis
@@easymathematik thank you
@@thorbynumbers5368 Your welcome.
@@noelani976 Because I have a brain.
@@noelani976 sometimes it is nice to wonder for a while and think about something, rather than immediately looking "the answer".
A luxurious explanation.
great timing! just started learning this
Hi,
The lenght of the arc of cercle is equal to x, is the length of the arc of hyperbola also equal to x?
Just had a look, the arclength of a hyperbola is given by an elliptic integral evaluated using the point x
1:01 that hyperbolic sine is written as a regular sine
The "x" argument of cosine and sine are fairly obvious with what it means, as relates to a unit circle. I'm a bit confused about what the argument of cosh and sinh *means*. Everything I find in my books or on the web just says that it's twice the shaded area which is fine, I guess, but in context here, there's a strong relationship between that sector area and the angle in the unit circle; is there a similar relationship between the corresponding angle for the hyperbola?
the argument of cosh and sinh , is actually the natural log ln(x)
you can check this page：en.wikipedia.org/wiki/Hyperbolic_angle
i think, half of the argument is the shaded area, is the right way to explain it, instead of saying the argument is twice the shaded area, which might mislead you to consider the below half part of the hyperbolic sector can be added to make it two times larger, but it is not the true meaning.
why do i say, half of the argument? because y=1/x is actually a hyperbola with radius sqrt2, and counterclockwise turned 45 degree
on the wiki page above, you will know ,''The magnitude of this angle is the area of the corresponding hyperbolic sector, which turns out to be lnx''
when you go to unit hyperbola from y=1/x, you need to shrink all length by sqrt2, and consequently, all area will shrink by 2
that's the true reason, why half of the argument is the shaded area,
I will make a video on this topic , about the origin of cosh and sinh, maybe a few months later, i hope somebody will like it.
and i hope to receive your reply, i would like to make friends with some who likes math, maybe you can help me with the translation work of my video if you like to , i'm not an English speaker. haha just kidding, don't take it too seriously
Couldn't the hyperbolic functions be defined using the y=1/x hyperbola, which is simpler than the x^2-y^2=1 hyperbola?
Hyperbolics are just like trigonometric functions without the i in the exponents.
@@markv785 Fackn lol!
What about parabolic trigonometric functions?
@0:56 Just FYI, there is an h missing in the equation as written.
Lost me the second he said double integral to find the area lmao
okay your t-shirt has got me in a frenzy what's the deal with those sums
I think there is an easier way to understand that the sector area is x/2. We know that the circle area is pi and it's a sector of size 2 pi. so for x it should be pi / (2pi) * x = x/2
brilliant and very informative
Delightful and excellent!
Even though I've haven't done calc 1 yet, somehow this makes sense
Some of the most beautiful math in the world. Fun to learn, more fun to reach
Just fantastic 👏 explanation
Does parabolic trig have a similar area?
Why hyperbolic. Sin cos for waves or curves in planes. Hyperbolic for 3D curves volume calculations.
please do more hyperbolic trig
During watching the video, I was trying to prove the equality on your T-shirt.
Awesome video
Can anyone explain to me, in simple terms, how the first integral works? Mostly in terms of what dA is and why the switch to polar coordinates give rise to rdrdtheta? Sry if it’s really obvious but that part completely escapes me :))
dA is simply the infinitesimal area (or area element) that is being integrated over (with respect to), unlike dx. It is like a notation convention too because you can't calculate it until you actually express the A in terms of what describes the actual area/surface. In this case to calculate the area of the sector (part of circle) it can be done with polar coordinates as those relate to circles. There are other ways too. With circles you have usual variables: r (radius) and theta (angle) which describe a sector, where theta 2pi is full circle of course. When he solves this integral this whole procedure becomes more clear. hope this helps a bit. It can still be a bit confusing to wrap your head around double integrals sometimes, and the problems can get very tricky
These functions are predominantly present in PDEs
Eh. If someone understands exp(ix) (aka e^x) you just write down sin and cos in terms of these, then get rid of the i's and you've got the hyperbolic namesakes.
sin(x) = 1/(2i)*(exp(ix) - exp(-ix))
sinh(x) = 1/2* (exp(x) - exp(-x))
cos(x) = 1/2*(exp(ix) + exp(-ix))
cosh(x) = 1/2*(exp(x) + exp(-x))
Most of the symmetries you highlight are plain to see in these definitions.
The above are my favorite equalities of sin and cos, with the knowledge of the fairly simple infinite series definition of exp you can fairly simply get the infinite series definitions of sin and cos, and as I said, pretty much all the symmetries are fairly simple to see, because exp(ix) is so beautifully simple in terms of Derivates, integrations, ..... Everything but the value I admit.
Thanks
So what does 'h' do?
the area of a sector is theta*r^2/2 tho?
So *_that's_* what the hyperbolic functions are. I see. Thank you!
I spent a whole day to understand this video because 2 looks like alpha or ∂ for me.
But I had a good day.
that's why i hate my name, it's look like a Hyperbolic trig functions
I'm shaking right now
Same I did not expect that
Why?
"Jakob"ian elliptic trig functions?
So the idea is that the area is the same in both cases, i.e half the argument?
8:44 - nice notice =) It's better late than never.
i wonder if theres some kind of coordinate system which allows us to quickly calculate the double integral for A2 in the same way that a polar coordinate system allowed us to quickly calculate the double integral for A1
apparently there is. the transformation x=r•cosh(θ), y=r•sinh(θ) parametrizes the area well and we see that A2 is [(integral from 0 to 1) (integral from 0 to x) r dθ dr] after calculating the jacobian, which is exactly the same as the formula for A1.
My English teacher made this a much bigger deal than I thought it was
Wasn’t until grad school engineering did I “get” the hyperbolic functions. They were solutions to some 4th order differential equations such as the deflection of a beam. They are equal to their second derivative while regular sin and cos are equal to their 4th derivative. This gives you the options you need to solve these equations.
i always thought the h just a constant for x
You are the best!
Hi Michael! I really like your videos :) is there any chance you've seen my suggestion for the problem for your Overkill series?
Amazing genius
Why use a double integral when you can use HS Geometry?
Area of Sector
= ((angle)/(2pi))(pi)r^2
=(x/(2pi))(pi)(1)^2
=x/2
For fun
Because he's setting up the method of calculating the area for the area of the hyperbolic sector.
So is there a parabolic sine and cosine?
Or non-circular elliptical?
You can define them. (parabolic)
While sin,cos,sinh, cosh are trancendental functions (not a root of a polynomial and in general not expressible with radicals), parabolic sin and parab. cos are zeros of third degree polynomials.
Also the relations of sinp and cosp are not so "nice".
For elliptical sin/cos see "Jacobian elliptic function".
They generalize sin/cos/sinh/cosh.
Well it is an "h" so we obviously take the limit as h approaches 0.
:)
A very nice proof to blackpenredpen's old video
6:07 I do the same. It's much simpler just getting back to original variable in the end and putting original limits to the integral.
I hate hyperbolic equations- I get they are necessary but I share them for being so cumbersome
Is it correct to say that A equals "double integral of region"? Isn't it just "integral over region"? 🤔 Anyway integral over 2d-region is automatically double...
Just a semantic difference
So, what's the deal?
Sentences do not start with 'ok'!
Yes finally!!
T-shirts design is very impressive.🤓
Thank you, profesor very cool
Does x have to be less than π/4 in the second area?
No, because x does not refer to an angle on the hyperbola, it only refers to the area. ‘x’ refers to the angle and area only for the circle.
@@pacolibre5411 i understand now thanks!
huh that's insane
cosh2x - sinH2x = 1 forgotten the h in the formula... and he corrected it at the end...
The '2' there is an exponential power and not coefficient to the argument the function takes. Something like this: (coshx)^2 - (sinhx)^2 = 1
@@noelani976 i know, I just typo'd.
I cant wait until I finish my degree, they take me out back and finally let me in on the joke that all of this stuff actually is complete gobbledy-guk designed to test ones dedication
That? That is just decoration.
"So I go up to the cashier, and I ask: what's the deal with hyperbolic trig functions?"
am I the only one who is surprised that there is no 'r' in the area of the circle @3:50 ??
He’s considering the unit circle
I think you should represent the major push for WISE: hotter teachers!
I read the thumbnail in jerry Seinfelds voice
Uh, you wrote "cosh² x - sin² x = 1".
Bravo!