I've been a Brilliant member for a year and a half now, and it all began with your special offer. I'm delighted with the app! I've learnt a lot and I've enjoyed it so much! So thank you for introducing me to Brilliant and thank Brilliant for sponsoring you!
Easy way: exp(it)=cos(t)+i*sin(t) But also exp(it)=cosh(it)+sinh(it) Pairing up the odd part with odd part, and even with even we get: cosh(it) = cos(t) sinh(it) = i*sin(t)
An easy way to remember this is that, e^(ix)=cosx+isinx on one hand, on the other, e^ix=cosh(ix)+sinh(ix). Match the even part of one side with the even part of the other side, and do the same with the odd part. You get that cosh(ix)=cosx and sinh(ix)=isinx. Now evaluate these functions at x=it and you get the rest ;)
I actually discovered this when I noticed that cosh and sinh had a structure similar to the inner and outer products of geometric algebra, which are defined as the symetric and antisymetric components of the full product. But the inner and outer products are usually defined with sin and cos... along with i. This is what led me to realize the relation between them and the real and imaginary / even and odd parts of the exponential. The only reason one relation is x^2 + y^2 = 1 and the other is x^2 - y^2 = 1 is because the imaginary factor of y flips the sign when squared. It felt so awesome to find that on my own. Now I kind of want to make a visualization of the complex exponential's even and odd parts to try and get the hyperbolic and spherical trig functions to appear on different axes of the same graph.
I always asked myself why the hyperbolic trigs functions and the complex trigs functions looked so similar. Even my teacher didn't showed this relation. Now i have my answer ! Thanks BPRP !
You solve mathematics like you are hanging out with Ur friends😜😜 And Ur excitement after solving is just awesome. Just because of teacher like u I'm happy of being a mathematic student. Thank you🙏
Plot them! The hyperbolic sin and cos “jump” off the tops and bottoms of the sin and cos at right angles in the y-i plane. Likewise, when sinh and cosh are real, sin and cos are at right angles in the y-i plane. Etc…. It is beautiful. Next extend to tan and tanh, sec and sech …. Then extend to Bessel and J functions!
Very interesting. I was wondering about the relation between the hyperbolic trig functions and the complex definitions of the trig functions after seeing one of your videos, and you explained these concepts so clearly.
Rational paramerization of hyperbola is based on observation (1-t^2)^2+(2t)^2=(1+t^2)^2 (2t)^2=(1+t^2)^2-(1-t^2)^2 1=\left(\frac{1+t^2}{2t} ight)^{2} -\left(\frac{1-t^2}{2t} ight)^{2}
Before 1 min: There's a 3rd way to interpret the angle - the arc length subtended on a unit circle, whose equation you've written: x² + y² = 1. This may or may not work for the unit rectangular hyperbola; I'm checking into that. It does have the right behavior near 0, and it does go to ∞, but those are no guarantee... Fred
I was searching for these formulae (didn't need the proofs, but they were cool too) for about an hour until I found it here and was able to answer my question.
Great one, thanks! If you could do more parameterization videos it would be great since finding them is always so confusing.. also integrations along a curve with parameterization
What I think is cool is if you were to somehow create a 4D graph and declare your x, y, z, and t axis, and call the z axis the imaginary input and call the t axis the imaginary output, the function x^2+y^2=1 on the t axis looks like x^2-y^2=1 and vice versa. So I kind of think of the hyperbolic function as a complex version of the circle function and vice versa
I noticed that to. But I went the arctan route. cos(i*arctan(3/4))-i*sin(i*arctan(3/4))=1.9031323020709 cos(arctan(i*(3/4)))+sin(arctan(i*(3/4))) =sqrt(7)(4/7+3/7i) for tan= i*3/4 Works just fine this way. You can do geometry with it. It's just pythagorous' theorem with an 'i' in it. [x/sqrt(x^2+y^2), y/sqrt(x^2+y^2)]
makes me want to say.....this is wonderful short video.... beautiful work...so ...so ..excellent...but i think you need to add a quick physic conclusion to your video.........this certainly is one of the slickest short video in circulation....why?....because you connects non-Euclidean equilateral triangle's surface area(excess/deficit) change to a Euclidean triangle's total energy change and the triangle's inertial mass change dependent on (a function off) the average of the total number of summed ''-' , '+' and '0' Gaussian curved triangle edges counted ......
Cool!!!! So one can get the derivative of sinh and cosh using chain+product rule from the equal sin/cos statement, never thought on that :-O I always did that from the definition of sinh/cosh only ("e-stuff").
ellipse eqn in complex extended x-plane-y-axis will be a hyperbola in the Im-x-side. This so parametric forms cos and cosh are complex and real counterparts
Tbh I've always felt like this is true because I can say integral of 1/1-x^2 dx = integral of 1/1+(ix)^2 and then use u-sub. But at the same time, the integral is tanh(x)
0:57 Shouldn't the area = t? The area 0f a unit circle is A = 2π and the area of a sector of a circle is A*(corresponding angle of sector/2π), assuming the angle is in radians. Hence, the area in the diagram should be 2π•t/2π = t.
to simplify the work or notation of multiple real life problems, instead of putting an enormous amount of digits you simply use hyperbolic functions, same as trigonometry in general
Sir at 6:50 u said imaginary looking Theta = it So t is also imaginary so that they can be real Sin theta = real function Sin h x= imaginary function ?????? Sir please please clear this doubt Thank you
cosh(it) = cos(t) Like, it's simply inserting i*t into x for cosh(x) and get cos(t) as the result just like as if you have inserted i*t for cos(x) and get cosh(t) as a result.
hey bprp, there was an integral video involving cos's and sin's I think and you solved it with a creative way of adding 2 solutions of 2 integrals together and I can't find that video, any ideas? thank you :D
Still unsatisfied how can you say that this equation can make the area t/2 and is there a way to come up with this formula using calculus like what you can do with sin and cosine by knowing the derivatives first then using taylor then coming up with a formula like eulers identity
9:27 isn't it? or isin(it)? ;)
damn
Oh my god that was pure gold.
Best comment
you should be the winner
Loool
WoW! I haven't been this blown away since when I was shown Euler's identity!
I've been a Brilliant member for a year and a half now, and it all began with your special offer.
I'm delighted with the app! I've learnt a lot and I've enjoyed it so much! So thank you for introducing me to Brilliant and thank Brilliant for sponsoring you!
Thank you Andy! Glad to hear that you like it!!!
I love this kind of videos. I love all the proof videos! Thanks!
Fernando Garay thank you
looking at the series expansions for exp(x) cosh(x) and sinh(x) is what really drove this point home for me.
this is by far the best take on hyperbolic functions I found on youtube so far. And I looked far and wide too!
complex relationship
well played, good sir, well played
I've been fascinated by these patterns for a while, and yours is an excellent explanation. Thanks!
Easy way:
exp(it)=cos(t)+i*sin(t)
But also
exp(it)=cosh(it)+sinh(it)
Pairing up the odd part with odd part, and even with even we get:
cosh(it) = cos(t)
sinh(it) = i*sin(t)
cos(it) + cosh(t) = cosh(it)
that implies cosh(t) = cos(t)/2
So 2cosh(t) = cosh(it)?
@@cringy7-year-old5that is seriously wrong…
Cos(it)=0?
An easy way to remember this is that,
e^(ix)=cosx+isinx on one hand, on the other, e^ix=cosh(ix)+sinh(ix). Match the even part of one side with the even part of the other side, and do the same with the odd part. You get that cosh(ix)=cosx and sinh(ix)=isinx. Now evaluate these functions at x=it and you get the rest ;)
I actually discovered this when I noticed that cosh and sinh had a structure similar to the inner and outer products of geometric algebra, which are defined as the symetric and antisymetric components of the full product. But the inner and outer products are usually defined with sin and cos... along with i. This is what led me to realize the relation between them and the real and imaginary / even and odd parts of the exponential. The only reason one relation is x^2 + y^2 = 1 and the other is x^2 - y^2 = 1 is because the imaginary factor of y flips the sign when squared.
It felt so awesome to find that on my own. Now I kind of want to make a visualization of the complex exponential's even and odd parts to try and get the hyperbolic and spherical trig functions to appear on different axes of the same graph.
Give this problem a try and when you’re ready, continue the video.
Did *You* figure it out?
Reece 5..4..3..2..1
Math meanies 😡
Hey guys it's presh talwaker making sure you mind your decisions
I always asked myself why the hyperbolic trigs functions and the complex trigs functions looked so similar. Even my teacher didn't showed this relation. Now i have my answer ! Thanks BPRP !
You solve mathematics like you are hanging out with Ur friends😜😜
And Ur excitement after solving is just awesome. Just because of teacher like u I'm happy of being a mathematic student. Thank you🙏
Plot them! The hyperbolic sin and cos “jump” off the tops and bottoms of the sin and cos at right angles in the y-i plane. Likewise, when sinh and cosh are real, sin and cos are at right angles in the y-i plane. Etc…. It is beautiful. Next extend to tan and tanh, sec and sech …. Then extend to Bessel and J functions!
Omg.. The biggest problem of my life.. Finally solved 😱😱😱😱😱impressed
DD
Yup!!!! : )
@@blackpenredpen I have plenty of calculus books and have never seen that one around, weird :)
Very interesting. I was wondering about the relation between the hyperbolic trig functions and the complex definitions of the trig functions after seeing one of your videos, and you explained these concepts so clearly.
It's like you can read my mind comrade! second time I was studying some maths and you made a vid exactly about what I was studying, great vid!
Rational paramerization of hyperbola is based on observation
(1-t^2)^2+(2t)^2=(1+t^2)^2
(2t)^2=(1+t^2)^2-(1-t^2)^2
1=\left(\frac{1+t^2}{2t}
ight)^{2} -\left(\frac{1-t^2}{2t}
ight)^{2}
Before 1 min: There's a 3rd way to interpret the angle - the arc length subtended on a unit circle, whose equation you've written: x² + y² = 1.
This may or may not work for the unit rectangular hyperbola; I'm checking into that. It does have the right behavior near 0, and it does go to ∞, but those are no guarantee...
Fred
Did you saw the joke isn't it?
So the similarity of "i sin(it)" to isn't it.
der Ultrahero
Nice catch!!!
Or "I sign it"?
I was searching for these formulae (didn't need the proofs, but they were cool too) for about an hour until I found it here and was able to answer my question.
Great one, thanks!
If you could do more parameterization videos it would be great since finding them is always so confusing.. also integrations along a curve with parameterization
Is there a geometrical representation of tanh(t), coth(t) etc, just like cosh(t) and sinh(t) are the x and y values of the points of the hyperbola?
Putting this comment just so if someone else finds it. Right now its late where Im from so I'll try and see if there is one in the morning
@@cuzeverynameistaken1283 did you find one?
@@filyb he’s still working on it
@@joea-497kviews2 lmao
@@joea-497kviews2 eta perhaps?
What I think is cool is if you were to somehow create a 4D graph and declare your x, y, z, and t axis, and call the z axis the imaginary input and call the t axis the imaginary output, the function x^2+y^2=1 on the t axis looks like x^2-y^2=1 and vice versa. So I kind of think of the hyperbolic function as a complex version of the circle function and vice versa
Your videos should be required viewing for most math classes. Do you do anything for dicrette algebra?
Throw in a bit of an explanation of Eulers formula in terms of the Taylor series of e^x polynomial... love your passion...😊
I noticed that to. But I went the arctan route.
cos(i*arctan(3/4))-i*sin(i*arctan(3/4))=1.9031323020709
cos(arctan(i*(3/4)))+sin(arctan(i*(3/4))) =sqrt(7)(4/7+3/7i) for tan= i*3/4
Works just fine this way. You can do geometry with it.
It's just pythagorous' theorem with an 'i' in it.
[x/sqrt(x^2+y^2), y/sqrt(x^2+y^2)]
I just discovered your channel. Your videos are brilliant! Good thing they're your sponsor :D
I hope you're gonna be a math teacher because yours videos are so clear and precises
Excellent video! This is super interesting! Thanks for making these videos!
Look at this cute face he is blushing while playing with Maths 😍 ,maths must be his love.
Your videos are pretty amazing man. Keep going. 👌
makes me want to say.....this is wonderful short video.... beautiful work...so ...so ..excellent...but i think you need to add a quick physic conclusion to your video.........this certainly is one of the slickest short video in circulation....why?....because you connects non-Euclidean equilateral triangle's surface area(excess/deficit) change to a Euclidean triangle's total energy change and the triangle's inertial mass change dependent on (a function off) the average of the total number of summed ''-' , '+' and '0' Gaussian curved triangle edges counted ......
please continue this series!
sinh(x) = -i sin(ix)
cosh(x) = cos(ix)
tanh(x) = -i tan(ix)
sinh(ix) = i sin(x)
cosh(ix) = cos(x)
tanh(ix) = i tan(x)
Wow!! Thank you for the video!
9: 26 "isin(it)"?
nice one :)
Cool!!!! So one can get the derivative of sinh and cosh using chain+product rule from the equal sin/cos statement, never thought on that :-O I always did that from the definition of sinh/cosh only ("e-stuff").
Thank you sooooo much!!!
ellipse eqn in complex extended x-plane-y-axis will be a hyperbola in the Im-x-side. This so parametric forms cos and cosh are complex and real counterparts
Oh You mentioned it? Just now say it.
Nice lecture👍
RIGHT HERE, RIGHT HERE, RIGHT HERE
thanks
Then there is Osborn's Rule, a very useful relationship between trigonometric and hyperbolic functions and identities.
Sir you are awesome....!!!!
You don't even need Euler's formula to show that cos(it) = cosh(t). You can also show that their power series are the same.
Yeah, Tibees just did that Bob Ross style!
You mean her newest video? I don't see any cosh in there.
@@artey6671 yeah guess you are right. May have misremembered.
great video
after all i've been through in last year , "Imaginary" is a inappropriate title.
Beautiful!
Damn clickbait title! I wish professors can use clickbait to make lectures more interesting
Are there other conic section analogues of the trigonometric functions? Parabolic sine? Elliptical cosine?
Well to some extent you can also use x(t)=sqrt(1-t) and y(t)=sqrt(t) I mean (?)
Tbh I've always felt like this is true because I can say integral of 1/1-x^2 dx = integral of 1/1+(ix)^2 and then use u-sub. But at the same time, the integral is tanh(x)
great video. but try starting from first principles and proving that for a hyperbola x=cosht and y=sinht and see how long that takes you!!
3^2 - 2^2 = 1 too. 😎
Can you please make a video relating tan and tanh
Yes it's very cool
Tengo sueño ... pero igual veo estos videos aunque hayan sido contenidos que vi hace muchos años!....
QUESTION: With -i out in front of sin(it), [-isin(it)], doesn't the proof fail?
I'm headed back to your channel to find the link to "even" and "odd" parts of e^t, described at 15:18. Not sure where to look....
Ah, found it here. ua-cam.com/video/oLZoGEcJ2YE/v-deo.html
It's here: ua-cam.com/video/oLZoGEcJ2YE/v-deo.html
Very cool
Yay!
We just went over sinh and cosh in my calc class today. What are the chances? This is a much more complete explination than we got.
How's your stock of whiteboard pens ? 😊
lim x → 0 sin2x^(tan2x) ²
0:57 Shouldn't the area = t? The area 0f a unit circle is A = 2π and the area of a sector of a circle is A*(corresponding angle of sector/2π), assuming the angle is in radians. Hence, the area in the diagram should be 2π•t/2π = t.
Area of circle with r = 1: pi * r ^ 2 = pi. Not 2pi.
Wait... Are the trig functions C -> N? Or can some input a+bi give imaginary output?
they're C->C. Also, you put C->N, pretty sure you meant C->R or C-> [-1,1].
@@justacutepotato2945 yeah, you're right. And I meant C->R.
What do you need hyperbolic functions for in math? Of course, except defining them and solving equations with them?
to simplify the work or notation of multiple real life problems, instead of putting an enormous amount of digits you simply use hyperbolic functions, same as trigonometry in general
What we do with the part of the hyperbola on the left side
What about x=sec(t) and y=tan(t) for 0
Why you don't have spanish subtitles ?? Its so interesting
Sir at 6:50 u said imaginary looking
Theta = it
So t is also imaginary so that they can be real
Sin theta = real function
Sin h x= imaginary function
??????
Sir please please clear this doubt
Thank you
Ever heard of gd(x), the Gudermannian function?
You probably don't know this but you made a pun at 5:49
Bruh😍awesome.... Love ot
i love you man 💕💕💕💕
Notice how close they are to co shit.
Could you make a vid about Lobachevsky space pleaseee? don't make me beg
cosh(it) --> ohsh(it)
oh no
cos(it)=cosh(t) is interesting. cosh(it)=? is a more interesting question though.
cosh(it) = cos(t)
Like, it's simply inserting i*t into x for cosh(x) and get cos(t) as the result just like as if you have inserted i*t for cos(x) and get cosh(t) as a result.
*slaps theta* TAG YOU'RE (it)
¡¡Lo máximo!!
hey bprp, there was an integral video involving cos's and sin's I think and you solved it with a creative way of adding 2 solutions of 2 integrals together and I can't find that video, any ideas? thank you :D
We have theta be real and t be real also. So, how can we put theta = it ??
We're extending the theta and t to complex world
That's in my textbook xD
What's the usage of sinh and cosh？
ALGEBRAIC EXPRESSIONS HATE HIM.
What about cosh(it)
so Mθ=Mit ....YAY!
Can u Integrate xtan(x)? 😛 Help me if u can. I really love ur Videos. I learn a lot from them. Thanks
you have 99% of like and 1% of dislike... Not too bad... I like ;-)
cosh(it)
OMG!!!! @&@&&@&@&@; THIS IS SO EXTREME LIKE THE TITLE!!!
Why you don't advatise for patrion
Still unsatisfied how can you say that this equation can make the area t/2 and is there a way to come up with this formula using calculus like what you can do with sin and cosine by knowing the derivatives first then using taylor then coming up with a formula like eulers identity
Jim Allyson Nevado as I said in the video. I will do a proof for that. So stay tuned!
blackpenredpen waiting for that
blackpenredpen oops i apologize for not listening carefully
Btw, is this a newly discovered relation?
Yes, considering that the history of mathematics goes back millennia. Wikipedia dates them to the 1760s.
The return of black shirt red shirt.
Gorgeous
Wow