Question

Interesting!!

Answer 1

Just found out these formulas, the proofs for them are very easy.
\[\cosh(ix)=\cos(x)\]
and\[\cos(ix)=\cosh(x)\]
where cosh(x) is the hyperbolic cosine function
given by
\[\cosh(x)=\frac{e^x+e^{-x}}{2}\]

Now this is easily proved using euler's identity
\[e^{ix}=\cos(x)+i \sin(x)\]
and consider
\[e^{i(-x)}=\cos(-x)+i \sin(-x)\]
\[\implies e^{-ix}=\cos(x)-i \sin(x)\]
\[\cosh(ix)=\frac{e^{ix}+e^{-ix}}{2}=\frac{\cos(x)+i \sin(x)+\cos(x)-i \sin(x)}{2}=\frac{2\cos(x)}{2}=\cos(x)\]
Now consider the argument ix and -ix
\[e^{i(ix)}=\cos(ix)+i \sin(ix)\]
\[\implies e^{-x}=\cos(ix)+i \sin(ix)\]
\[e^{i(-ix)}=\cos(-ix)+i \sin(-ix)\]
\[\implies e^x=\cos(ix)-i \sin(ix)\]
Now for the hyperbolic function
\[\cosh(x)=\frac{e^x+e^{-x}}{2}=\frac{\cos(ix)-i \sin(ix)+\cos(ix)+i \sin(ix)}{2}=\frac{2\cos(ix)}{2}=\cos(ix)\]

Answer 2

I found them very interesting, is there a name for such pairs?

Answer 3

I used to know a name...but I forgot... I had to use this in PDEs though.

Answer 4

the trigonometric functions and the hyperbolic trigonometric functions are defined for real values only.
For example the writing
cos(ix) has no meaning

Answer 5

You've found something quite interesting, it turns out both of these functions are intimately related if you look into the complex plane! In fact, @Nishant_Garg the two formulas you have given are actually basically the same thing, check this out:

\[\cosh(ix) = \cos(x)\]
We can then plug in \(x \to ix\) to get:
\[\cosh(i *i x) = \cos(ix)\]
\[\cosh(-x) = \cos(ix)\]

However cosh is an even function, so:
\[\cosh(x) = \cos(ix)\]

You also might have noticed there is a similar situation with sine and sinh. I'll leave that to you to figure out, it's a little trickier.

A more general form of this would be to combine both as:

\[e^x = \cosh x + \sinh x\]
\[e^{iy} = \cos y + i \sin y\]
into
\[e^{x+iy} = ( \cosh x + \sinh x) ( \cos y + i \sin y)\]
To get one big bad function of a complex variable!

Answer 6

I think that if we are in complex analysis, then we we are not in real analysis and vice versa

Answer 7

amazing, hyperbolic functions are mind blowing

Answer 8

all real functions can be defined in complex plane, nevertheless when we make that transition we go from real analysis into the complex one

Answer 9

I think it's a good question to ask, why is Euler's constant "e" related to the quadratic forms at all in the first place @Nishant_Garg ? Weird huh? :D

Answer 10

for example:
cos(ix) is a complex function, so it is represented by a complex number
whereas cosh(x) is a real function, so has the subsequent writing meaning:
cosh(x)=cos(ix) ?

Answer 11

@Michele_Laino I think the fact that we've used "i" from the start says we're out of the realm of real analysis to begin with.

Answer 12

ok! nevertheless what do you think about the subsequent formula:
cosh(x)=cos(ix) ? @Empty

Answer 13

I understand that cos(ix) is a real number for each value of x?

Answer 14

cos(x) is a real number for each value of x, however cos(ix) is not, we can consider an imaginary circular angle to be a hyperbolic angle!

This is very much like when you look at a conic section and see that the hyperbola is a 90 degree cut perpendicular to the cut that creates the circle! https://dj1hlxw0wr920.cloudfront.net/userfiles/wyzfiles/60cbe5b1-6b6c-48bd-9208-df73201f498a.gif

Answer 15

I tried for the sine
let's see
\[\sinh(ix)=\frac{e^{ix}-e^{-ix}}{2}=\frac{\cos(x)+i \sin(x)-(\cos(x)-i \sin(x))}{2}\]\[\sinh(ix)=\frac{\cos(x)+i \sin(x)-\cos(x)+i \sin(x)}{2}=\frac{2i \sin(x)}{2}=i \sin(x)\]
and
\[\sinh(x)=\frac{e^x-e^{-x}}{2}=\frac{\cos(ix)-i \sin(ix)-(\cos(ix)+i \sin(ix))}{2}\]\[\sinh(x)=\frac{\cos(ix)-i \sin(ix)-\cos(ix)-i \sin(ix)}{2}=\frac{-2i \sin(ix)}{2}=-i \sin(x)\]
\[\sinh(x)=-i \sin(ix)\]\[i \sin(ix)=-\sinh(x)\]\[i \sin(ix)=i^2 \sinh(x)\]\[\sin(ix)=i \sinh(x)\]

Answer 16

right! nevertheless cosh(x) is a real number for each value of x @Empty

Answer 17

Wait what have I said haha I agree with you, cosh(x) and cos(ix) will both have a completely real range if x is completely real domain, because they're equal.

Answer 18

@Nishant_Garg Perfect! :D

So is there an elliptic or parabolic pair of trig functions hiding out there somewhere? :P

Answer 19

No idea, that would be interesting though!! a trig functions for every conic section

Answer 20

Why is it that these quadratic forms are represented by e^x but what about cubic forms? It seems sort of weird and special for some reason and I don't know why.