Prove derivative of hyperbolic trig functions ... without using cosh x = [ e^x + e^-x ] / 2?

Question

Im looking for a proof of the derivative of the trig functions... FROM WHICH I can derive cosh x = [ e^x + e^-x ] / 2

Can anyone help with this?

The hyperbolic functions existed independent from the normal trig functions, and prior to complex analysis and their subsequently proven relationship.

Im looking for a way to prove:
cos ix = [ e^i(ix) + e^-i(ix) ] / 2 = [ e^x + e^-x ] / 2 = cosh x

... without merely defining cosh x = [ e^x + e^-x ] / 2.

My point is that how do we know cosh x = [ e^x + e^-x ] / 2 ?

cos ix = [ e^x + e^-x ] / 2... but how does that happen to be cosh also?

Please understand that my concern is the progressive concept tree. I dont want the mere demonstration of the relationship, I want an explanation of how we ever arrived at this relationship from simpler concepts as a basis.

I used to be satisfied with the mere definition of cosh x = cos ix.... or cosh x = [ e^x + e^-x ] / 2... but that was until two days ago when I discovered that the hyperbolic functions existed prior to complex analysis and is used in its own right to describe certain geometric phenomenon.

Now, even though I understand how we managed to arrive at
cos ix = [ e^x + e^-x ] / 2,
I dont see how that was proven to be equal to cosh x

We would have first needed to know that
cosh x = [ e^x + e^-x ] / 2 in order to make that comparison.

Taylor expansion uses derivatives... to create an infinite series... to create Euler's formula cis x = e^ix... to create cosine and sine in terms of e... where we apply imaginary angles... to create sin ix and cos ix in terms of e.

But where do cosh and sinh fall in to place? We cannot equate them without first knowing what they are in terms of e. The only way I can conceive of doing this is via Taylor expansion... which requires differentiation.

I used to think that hyperbolic functions were created solely to evaluate normal trig values for complex angles. cos ix = [ e^x + e^-x ] / 2 would have been sufficient, as defining cosh x = cos ix would work for me. But cosh preexisted cos ix.

How did we ever equate the two?

My concept issue is solely a matter of finding a proof cos ix = cosh x that isnt based on the consequence of it being true. Circle logic is not a valid proof.

The historical and conceptual evolution of math, from simpler concepts to more complex ones.

Is there not a single mathematician out there that can answer this simple problem?

I find it rather sad that math geniuses out there know all this crap about math but fail to understand their proofs.

Scratch my header question. I dont care how you prove the derivative of the hyp functions.... just dont use higher concepts that result from the fact.

Either find a proof for the derivative using using cosh x = [ e^x + e^-x ] / 2
Or prove cosh x = [ e^x + e^-x ] / 2 without using cosh x = cos ix

Either way, this circle of self proving proofs needs to find an out.

(Ω)kaksi...
That is confusing... as its not clearly written.

But I like how you reference an answer you made under a separate account.

In any case, that looks more like a demonstration of the relationship rather than a proof of it.

You seem to be defining hyp trig functions in terms of the imaginary unit and the normal trig functions.

But the hyp functions preexisted complex analysis and this supposed definition. I still fail to see the relationship between the two.

If you define, first, that (cosh a, sinh a) as the (x, y) coordinate pair of a unit hyperbola... show the relationship now.

I concede everything you all said about complex numbers in normal trig functions...and all that about integration and anti-derivatives.

But how does that happen to equal the coordinate pair of a unit hyperbola?

What realization allows us to stick cos ix on one side and cosh x on the other? Besides the assumption that cosh x = [ e^x + e^-x ] / 2 ?

I mean... from my perspective... cosh x is a separate entity of math from cos ix.

So what that cos ix = [ e^x + e^-x ] / 2 ?

We can evaluate the cosine of imaginary values. We can evaluate hyperbolic trig. What proof equates them?

The equation cosh x = [ e^x + e^-x ] / 2 is based on cos ix... its a consequence, not a proof.

Answer 1

Let (x, y) be a point on the unit hyperbola x² - y² = 1 in the first quadrant. Define the hyperbolic sector of this point to be the region bounded by the x-axis, the unit hyperbola, and the line segment from (0, 0) to (x, y). The hyperbolic angle α may then be defined as twice the area of the hyperbolic sector. Now, note that the area of the hyperbolic sector is the area of the right triangle with vertices (0, 0), (x, 0), and (x, y) minus the area under the hyperbola from 1 to x. The top half of the hyperbola is given by y=√(x²-1). So the hyperbolic angle α is:

α = xy - 2 [1, x]∫√(t² - 1) dt

Solving the integral, we make the substitution u=arcsec (t), so that sec u = t. Then dt = sec u tan u du. Note that when t=1, u=0 and when when t=x, u=arcsec x, so we have:

α = xy - 2 [0, arcsec (x)]∫√(sec² u - 1) sec u tan u du

Simplifying this integral:

α = xy - 2 [0, arcsec (x)]∫√(tan² u) sec u tan u du
α = xy - 2 [0, arcsec (x)]∫sec u tan² u du (since tan u is positive)

Now, in order to find [0, arcsec (x)]∫sec u tan² u du, we break this off from the main problem and integrate by parts -- let v = tan u and dw = sec u tan u du. Then dv = sec² u du and w = sec u. So we have:

[0, arcsec (x)]∫sec u tan² u du = sec u tan u|[0, arcsec x] - [0, arcsec x]∫sec³ u du

[0, arcsec (x)]∫sec u tan² u du = x tan (arcsec x) - [0, arcsec x]∫sec u (1+tan² u) du

[0, arcsec (x)]∫sec u tan² u du = x √(sec² (arcsec x) - 1) - [0, arcsec x]∫sec u du - [0, arcsec x]∫sec u tan² u du

[0, arcsec (x)]∫sec u tan² u du = x √(x² - 1) - [0, arcsec x]∫sec u du - [0, arcsec x]∫sec u tan² u du

Adding [0, arcsec x]∫sec u tan² u du to both sides:

2 [0, arcsec (x)]∫sec u tan² u du = x √(x² - 1) - [0, arcsec x]∫sec u du

So substituting this into the original equation:

α = xy - x √(x² - 1) + [0, arcsec x]∫sec u du

But note that y=√(x²-1):

α = xy - xy + [0, arcsec x]∫sec u du
α = [0, arcsec x]∫sec u du

Now make the substitution v = sec u + tan u, dv = sec u tan u + sec² u du = v sec u du. Note that when u=0, v=1, and when u=arcsec x, v = x + tan (arcsec x) = x + √(sec² (arcsec x) - 1) = x + √(x² - 1). So we have:

α = [1, x + √(x² - 1)]∫1/v dv
α = ln v | [1, x + √(x² - 1)]
α = ln (x + √(x² - 1)) - ln 1
α = ln (x + √(x² - 1))

From this we may derive an expression for x in terms of the hyperbolic angle α. By exponentiating this equation, we obtain:

e^α = x + √(x² - 1)

Squaring:

e^(2α) = x² + 2x√(x² - 1) + x² - 1
e^(2α) = 2x(x + √(x² - 1)) - 1

But note that e^α = x + √(x² - 1), so we have:

e^(2α) = 2xe^α - 1

Dividing by e^α and simplifying:

e^α = 2x - e^(-α)
2x = e^α + e^(-α)
x = (e^α + e^(-α))/2

If we want a similar expression for y, we may simply note that:

y
= √(x² - 1)
= √(((e^α + e^(-α))/2)² - 1)
= √((e^(2α) + 2 + e^(-2α))/4 - 1)
= √((e^(2α) - 2 + e^(-2α))/4)
= (e^α - e^(-α))/2

So, if we define cosh α to the the x-coordinate of a point that subtends a hyperbolic angle α and sinh α to be the y-coordinate of the same, the above serves as a proof that cosh α = (e^α + e^(-α))/2 and sinh α = (e^α - e^(-α))/2.

There is also a rather interesting article on the history of the hyperbolic functions here: http://ww2.hyperbolicsliderules.com:8110/Documents/Article_12042006.doc (MS word format).

Answer 2

The gist of the discussion out there (sorry, there are too many pages to provide all the links) is that wands help to focus the magic resident in the witch or wizard, but wandless magic is still possible, particularly for very talented people like Snape and Dumbledore. Some examples ... Apparition Assuming one's Animagus form (or Metamorphpagus) Accio (the Summoning Charm) Elves can do magic without using wands Lumos

Answer 3

Im looking for a proof of the derivative of the trig functions... FROM WHICH I can derive cosh x = [ e^x + e^-x ] / 2 Can anyone help with this?


Assume cos(ix), or e^ix can be written in complex form as A(x)+iB(x), Where A and B are real value functions.
Lets find cos(ix),
Assuming cos(ix) is continuous, We have

cos(ix)=A+iB
-isin(ix)=A'+iB'
cos(ix)=A"+iB"

Equating coefficients we have A=A", B=B" => A=C1e^x+C2e^-x, B=iC3e^x+iC4e^-4

So
I. cos(ix) = C1e^x+C2e^-x+iC3e^x+iC4e^-x
II. isin(ix) = C1e^x-C2e^-x+iC3e^x-iC4e^-x

Evaluating equations 1 and II at x=0

I. cos(i0) = 1= C1+C2+iC3 + iC4
II. isin(i0) = 0 =C1 -C2+iC3 - iC4

Once again equating coefficients:
C1+C2=1, C1-C2=0 => C1=C2=1/2
C3+C4=0, C3-C4=0 => C3=C4=0

Hence cos(ix)=(e^x+e^-x)/2.

One can derive the representation of e^(ix) in polar form following the same approach as cos(ix). Here you find A=-A" and B=-B". Both solutions for A and B requires recognition of the behavior of the nth derivative of sin(x) and cos(x). You'll find A=C1cos(x)+C2sin(x) B= C3cos(x)+C4sin(x).

e^ix=C1cos(x)+C2sin(x) +iC3cos(x) + iC4sin(x)
ie^ix=-C1sin(x)+C2cos(x) -iC3sin(x) + iC4cos(x)
Applying BC's for e^ix, and ie^x at (x=0)

e^0=1=C1(1)+C2(0)+iC3(1) +iC4(0)
ie^ix=i=-C1(0)+C2 -iC3(0) + iC4(1)

Hence C1=1,C2=0, C3=0, C4=1

e^ix=cos(x)+isin(x)

Answer 4

I hope this can help

e^x = ∑(n 0-->∞) [ x^n / n! ]

cos x =∑(n 0-->∞) [ (-1)^n x^(2n) /(2n)! ]

get (e^ix+ e^-ix )/2
& cos x

but if u want to prove that cosh x = e^x + e^-x
so prove that tan x is rise/run ?
I think this is the definition of cosh x it equals e^x + e^-x
anyhow I will try to find another definition

Answer 5

http://answers.yahoo.com/question/index;_ylt=AmhY7dqfsAoz1cVX8JXRcHQCxgt.;_ylv=3?qid=20071124172533AALL0mt

♠ noli audis ergo noli cogitas;
♣ well exp(jb) = cos(b) +j*sin(b) right?
exp(x) = exp(j*(-jx)) = cos(jx) –j*sin(jx);
exp(-x) = exp(j*jx) = cos(jx) +j*sin(jx);
♦ exp(x) + exp(-x) =2cos(jx), hence
cos(jx) =(exp(x) + exp(-x))/2,
which was named as ch(x);