How Do Hyperbolic Functions Relate to Trigonometric Functions?

[Oblio]

The hyperbolic functions are defined as follows:

coshz = e^{z} + e^{-z} /2

sinhz = e^{z} - e^{-z} /2

a.)Show that coshz = cos (iz). What is the corresponding relationship for sinhz?
b.)What are the derivatives of coshz and sinhz? What about their integrals?
c.)Show that cosh^2z - sin^2 =1
d.)Show that the integral of dx/sqrt[1+x^2 = arcsin x.

Hint : substitution x = sinhz.
I'd LOVE starters on showing this, we're told to assume z is real.
I get the idea that there's an imaginery aspect to hyperbolic functions, since coshz = cos(iz) ?

 

[learningphysics]

depends on what you're allowed to start with.

can you use e^{iz} = cos(z) + isin(z)

 

[Oblio]

I'm not sure.
Beforehand these values were sketched over a range of real values of z.
I don't know if that answers whether 'were allowed' ...

 

[Oblio]

Correction: part a.) isn't necessary. My bad.

 

[Oblio]

I have really no idea how to derive hyperbolic functions.
Is d/dx of cosh = -sinh?

 

[learningphysics]

I have really no idea how to derive hyperbolic functions.
Is d/dx of cosh = -sinh?

No. Get the derivative of \frac{e^z + e^{-z}}{2} with respect to z.

 

[Oblio]

=2e^z + 2e^-z / 4

?

 

[learningphysics]

=2e^z + 2e^-z / 4

?

check your work...

 

[Oblio]

is d/dx of e^z + e^-z the same thing or not?

 

[Oblio]

nm reading now, i know that's wrong

 

[Oblio]

is d/dx 0?

 

[Oblio]

orrr..

e^z - e^-z / 4
?

 

[learningphysics]

orrr..

e^z - e^-z / 4
?

Show your steps...

 

[Oblio]

Show your steps...

e^{z} + e^{-z} / 2

d/dx e^{z} = e^{z} * d/dx (z)
z=real number, d/dx = 1

d/dx e^{-z} = e^{-z} * d/dx (-z)
z=real number, d/dx = -1

Denominator ^2 by quotient rule...

=e^{z} - e^{-z} / 4

?

 

[learningphysics]

e^{z} + e^{-z} / 2

d/dx e^{z} = e^{z} * d/dx (z)
z=real number, d/dx = 1

d/dx e^{-z} = e^{-z} * d/dx (-z)
z=real number, d/dx = -1

Denominator ^2 by quotient rule...

=e^{z} - e^{-z} / 4

?

just do d/dz...

it's just (1/2)(e^z + e^-z)

taking d/dz you get \frac{1}{2}(\frac{d}{dz}e^z + \frac{d}{dz}e^{-z})

so the answer is just (e^z - e^-z)/2

if you do it using the quotient rule... you need to do derivative of the numerator by the denominator, minus the numerator*deriavative of the denominator divided by the denominator squared so...

\frac{(e^z - e^{-z})2 - (e^z + e^{-z})(0)}{2^2}

and you get the same result.

 

[Oblio]

I don't know the rule of putting a half there...

 

[learningphysics]

I don't know the rule of putting a half there...

It's just taking out the constant.

if z = A*y

then taking the derivative of both sides with respect to x...

dz/dx = A*(dy/dx)

For example... the derivative of 5e^(2x) = 5*d/dx(e^(2x)) = 5*2e^(2x) = 10e^(2x)

 

[Oblio]

alrighty, i think i get it.
basically what i get is that d/dx of cosh is sinh and vice versa right?

 

[Oblio]

I forget integrating quotients.. and e...

 

[learningphysics]

alrighty, i think i get it.
basically what i get is that d/dx of cosh is sinh and vice versa right?

Yeah.

 

[Oblio]

all i have left to do is prove that arcsinhx = integral dx 1/ sqrt[1+x^2] and I can't figure this out...

 

[learningphysics]

all i have left to do is prove that arcsinhx = integral dx 1/ sqrt[1+x^2] and I can't figure this out...

did you do the substitution x = sinhz ?

 

[Oblio]

i may have gotten somewhere...

can you explain why e^2x + e^-2x cancel out?

 

[learningphysics]

i may have gotten somewhere...

can you explain why e^2x + e^-2x cancel out?

Not sure... use the identity in part c) for the integral...

 

[Oblio]

but here its in a square root and added to one

 

[learningphysics]

but here its in a square root and added to one

yeah... what is sinh^2z + 1 using that identity?

 

[learningphysics]

Are you sure you have the question posted correctly? I think the question should be to prove that integral equals arcsinhx

 

[Oblio]

lol my bad.
for convenience on here i didnt match letters to the actual letters, and when you said c i looked at the wrong one.

i figured it out though, when you directed me to the right one with your last comment.

cosh^2 = 1 + sinh^2

integral dx 1/sqrt[cosh^2
=integral dx 1/cosh..

now I know the integra of cosh is sinh, but do i have to do something funky since its a quotient?

 

[Oblio]

It seems wrong that I can say
the integral of 1/cosh is 1/sinh

 

[Oblio]

Are you sure you have the question posted correctly? I think the question should be to prove that integral equals arcsinhx

thats what i meant, sorry didnt notice that typo.
I should have it right in my last posts...

 

[learningphysics]

It seems wrong that I can say
the integral of 1/cosh is 1/sinh

you need to change the dx... if x = sinhz, what is dx?

 

[Oblio]

coshz..

 

[learningphysics]

coshz..

yes dx = coshz*dz.

so what do you get from the integral?

 

[Oblio]

do i just do the integral of each the numerator and denominator?

 

[Oblio]

either way i don't see how i won't just get 1/1 = 1...

 

[learningphysics]

either way i don't see how i won't just get 1/1 = 1...

Right... you get the integral of dz, which is just z. z = arcsinh(x), since x = sinh(z).

 

[Oblio]

Right... you get the integral of dz, which is just z. z = arcsinh(x), since x = sinh(z).

I had accidentally switched some z's for x's... that's a bad thing. lol

I get it.
The only thing that seems odd now, is
WHY can one just say x=sinhz? Was that shown in the question somehow already that I'm not seeing?

 

[learningphysics]

I had accidentally switched some z's for x's... that's a bad thing. lol

I get it.
The only thing that seems odd now, is
WHY can one just say x=sinhz? Was that shown in the question somehow already that I'm not seeing?

I agree with you that it is odd the order we did it...

The better way to approach is to let z = arcsinh(x) (so we're introducing a new variable z... x is already given.)... and then from there saying x = sinh(z). I think that makes more sense.

 

[Oblio]

So, am I correct in thinking though, that this is only true SINCE x =sinhx.
I mean, they made it a 'Hint', but that definition of x is completely necessary to solve it, isn't it?

Without the hint, it couldn't be done?

 

[Oblio]

I guess I mean, it should written that the statement is true WHEN x =sinhz. (made a typo above i see)

 

[Oblio]

agree or am i missing something still? lol

 

[learningphysics]

So, am I correct in thinking though, that this is only true SINCE x =sinhx.
I mean, they made it a 'Hint', but that definition of x is completely necessary to solve it, isn't it?

Without the hint, it couldn't be done?

No, that definition wasn't necessary...

We need to define z = arcsinh(x). we're not defining x... x is already given... we define z = arcsinh(x), and then we get x = sinh(z). but this isn't a definition... just a relationship... z is the variable being defined.

given any variable... say u...

I can simply define v = arcsinh(u). now I have defined v not u... but u = sinh(v), because of the way v has been defined.

 

[learningphysics]

I guess I mean, it should written that the statement is true WHEN x =sinhz. (made a typo above i see)

z is defined in such a way that x = sinhz... in other words we create a variable z that is defined as arcsinh(x).

so z is being defined, not x.

 

[Oblio]

So, am I correct in thinking though, that this is only true SINCE x =sinhx.
I mean, they made it a 'Hint', but that definition of x is completely necessary to solve it, isn't it?

Without the hint, it couldn't be done?

Ok, switch x for z.
Was it necessary for the question to define z?

 

[learningphysics]

Ok, switch x for z.
Was it necessary for the question to define z?

The question didn't have to define it... the question suggested it as a hint to help you solve the problem...

Even if the question didn't define it... you could define a variable as arcsinh(x), to help you solve the problem.

The question is simply prove that arcsin(x) = integral(...)

The z = arcsinh(x) is only a hint to solve the problem... the question is complete even without that hint.