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...