Sin hx, cos hx, tan hx

  • #1
sin hx, cos hx, tan h x.... What are they? What does the "h" mean here?

Any explanations or websites would be appreciated. :smile:
 

Answers and Replies

  • #2
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Sinh, cosh and tanh are the hyperbolic sine, hyperbolic cosine, and hyperbolic tangent functions. You guessed it, the h means hyperbolic.

sinh x = (ex-e-x)/2
cosh x = (ex+e-x)/2
tanh x = sinhx/coshx

They have a similar relation to a hyperbola as the ordinary trig. functions do to a circle.

Also:

sinh z = -i*sin(i*z)
cosh z = cos(i*z)


Where i is sqrt(-1)
 
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  • #3
Originally posted by Lonewolf
Also:

sinh x = -i*sin(i*z)
cosh x = cos(i*z)


Where i is sqrt(-1)
I guess z is a complex number, right?

sinh x = -i*sin(i*z) ...........(1)
sinh x = (ex-e-x)/2...........(2)
Differentiate sin hx = ex (interesting :smile: )

but I don't think a function involving complex number, -i*sin(i*z), is differentiable, or am I wrong?
Why -i*sin(i*z) = ex-e-x)/2 = sin hx ?
 
  • #4
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Sorry, those 'x's should have been 'z's

Yes, you can differentiate functions with a complex variable, in almost the same way as a real function.

To answer, or not, the second part of your question...I don't know. Maybe someone else does.
 
  • #5
Tom Mattson
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Originally posted by KL Kam
I guess z is a complex number, right?
Lonewolf meant to use x on both sides, I think.

Differentiate sin hx = ex (interesting :smile: )
No, you get:

d(sinh x)/dx=(1/2)d/dx(ex-e-x)
d(sinh x)/dx=(1/2)(ex+e-x)
d(sinh x)/dx=cosh x

but I don't think a function involving complex number, -i*sin(i*z), is differentiable, or am I wrong?
Yes, the above holds for complex variables, too.

Why -i*sin(i*z) = ex-e-x)/2 = sin hx ? [/B]
Again, I think Lonewolf meant to put x on both sides.
 
  • #6
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Lonewolf meant to use x on both sides, I think.
#

Yeah, I did. Sorry about the confusion.
 
  • #7
d(sinh x)/dx=(1/2)(ex+e-x)
oops, I swollowed the "-" sign when I differentiated e^(-x)

Yes, you can differentiate functions with a complex variable, in almost the same way as a real function.
Is it because we can represent a function with complex variables by using an Argand(sp?) Diagram, just as we use a Cartesian plane to represent functions with real variables?


(1/2)(ex+e-x)=cosh x
Is it the defination of cos hx ? (similarly we can find the defination of sin hx)

I've heard of Euler's formula eix = cos x + i sin x
are the hyperbolic functions somehow related to Euler's formula?

PS. I haven't met hyperbolic functions and Euler's forumla in school's syllabus yet
 
  • #8
Tom Mattson
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Originally posted by KL Kam
Is it because we can represent a function with complex variables by using an Argand(sp?) Diagram, just as we use a Cartesian plane to represent functions with real variables?
It is because the "limit" definition of a derivative holds equally well for complex numbers.

f'(z)=limh-->∞(f(z+h)-f(z))/h

I think your sentence should be put the other way around: We can draw an Aargand diagram for a differentiable function of a complex variable because it is differentiable.

I am probably missing some rigor there, but we have enough mathematicians here to fix me if I am wrong.

(1/2)(ex+e-x)=cosh x
Is it the defination of cos hx ?
Yes.

I've heard of Euler's formula eix = cos x + i sin x
are the hyperbolic functions somehow related to Euler's formula?
There is an analog. You can invert the system of equations:

sinh(x)=(1/2)(ex-e-x)
cosh(x)=(1/2)(ex+e-x)

and solve for ex, which would be the analog of Euler in for real exponents.

The result is:

ex=sinh(x) + cosh(x)

edit: typo
 
  • #9
Hurkyl
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While it is an elementary fact of complex analysis that functions like

f(z) = -i sin (iz)

are differentiable, it is not a trivial thing to prove from scratch. One has to define complex derivatives in a manner analogous to real derivatives and derive all of the existance theorems from that.


As for the why...


Euler's identity, I think, was the first connection between complex numbers and the trigonometric functions. It states that for any positive integer n:

(cos θ + i sin θ)n = cos nθ + i sin nθ

Any complex number z can be written, for some r and θ, as:

z = r (cos θ + i sin θ)

So one can write a formula for integer exponentiation:

zn = rn (cos nθ + i sin nθ)

This is the first inkling that the trig functions have something to do with exponents.


As to actually getting the formula for the trig functions in terms of exponentials, there are two ways to do it. One is to look at the Taylor series. The other way is to differentiate to make a differential equation and solve it. The formula for the trig functions are:

cos &theta = (1/2) (eiθ + e-iθ)
sin &theta = (1/2i) (eiθ - e-iθ)

The hyperbolic trig functions had been previously computed as:

cosh u = (1/2) (eu + e-u)
sinh u = (1/2) (eu - e-u)

So you just need to plug in the imaginary values to confirm the identities like:

cos ix = cosh x
 
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  • #10
Tom Mattson
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Hurkyl,

On the home row of your keyboard, where your right pinkie finger should be, is the "semicolon" key.

It makes all the difference between &theta and θ

:wink:
 
  • #11
x = a sin Θ
Θ = sin-1 (x/a)

X = A sin h θ
how can we write θ in terms of X and A ?
 
  • #12
Hurkyl
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On my computer, &theta is sufficient by itself to produce a theta. Until just now, I had no idea it wasn't sufficient for everyone else. :frown:
 
  • #13
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X = A sin h θ
how can we write θ in terms of X and A ?
In almost exactly the same way. We use the inverse function sinh-1θ

So, θ = sinh-1(X/A);

Where sinh-1θ = ln(θ+√(θ2+1))
 

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