Sin hx, cos hx, tan hx

In summary: Originally posted by Lonewolf 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:Same here. I thought it was like that for everyone else too.Originally posted by Lonewolf So, θ = sinh-1(X/A);Where sinh-1θ = ln(θ+√(θ2+1))I think you mean:θ = sinh^-1(X/A);Where sinh^-1θ = ln(θ+√(θ^2+1))In summary, sinh, cosh and tanh
  • #1
KLscilevothma
322
0
sin hx, cos hx, tan h x... What are they? What does the "h" mean here?

Any explanations or websites would be appreciated. :smile:
 
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  • #2
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
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
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
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
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
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 existence 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
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
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
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))
 

What is the difference between sin hx, cos hx, and tan hx?

Sin hx, cos hx, and tan hx are all trigonometric functions that are used to calculate the relationship between the sides and angles of a right triangle. The main difference between these functions is the ratio that they represent. Sin hx is the ratio of the opposite side to the hypotenuse, cos hx is the ratio of the adjacent side to the hypotenuse, and tan hx is the ratio of the opposite side to the adjacent side.

How are sin hx, cos hx, and tan hx used in science?

Trigonometric functions like sin hx, cos hx, and tan hx are used in various scientific fields such as physics, engineering, and astronomy. They are used to calculate distances, angles, and forces in different systems. For example, in physics, these functions are used to calculate the trajectory of a projectile or the amplitude of a wave.

What is the unit circle and how is it related to sin hx, cos hx, and tan hx?

The unit circle is a circle with a radius of 1 unit, centered at the origin of a coordinate system. It is used to understand the values of sin hx, cos hx, and tan hx at different angles. The x-coordinate of a point on the unit circle represents cos hx, and the y-coordinate represents sin hx. The tangent of an angle can be calculated by dividing the y-coordinate by the x-coordinate at that angle.

What are the important properties of sin hx, cos hx, and tan hx?

Some important properties of trigonometric functions include periodicity, symmetry, and monotonicity. Sin hx and cos hx are periodic functions with a period of 360 degrees or 2π radians. They also exhibit symmetry about the origin. Tan hx is a monotonic function, meaning it continuously increases or decreases without repeating any values.

How can I use the inverse trigonometric functions to find the angle given the ratio of sides?

The inverse trigonometric functions, such as sin^-1 hx, cos^-1 hx, and tan^-1 hx, can be used to find the angle given the ratio of sides. For example, if we know the ratio of the opposite side to the adjacent side is 0.5, we can use the inverse tangent function to find the angle. tan^-1 (0.5) = 26.57 degrees or 0.463 radians.

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