another question on hyperbolic function?

Question

so i understood what's meant by hyperbolic fuction
i need to ask about something else
if cos x= adj/hyb i know from where it comes but i want to know from where cosh x= (e^x+e^-x)/2

Answer 1

There isn't an IMMEDIATE direct physical/geometrical connection as there is with the traditional trigonometric functions. (I'll outline how a not so immediate physical interpretration arises, below.) There are however several ways of seeing interesting connections that make them very useful in practical applications, and that therefore may help justify one's interest in them other than asking questions like : "What happens if I replace x by -ix?"

You know that y (x) = cos (x) = [e^(ix) + e^(-ix)] / 2, ......(T1)

and that : z (x) = cosh (x) = [e^x + e^(-x)] / 2, ......(H1)

which shows that cosh (ix) = cos x, and cosh w = cos (-iw) = cos (iw). ......(HT)

These relationships have several consequences; alternatively, you can view them as the reason why the hyperbolic functions were physically desirable to invent in the first place.

Simple harmonic motion for a small perturbation y (from equilibrium) is given by differential equations of the form:

d^2 y / dx^2 + m^2 x = 0, with solutions y = A cos x + B sin x. ......(T2)

The solutions to these equations are periodic and bounded. Physically, they're involved with locally STABLE physical problems. (I'll only discuss these simplest, first order analyses.)

But what about UNSTABLE physical problems? For them, in the (simplest) local linear regime, the corresponding D.E. takes the form:

d^2 z / dx^2 + m^2 x = 0, with solutions z = A cosh x + B sinh x. ......(H2)

The solutions to these equations are NOT periodic. In general they're UNBOUNDED. That's why, physically, they're involved with locally UNSTABLE physical problems.

You also know that whereas cos^2 x + sin^2 (x) = 1, ......(T3)

and: cosh^2 w - sinh^2 (w) = 1. ......(H3)

However, you may not know the following : When one rotates REAL coordinate axes (by an angle +/- theta; don't tie me down to a specific sign attribution!) , the transformation takes the form:

x' = x cos theta +/- y sin theta, ......(T4)

and a similar one for y'.

These transformations have an INVARIANT, effectively Pythagoras's Theorem:

x'(^2) + (y')^2 === x^2 + y^2. ......(T5)

(I put === to emphasize this invariance; I don't have access to an "identically equals' sign, with three dashes vertically.)

In Special Relativity (S.R.), instead of the transformation (T4), one has:

x' = x cosh theta +/- ict sinh theta, ......(T4)

and a similar one for (ict').

The natural consequence of THIS is the relationship made famous by Einstein, Lorentz, and more particularly Einstein's colleague, the tragic Hermann Minkowski:

x'(^2) - (ct')^2 === x^2 - (ct)^2. ......(T5)

(Pythagoras's Theorem for an imaginary angle!)

For that reason, the classic S.R. transformation can be viewed as ROTATION THROUGH AN IMAGINARY ANGLE! This MAY (or MAY NOT) help you to "understand" (or at least accept the possiblity) that whereas anything rotated in REAL 3-D space looks longest when viewed perpendicularly, and ALWAYS looks shorter at any other angle, the corresponding thing isn't true in S.R.: the "projections" in its case can be LARGER on "rotation" !

In one way of trying to visualize such things, the circle, so useful in real considerations, can be replaced by a hyperbola, which now plays the analogous role in S.R. THEN, it IS possible to appreciate a physical/geometrical interpretation of cosh and sinh.

For more investigation of this, an intriguing source is Burke, W.L. (1980) "Spacetime, Geometry, Cosmology."

Answer 2

If you remember the definitions of hyperbolic sine and hyperbolic cosine functions then then all do is substitute them in and solve. Here are the definitions of the functions in case you do not know them. sinhx = (e˟ - eˉ˟) / 2 coshx = (e˟ + eˉ˟) / 2 6sinhɸ - 4coshɸ - 3 = 0 6sinhx - 4coshx - 3 = 0 6(e˟ - eˉ˟) / 2 - 4(e˟ + eˉ˟) / 2 - 3 = 0 3(e˟ - eˉ˟) - 2(e˟ + eˉ˟) - 3 = 0 3e˟ - 3eˉ˟ - 2e˟ - 2eˉ˟ - 3 = 0 e˟ - 3 - 5eˉ˟ = 0 e ^ (2x) - 3e˟ - 5 = 0 (e˟)² - 3e˟ - 5 = 0 4(e˟)² - 12e˟ - 20 = 0 (2e˟ - 3)² - 29 = 0 (2e˟ - 3)² = 29 2e˟ - 3 = ±√29 2e˟ = 3 ± √29 e˟ = (3 ± √29) / 2 x = ln[(3 ± √29) / 2 x = ln(3 ± √29) - ln2 ɸ = ln(3 ± √29) - ln2 We reject the negative answer and are left with ɸ = ln(3 + √29) - ln2.

Answer 3

Draw a set of coordinate axes.
Now draw the lines y=x and y= -x
These two lines are the asymptotes for a rectangular hyperbola
Now sketch in a hyperbola. (Only the right half is necesary).
Now draw a line from the origin intersecting the hyperbola at some point (x,y). Draw the reflection of this line intersecting the hyperbola at (x,-y).
Now let the area bounded by the two lines and the hyperbola be called x.

Then the perpendicular distance from (x,y) to the x-axis = sinh x and the perpendicular distance from (x,y) to the y-axis = cosh x.

I hope I expalined this OK. It's hard to do without a diagram.

Answer 4

We begin by observing that the argument given to prove part (iii) in Theorem 5.1 easily generalizes to the complex case with the aid of Definition 5.5. That is,

(5-30) ,

for all z, whether z is real or complex. Hence,

(5-31)

Adding the above two expressions and solving for gives

(5-32) ,

and subtracting (5-32) from (5-31) and solving for gives

(5-33) .

Answer 5

It came from applying the definitions of sine and cosine to the x²-y²=1 hyperbola. It looks like they created Taylor series and they simplfied down to the formulas above.

Check out the links below.

Answer 6

Differential equations.

cos(nx) and sin(nx) form a basis for the solutions to y"+n^2*y=0. cosh and sinh form a basis for y"-n^2*y=0

Answer 7

if x is imaginair then cosh(x) is real