# Why does the definition of cosh and sinh contain division by 2?

1. Feb 2, 2010

### Juwane

$$cosh = \frac{e^x + e^{-x}}{2}$$

In the above definition, why there is division by 2? Is it there so that when x=0, y could be 1?

2. Feb 2, 2010

### dextercioby

The 1/2 comes from the related function <cos>. The definition of <cos> in terms of exponentials has a 1/2. To relate the hyperbolic functions in the simplest way to their circular counterparts, the 1/2 must be in the definition of <cosh/ch>.

3. Feb 2, 2010

### HallsofIvy

There is a thread under "General Math" about even and odd functions in which I addressed exactly this point!

For any function f(x), we define its "even part" to be (f(x)+ f(-x))/2 and its "odd part" to be (f(x)- f(-x))/2. They are, of course, even and odd functions. The "2" in the denominator is so that their sum is (f(x)+ f(-x))/2+ (f(x)- f(-x))/2= (f(x)+ f(-x)+ f(x)- f(-x))/2= 2f(x)/2= f(x).

Cosh(x) and sinh(x) are the even and odd parts of ex: with the "2" in the denominator, cosh(x)+ sinh(x)= ex.

They are also, by the way, the "fundamental solutions" to the differential equation $d^2y/dt^2= y$. That is, cosh(x) is the solution to that equation such that y(0)= 1 and y'(0)= 0 and sinh(x) is the solution such that y(0)= 0, y'(1)= 0. If y is a solution to that differential equation with y(0)= A, y'(0)= B, then y(x)= A cosh(x)+ B sinh(x)- the coefficients are just the value of y and its derivative at 0.

Of course, you need the "2" in the denominator to make d sinh/dx(0)= cosh(0)= (e0+ e-0)/2= (1+ 1)/2= 1.

Last edited by a moderator: Feb 2, 2010