# Hyperbolic Functions

• Nov 17th 2008, 12:49 AM
U-God
Hyperbolic Functions
Show that $\displaystyle cosh(x-y) = cosh(x)cosh(y) - sinh(x)sinh(y)$

Everytime I try this with exponential it ends up being really messy and no doubt I'm cancelling something wrong or something similar. Is the exponential the only way to do it?

Show that $\displaystyle arccosh(x) = log(x+ \sqrt{x^2 - 1})$

Same as above, my exponentials just get hugely messy..

Write $\displaystyle tanh(arccosh(x))$ as an algebraic expression in x.
Is the only way to do this with the exponentials [do I have to use the answer above]? If so, same as the two above.

• Nov 17th 2008, 03:37 AM
mr fantastic
Quote:

Originally Posted by U-God
Show that $\displaystyle cosh(x-y) = cosh(x)cosh(y) - sinh(x)sinh(y)$

Everytime I try this with exponential it ends up being really messy and no doubt I'm cancelling something wrong or something similar. Is the exponential the only way to do it?

Mr F says: Read this for the derivation of a similar formula http://personal.maths.surrey.ac.uk/st/S.Gourley/hyperbolic.pdf

Show that $\displaystyle arccosh(x) = log(x+ \sqrt{x^2 - 1})$

Mr F says: Read this http://tartarus.org/gareth/maths/stuff/hyperbolic_functions.pdf

Same as above, my exponentials just get hugely messy..

Write $\displaystyle tanh(arccosh(x))$ as an algebraic expression in x.
Is the only way to do this with the exponentials [do I have to use the answer above]? If so, same as the two above.

Let $\displaystyle \text{arcosh} x = \alpha \Rightarrow \cosh \alpha = x$.

You need an expression for $\displaystyle \tanh \alpha$.

Note that $\displaystyle \cosh^2 \alpha - \sinh^2 \alpha = 1 \Rightarrow 1 - \tanh^2 \alpha = \frac{1}{\cosh^2 \alpha}$.
• Nov 17th 2008, 05:31 AM
Soroban
Hello, U-God!

Quote:

Show that: .$\displaystyle \cosh(x-y) \:=\: \cosh(x)\cosh(y) - \sinh(x)\sinh(y)$

The right side is: .$\displaystyle \frac{e^x+e^{-x}}{2}\cdot\frac{e^y + e^{-y}}{2} \;-\; \frac{e^x-e^{-x}}{2}\cdot\frac{e^y-e^{-y}}{2}$

. . $\displaystyle = \;\frac{e^{x+y} + e^{x-y} + e^{-x+y} + e^{-x-y}}{4}\;-\;\frac{e^{x+y} - e^{x-y} - e^{-x+y} + e^{-x-y}}{4}$

. . $\displaystyle = \;\frac{{\color{red}\rlap{////}}e^{x+y} + e^{x-y} + e^{-x+y} + {\color{green}\rlap{/////}}e^{-x-y} - {\color{red}\rlap{////}}e^{x+y} + e^{x-y} - e^{-x+y} - {\color{green}\rlap{/////}}e^{-x-y}}{4}$

. . $\displaystyle = \;\frac{2e^{x-y} + 2e^{-x+y}}{4} \;\;=\;\;\frac{e^{x-y} + e^{-(x-y)}}{2} \;\;=\;\;\cosh(x-y)$

Quote:

Show that: .$\displaystyle \text{arccosh}(x) \:=\:\ln(x+ \sqrt{x^2 - 1})$

We have: .$\displaystyle y \:=\:\text{arccosh}(x) \quad\Rightarrow\quad \cosh y \:=\:x \quad\Rightarrow\quad \frac{e^y + e^{-y}}{2} \:=\:x$

Multiply by $\displaystyle 2e^y\!:\;\;e^{2y} + 1 \:=\:2xe^y\quad\Rightarrow\quad e^{2y} - 2xe^y + 1 \:=\:0$

We have a quadratic: .$\displaystyle (e^y)^2 - 2x(e^y) + 1 \:=\:0$

Quadratic Formula: .$\displaystyle e^y \;=\;\frac{2x\pm\sqrt{4x^2-4}}{2}$

. . We have: .$\displaystyle e^y\;=\;x + \sqrt{x^2-1}$ .
(We use the positive square root.)

. . Then: .$\displaystyle y \;=\;\ln(x + \sqrt{x^2-1})$

Therefore: .$\displaystyle \text{arccosh}(x) \;=\;\ln(x + \sqrt{x^2-1})$