# Math Help - Hyperbolic Functions

1. ## Hyperbolic Functions

Show that $cosh(x-y) = cosh(x)cosh(y) - sinh(x)sinh(y)$

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

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

Write $tanh(arccosh(x))$ 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.

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

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

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

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

Write $tanh(arccosh(x))$ 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 $\text{arcosh} x = \alpha \Rightarrow \cosh \alpha = x$.

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

Note that $\cosh^2 \alpha - \sinh^2 \alpha = 1 \Rightarrow 1 - \tanh^2 \alpha = \frac{1}{\cosh^2 \alpha}$.

3. Hello, U-God!

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

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

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

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

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

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

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

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

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

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

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

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

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