Hyperbolic Functions

**U-God** · November 17th 2008, 12:49 AM

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.

Thanks in advance,

**mr fantastic** · November 17th 2008, 03:37 AM

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.

Thanks in advance,

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

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

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

**Soroban** · November 17th 2008, 05:31 AM

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

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