# From cosh2A = 1+2(sinh^2)A, how do you prove that (sinh^4)A + (cosh^4)A=(cosh4A + 3)/4 and also (cosh^4)A - (sinh^4)A = cosh2A?

Feb 7, 2017

see below

#### Explanation:

Part I .
${\sinh}^{4} A + {\cosh}^{4} A = \frac{\cosh 4 A + 3}{4}$

Use the formulas:
$\cosh 2 A = 2 {\cosh}^{2} A - 1$ -->Solve for ${\cosh}^{2} A$

$\cosh 2 A = 1 + 2 {\sinh}^{2} A$ --->Solve for ${\sinh}^{2} A$

Left Hand Side:

${\sinh}^{4} A + {\cosh}^{4} A = {\left({\sinh}^{2} A\right)}^{2} + {\left({\cosh}^{2} A\right)}^{2}$

$= {\left(\frac{\cosh 2 A - 1}{2}\right)}^{2} + {\left(\frac{\cosh 2 A + 1}{2}\right)}^{2}$-->FOIL

$= \frac{{\cosh}^{2} 2 A - 2 \cosh 2 A + 1}{4} + \frac{{\cosh}^{2} 2 A + 2 \cosh 2 A + 1}{4}$

$= \frac{{\cosh}^{2} 2 A - 2 \cosh 2 A + 1 + {\cosh}^{2} 2 A + 2 \cosh 2 A + 1}{4}$

$= \frac{{\cosh}^{2} 2 A - \cancel{2 \cosh 2 A} + 1 + {\cosh}^{2} 2 A + \cancel{2 \cosh 2 A} + 1}{4}$

$= \frac{{\cosh}^{2} 2 A + 1 + {\cosh}^{2} 2 A + 1}{4}$

$= \frac{2 {\cosh}^{2} 2 A + 2}{4}$

Note: $\cosh 4 A = 2 {\cosh}^{2} 2 A - 1$ $\therefore$ $\cosh 4 A + 1 = 2 {\cosh}^{2} 2 A$

$= \frac{\cosh 4 A + 1 + 2}{4}$

$= \frac{\cosh 4 A + 3}{4}$

$\therefore =$ Right Hand Side

Part II
${\cosh}^{4} A - {\sinh}^{4} A = \cosh 2 A$

Use the properties :
${\cosh}^{2} A - {\sinh}^{2} A = 1$ and $\cosh 2 A = 1 + 2 {\sinh}^{2} A$

Left Hand Side :

${\cosh}^{4} A - {\sinh}^{4} A = \left({\cosh}^{2} A + {\sinh}^{2} A\right) \left({\cosh}^{2} A - {\sinh}^{2} A\right)$

$= \left({\cosh}^{2} A + {\sinh}^{2} A\right) \cdot 1$

$= {\cosh}^{2} A + {\sinh}^{2} A$

$= 1 + {\sinh}^{2} A + {\sinh}^{2} A$

$= 1 + 2 {\sinh}^{2} A$

$= \cosh 2 A$

$\therefore =$ Right Hand Side