cosh left(sinh ^{-1}(sqrt{8})+cosh ^{-1} 5rig | vedclass

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Question

(B) Let $x = \sinh^{-1}(\sqrt{8})$ and $y = \cosh^{-1}(5)$.
Then $\sinh x = \sqrt{8}$,so $\cosh x = \sqrt{1 + \sinh^2 x} = \sqrt{1 + 8} = 3$.
Thus $

Vedclass · Accepted Answer

$15+8 \sqrt{3}$

Vedclass · Answer

(B) Let $x = \sinh^{-1}(\sqrt{8})$ and $y = \cosh^{-1}(5)$.
Then $\sinh x = \sqrt{8}$,so $\cosh x = \sqrt{1 + \sinh^2 x} = \sqrt{1 + 8} = 3$.
Thus $e^x = \cosh x + \sinh x = 3 + \sqrt{8} = 3 + 2\sqrt{2}$.
For $y = \cosh^{-1}(5)$,$\cosh y = 5$ and $\sinh y = \sqrt{\cosh^2 y - 1} = \sqrt{25 - 1} = \sqrt{24} = 2\sqrt{6}$.
Thus $e^y = \cosh y + \sinh y = 5 + 2\sqrt{6}$.
We need to evaluate $\cosh(x + y) = \frac{e^{x+y} + e^{-(x+y)}}{2} = \frac{e^x e^y + e^{-x} e^{-y}}{2}$.
Note that $e^{-x} = \frac{1}{3 + 2\sqrt{2}} = 3 - 2\sqrt{2}$ and $e^{-y} = \frac{1}{5 + 2\sqrt{6}} = 5 - 2\sqrt{6}$.
$e^x e^y = (3 + 2\sqrt{2})(5 + 2\sqrt{6}) = 15 + 6\sqrt{6} + 10\sqrt{2} + 8\sqrt{3}$.
$e^{-x} e^{-y} = (3 - 2\sqrt{2})(5 - 2\sqrt{6}) = 15 - 6\sqrt{6} - 10\sqrt{2} + 8\sqrt{3}$.
$\cosh(x + y) = \frac{(15 + 6\sqrt{6} + 10\sqrt{2} + 8\sqrt{3}) + (15 - 6\sqrt{6} - 10\sqrt{2} + 8\sqrt{3})}{2} = \frac{30 + 16\sqrt{3}}{2} = 15 + 8\sqrt{3}$.

$\cosh \left(\sinh ^{-1}(\sqrt{8})+\cosh ^{-1} 5\right)=$

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