cosh (alpha + ibeta ) - cosh (alpha - ibeta ) | vedclass

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Question

(C) We use the identity $\cosh(x \pm y) = \cosh x \cosh y \pm \sinh x \sinh y$. Also,$\cosh(i\beta) = \cos \beta$ and $\sinh(i\beta) = i \sin \beta$.

Vedclass · Accepted Answer

$2i \sinh \alpha \sin \beta $

Vedclass · Answer

(C) We use the identity $\cosh(x \pm y) = \cosh x \cosh y \pm \sinh x \sinh y$. Also,$\cosh(i\beta) = \cos \beta$ and $\sinh(i\beta) = i \sin \beta$. $\cosh (\alpha + i\beta ) - \cosh (\alpha - i\beta ) = (\cosh \alpha \cosh (i\beta ) + \sinh \alpha \sinh (i\beta )) - (\cosh \alpha \cosh (i\beta ) - \sinh \alpha \sinh (i\beta ))$ $= \cosh \alpha \cosh (i\beta ) + \sinh \alpha \sinh (i\beta ) - \cosh \alpha \cosh (i\beta ) + \sinh \alpha \sinh (i\beta )$ $= 2 \sinh \alpha \sinh (i\beta )$ Since $\sinh (i\beta ) = i \sin \beta$,the expression becomes $2i \sinh \alpha \sin \beta $.

$\cosh (\alpha + i\beta ) - \cosh (\alpha - i\beta )$ is equal to

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