If sinh x = frac{5}{12},then cosh frac{x}{2} | vedclass

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(D) We know that $\cosh x = \sqrt{1 + \sinh^2 x}$.Given $\sinh x = \frac{5}{12}$,we have $\cosh x = \sqrt{1 + (\frac{5}{12})^2} = \sqrt{1 + \frac{25}{

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$\frac{5}{2 \sqrt{6}}$

Vedclass · Answer

(D) We know that $\cosh x = \sqrt{1 + \sinh^2 x}$.Given $\sinh x = \frac{5}{12}$,we have $\cosh x = \sqrt{1 + (\frac{5}{12})^2} = \sqrt{1 + \frac{25}{144}} = \sqrt{\frac{169}{144}} = \frac{13}{12}$.Using the identity $\cosh x = 2 \cosh^2 \frac{x}{2} - 1$,we get:$2 \cosh^2 \frac{x}{2} - 1 = \frac{13}{12}$$2 \cosh^2 \frac{x}{2} = \frac{13}{12} + 1 = \frac{25}{12}$$\cosh^2 \frac{x}{2} = \frac{25}{24}$$\cosh \frac{x}{2} = \sqrt{\frac{25}{24}} = \frac{5}{2 \sqrt{6}}$.

If $\sinh x = \frac{5}{12}$,then $\cosh \frac{x}{2} =$

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