If cosh x = frac{sqrt{14}}{3},sinh x = cos th | vedclass

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(B) Given $\cosh x = \frac{\sqrt{14}}{3}$.Using the identity $\sinh^2 x = \cosh^2 x - 1$,we get:$\sinh^2 x = \left(\frac{\sqrt{14}}{3}\right)^2 - 1 =

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$\frac{2}{3}$

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(B) Given $\cosh x = \frac{\sqrt{14}}{3}$.Using the identity $\sinh^2 x = \cosh^2 x - 1$,we get:$\sinh^2 x = \left(\frac{\sqrt{14}}{3}ight)^2 - 1 = \frac{14}{9} - 1 = \frac{5}{9}$.Since $\sinh x = \cos 	heta$,we have $\cos 	heta = \frac{\sqrt{5}}{3}$ (taking the positive root as $\sinh x$ is positive for $x > 0$ and $\cos 	heta$ is positive in the given range).Now,$\sin^2 	heta = 1 - \cos^2 	heta = 1 - \frac{5}{9} = \frac{4}{9}$.Thus,$\sin 	heta = \pm \frac{2}{3}$.Given $-\pi In the second quadrant,$\sin 	heta$ is positive.However,checking the range $-\pi Wait,re-evaluating the quadrant: $-\pi  0$.Therefore,$\sin 	heta = \frac{2}{3}$.

If $\cosh x = \frac{\sqrt{14}}{3}$,$\sinh x = \cos \theta$ and $-\pi < \theta < -\frac{\pi}{2}$,then $\sin \theta =$

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