Which of the following expressions is always | vedclass

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(C) We use the hyperbolic identities: $1$. $\cosh x = 2 \cosh^2 \frac{x}{2} - 1 \implies \cosh \frac{x}{2} = \sqrt{\frac{1+\cosh x}{2}}$ $2$. $\cosh x

Vedclass · Accepted Answer

$\sinh \left(\frac{x}{2}\right)=\frac{\sinh x}{\sqrt{2(\cosh x+1)}}$

Vedclass · Answer

(C) We use the hyperbolic identities: $1$. $\cosh x = 2 \cosh^2 \frac{x}{2} - 1 \implies \cosh \frac{x}{2} = \sqrt{\frac{1+\cosh x}{2}}$ $2$. $\cosh x = 1 + 2 \sinh^2 \frac{x}{2} \implies \sinh \frac{x}{2} = \sqrt{\frac{\cosh x - 1}{2}}$ $3$. $	anh \frac{x}{2} = \frac{\sinh x}{1+\cosh x} = \frac{\cosh x - 1}{\sinh x}$ $4$. $\sinh x = 2 \sinh \frac{x}{2} \cosh \frac{x}{2}$ Evaluating option $C$: $\frac{\sinh x}{\sqrt{2(\cosh x+1)}} = \frac{2 \sinh \frac{x}{2} \cosh \frac{x}{2}}{\sqrt{2(2 \cosh^2 \frac{x}{2})}} = \frac{2 \sinh \frac{x}{2} \cosh \frac{x}{2}}{2 \cosh \frac{x}{2}} = \sinh \frac{x}{2}$. Thus,option $C$ is correct.

Which of the following expressions is always true?

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