`sinh^2(x) = (-1+cosh(2x))/2` Verify the identity.

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

proof:

LHS=>

$\displaystyle{{\sinh}}^{{2}}{\left({x}\right)}={{\left(\frac{{{{e}}^{{x}}-{{e}}^{{-{x}}}}}{{2}}\right)}}^{{2}}$

= $\displaystyle\frac{{{{\left({{e}}^{{x}}-{{e}}^{{-{x}}}\right)}}^{{2}}}}{{4}}$

= $\displaystyle\frac{{{{\left({{e}}^{{x}}\right)}}^{{2}}-{2}{{e}}^{{x}}{{e}}^{{-{x}}}+{{\left({{e}}^{{-{{x}}}}\right)}}^{{2}}}}{{4}}$

= $\displaystyle\frac{{{{e}}^{{{2}{x}}}+{{e}}^{{-{2}{x}}}-{2}}}{{4}}$

= $\displaystyle\frac{{-{2}+{{e}}^{{{2}{x}}}+{{e}}^{{-{2}{x}}}}}{{4}}$

= $\displaystyle\frac{{\frac{{-{2}+{{e}}^{{{2}{x}}}+{{e}}^{{-{2}{x}}}}}{{2}}}}{{2}}$

= $\displaystyle\frac{{-{1}+\frac{{{{e}}^{{{2}{x}}}+{{e}}^{{-{2}{x}}}}}{{2}}}}{{2}}$

= $\displaystyle\frac{{-{1}+{\cosh{{2}}}{x}}}{{2}}$

= RHS

as LHS=RHS

so,

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

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