Notes: `e^(2x) = sinh(2x) + cosh(2x)` Verify the identity.

Sunday, 27 July 2014

$\displaystyle{{e}}^{{{2}{x}}}={\sinh{{\left({2}{x}\right)}}}+{\cosh{{\left({2}{x}\right)}}}$ Verify the identity.

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

Take note that hyperbolic sine and hyperbolic cosine are defined as

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

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

Apply these two formulas to express the right side in exponential form.

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

Adding the two fractions, the right side simplifies to

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

$\displaystyle{{e}}^{{{2}{x}}}={{e}}^{{{2}{x}}}$

This proves that the given equation is an identity.

Therefore, $\displaystyle{{e}}^{{{2}{x}}}={\sinh{{\left({2}{x}\right)}}}+{\cosh{{\left({2}{x}\right)}}}$ is an identity.

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

Take note that hyperbolic sine and hyperbolic cosine are defined as

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

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

Apply these two formulas to express the right side in exponential form.

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

Adding the two fractions, the right side simplifies to

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

$\displaystyle{{e}}^{{{2}{x}}}={{e}}^{{{2}{x}}}$

This proves that the given equation is an identity.

Therefore, $\displaystyle{{e}}^{{{2}{x}}}={\sinh{{\left({2}{x}\right)}}}+{\cosh{{\left({2}{x}\right)}}}$ is an identity.

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