Notes: `sinh(x+y) = sinh(x)cosh(y) + cosh(x)sinh(y)` Verify the identity.

Saturday, 4 March 2017

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

$\displaystyle{\sinh{{\left({x}+{y}\right)}}}={\sinh{{\left({x}\right)}}}{\cosh{{\left({y}\right)}}}+{\cosh{{\left({x}\right)}}}{\sinh{{\left({y}\right)}}}$

Take note that hyperbolic sine and hyperbolic cosine are defined by

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

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

Applying these formulas to the right side of the equation, it becomes

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

Multiplying the fractions, the right side turns into

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

Combining the like terms, it simplifies to

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

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

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

And, express the right side in...

$\displaystyle{\sinh{{\left({x}+{y}\right)}}}={\sinh{{\left({x}\right)}}}{\cosh{{\left({y}\right)}}}+{\cosh{{\left({x}\right)}}}{\sinh{{\left({y}\right)}}}$

Take note that hyperbolic sine and hyperbolic cosine are defined by

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

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

Applying these formulas to the right side of the equation, it becomes

Multiplying the fractions, the right side turns into

Combining the like terms, it simplifies to

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

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

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

And, express the right side in terms of hyperbolic function again. Applying the definition of hyperbolic sine, the right side transforms to

$\displaystyle{\sinh{{\left({x}+{y}\right)}}}={\sinh{{\left({x}+{y}\right)}}}$

This verifies that the given equation is an identity.

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

Notes

Saturday, 4 March 2017

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

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How are race, gender, and class addressed in Oliver Optic's Rich and Humble?

Saturday, 4 March 2017

Verify the identity.

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Post a Comment

How are race, gender, and class addressed in Oliver Optic&#39;s Rich and Humble?

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

How are race, gender, and class addressed in Oliver Optic's Rich and Humble?