`cosh(x) + cosh(y) = 2cosh((x+y)/2)cosh((x-y)/2)` Verify the identity.

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

proof:

Taking RHS , let us solve the proof

RHS=> $\displaystyle{2}{\cosh{{\left(\frac{{{x}+{y}}}{{2}}\right)}}}{\cosh{{\left(\frac{{{x}-{y}}}{{2}}\right)}}}$

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

its like 2((A+B)*(C+D))=2(AC+AD+BC+BD)

on multilication

= $\displaystyle{2}{\left[{\left[{\left({{e}}^{{\frac{{{x}+{y}}}{{2}}}}\cdot{\left({{e}}^{{\frac{{{x}-{y}}}{{2}}}}\right]}+{\left[{\left({{e}}^{{\frac{{{x}+{y}}}{{2}}}}\cdot{\left({{e}}^{{-\frac{{{x}-{y}}}{{2}}}}\right]}+{\left[{\left({{e}}^{{-\frac{{{x}+{y}}}{{2}}}}\cdot{\left({{e}}^{{\frac{{{x}-{y}}}{{2}}}}\right]}+{\left[\frac{{{{e}}^{{-\frac{{{x}+{y}}}{{2}}}}\cdot{\left({{e}}^{{-\frac{{{x}-{y}}}{{2}}}}\right]}}}{{4}}\right.}\right.}\right.}\right.}\right.}\right.}\right.}\right.}$

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

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

similarly

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

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

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

so,

$\displaystyle{\left[{\left[{\left({{e}}^{{\frac{{{x}+{y}}}{{2}}}}\cdot{\left({{e}}^{{\frac{{{x}-{y}}}{{2}}}}\right)}\right]}+{\left[{\left({{e}}^{{\frac{{{x}+{y}}}{{2}}}}\cdot{\left({{e}}^{{-\frac{{{x}-{y}}}{{2}}}}\right]}+{\left[{\left({{e}}^{{-\frac{{{x}+{y}}}{{2}}}}\cdot{\left({{e}}^{{\frac{{{x}-{y}}}{{2}}}}\right]}+{\left[\frac{{{{e}}^{{-\frac{{{x}+{y}}}{{2}}}}\cdot{\left({{e}}^{{-\frac{{{x}-{y}}}{{2}}}}\right]}}}{{2}}\right.}\right.}\right.}\right.}\right.}\right.}\right.}$

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

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

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

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

And so , LHS=RHS

so ,

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

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Wednesday, July 15, 2009

$\displaystyle{\cosh{{\left({x}\right)}}}+{\cosh{{\left({y}\right)}}}={2}{\cosh{{\left(\frac{{{x}+{y}}}{{2}}\right)}}}{\cosh{{\left(\frac{{{x}-{y}}}{{2}}\right)}}}$ Verify the identity.

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Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

Wednesday, July 15, 2009

Verify the identity.

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Post a Comment

Thomas Jefferson&#39;s election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

$\displaystyle{\cosh{{\left({x}\right)}}}+{\cosh{{\left({y}\right)}}}={2}{\cosh{{\left(\frac{{{x}+{y}}}{{2}}\right)}}}{\cosh{{\left(\frac{{{x}-{y}}}{{2}}\right)}}}$ Verify the identity.

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?