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# Prove the following identity: $\sinh A+\sinh B=2\sinh \left( \dfrac{A+B}{2} \right)\cosh \left( \dfrac{A-B}{2} \right)$ 

Last updated date: 10th Aug 2024
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Hint: We recall the definition of sine and cosine hyperbolic function as $\sinh x=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}$ and $\cosh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$. We begin from right hand side of the given statement and simplify using the exponential identity ${{a}^{m+n}}={{a}^{m}}\cdot {{a}^{n}}$ to arrive at the left hand side. 

Complete step-by-step solution:
We know that hyperbolic functions are functions analogous to ordinary trigonometric functions defined for the hyperbola, rather than the circle which is means just $\left( \cos t,\sin t \right)$ with parameter $t$ represents a circle with unit radius, the point $\left( \cosh t,\sinh t \right)$ represent form the right half of the equilateral parabola.
The basic hyperbolic functions are sine hyperbolic function $\left( \sinh x: R\to R \right)$ and cosine hyperbolic function $\left( \cosh x:R\to R \right)$. All the other hyperbolic functions are derived from hyperbolic sine and hyperbolic cosine. 
The hyperbolic sine is defined in terms of exponential function ${{e}^{x}}$ as,
$\sinh x=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}$
The hyperbolic cosine is defined in terms of exponential function ${{e}^{x}}$ as,
$\cosh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$
We are asked to prove the following statement
$\sinh A+\sinh B=2\sinh \left( \dfrac{A+B}{2} \right)\cosh \left( \dfrac{A-B}{2} \right)$
We shall begin simplifying fro right hand side that is
$2\sinh \left( \dfrac{A+B}{2} \right)\cosh \left( \dfrac{A-B}{2} \right)$
We use the definition of sine hyperbolic function in terms of exponential for $x=\dfrac{A+B}{2}$ to have;
$\sinh \left( \dfrac{A+B}{2} \right)=\dfrac{{{e}^{\dfrac{A+B}{2}}}-{{e}^{-\dfrac{A+B}{2}}}}{2}$
We use the definition of cosine hyperbolic function in terms of exponential for $x=\dfrac{A-B}{2}$ to have;
$\cosh \left( \dfrac{A-B}{2} \right)=\dfrac{{{e}^{\dfrac{A-B}{2}}}+{{e}^{-\dfrac{A-B}{2}}}}{2}$
We put the $\sinh \left( \dfrac{A+B}{2} \right)\cosh \left( \dfrac{A-B}{2} \right)$ in the right hand side of the statement to have;
\begin{align} & \Rightarrow 2\sinh \left( \dfrac{A+B}{2} \right)\cosh \left( \dfrac{A-B}{2} \right) \\ & \Rightarrow 2\dfrac{{{e}^{\dfrac{A+B}{2}}}-{{e}^{-\dfrac{A+B}{2}}}}{2}\times \dfrac{{{e}^{\dfrac{A-B}{2}}}+{{e}^{-\dfrac{A-B}{2}}}}{2} \\ & \Rightarrow \dfrac{\left( {{e}^{\dfrac{A+B}{2}}}-{{e}^{{-}\dfrac{A+B}{2}}} \right)\left( {{e}^{\dfrac{A-B}{2}}}+e{^{{-}\dfrac{A-B}{2}}} \right)}{2} \\ & \Rightarrow \dfrac{{{e}^{\dfrac{A+B}{2}}}{{e}^{\dfrac{A-B}{2}}}+{{e}^{\dfrac{A+B}{2}}}{{e}^{-\dfrac{A-B}{2}}}-{{e}^{-\dfrac{A+B}{2}}}{{e}^{\dfrac{A-B}{2}}}-{{e}^{-\dfrac{A+B}{2}}}{{e}^{-\dfrac{A-B}{2}}}}{2} \\ \end{align}

We use the exponential identity ${{a}^{m}}\cdot {{a}^{n}}={{a}^{m+n}}$ in the above step to have;
\begin{align} & \Rightarrow \dfrac{{{e}^{\dfrac{A+B+A-B}{2}}}+{{e}^{\dfrac{A+B-A+B}{2}}}-{{e}^{\dfrac{-A-B+A-B}{2}}}-{{e}^{\dfrac{-A-B-A+B}{2}}}}{2} \\ & \Rightarrow \dfrac{{{e}^{\dfrac{2A}{2}}}+{{e}^{\dfrac{2B}{2}}}-{{e}^{\dfrac{-2B}{2}}}-{{e}^{\dfrac{-2A}{2}}}}{2} \\ & \Rightarrow \dfrac{{{e}^{A}}+{{e}^{B}}-{{e}^{-B}}-{{e}^{-A}}}{2} \\ & \Rightarrow \dfrac{{{e}^{A}}-{{e}^{-A}}}{2}+\dfrac{{{e}^{B}}-{{e}^{-B}}}{2} \\ \end{align}
We use the definition of sine hyperbolic for $x=A,B$ in the above step to have
$\Rightarrow \sinh A+\sinh B$
The above result is on the left hand side of the statement. Hence it is proved.

Note: We note that a statement becomes identity when the statement is true for all parameters. The given statement is true for all $A,B$ and hence it is an identity. The trigonometric equivalent of the given identity is $\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$. We can alternatively prove using identities sum and difference of arguments of sine hyperbolic $\sinh \left( x+y \right)=\sinh x\cosh y+\cosh x\sinh y$ and $\sin \left( x-y \right)=\sinh x\cosh y-\cosh x\sinh y$.