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

Last updated date: 10th Aug 2024
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Hint: We will use the expansion of hyperbolic functions and substitute the values in LHS and RHS of the equation separately.
* $\cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2}$ and $\sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2}$

First we solve Left hand side of the equation i.e. $\left( {\cosh A - \cosh B} \right)$
We know $\cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2}$
So, we find the values of $\cosh A,\cosh B$
$\Rightarrow \cosh A = \dfrac{{{e^A} + {e^{ - A}}}}{2}$
$\Rightarrow \cosh B = \dfrac{{{e^B} + {e^{ - B}}}}{2}$
Now substituting the values in LHS of the equation we get
$\Rightarrow \left( {\cosh A - \cosh B} \right) = \left( {\dfrac{{{e^A} + {e^{ - A}}}}{2}} \right) - \left( {\dfrac{{{e^B} + {e^{ - B}}}}{2}} \right)$
Now since the denominator is the same we can subtract the terms in the numerator.
$\Rightarrow \left( {\cosh A - \cosh B} \right) = \left( {\dfrac{{({e^A} + {e^{ - A}}) - ({e^B} + {e^{ - B}})}}{2}} \right) \\ \Rightarrow \left( {\cosh A - \cosh B} \right) = \left( {\dfrac{{{e^A} + {e^{ - A}} - {e^B} - {e^{ - B}}}}{2}} \right) \\$
Group the terms in the numerator.
$\Rightarrow \left( {\cosh A - \cosh B} \right) = \left( {\dfrac{{({e^A} - {e^B}) + ({e^{ - A}} - {e^{ - B}})}}{2}} \right)$ … (1)
Now we solve right hand side of the equation i.e. $\left( {2\sinh (\dfrac{{A + B}}{2})\sinh (\dfrac{{A - B}}{2})} \right)$
Since, we know $\sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2}$.
So we find the values of $\sinh (\dfrac{{A + B}}{2}),\sinh (\dfrac{{A - B}}{2})$
$\sinh (\dfrac{{A + B}}{2}) = \dfrac{{{e^{\dfrac{{A + B}}{2}}} - {e^{\dfrac{{ - A - B}}{2}}}}}{2} \\ \sinh (\dfrac{{A - B}}{2}) = \dfrac{{{e^{\dfrac{{A - B}}{2}}} - {e^{\dfrac{{ - A + B}}{2}}}}}{2} \\$
Substituting the values in RHS of the equation we get
$\Rightarrow \left( {2\sinh (\dfrac{{A + B}}{2})\sinh (\dfrac{{A - B}}{2})} \right) = 2\left( {\dfrac{{{e^{\dfrac{{A + B}}{2}}} - {e^{\dfrac{{ - A - B}}{2}}}}}{2}} \right)\left( {\dfrac{{{e^{\dfrac{{A - B}}{2}}} - {e^{\dfrac{{ - A + B}}{2}}}}}{2}} \right)$
Cancel out the common factor 2 from both numerator and denominator.
$\Rightarrow \left( {2\sinh (\dfrac{{A + B}}{2})\sinh (\dfrac{{A - B}}{2})} \right) = \dfrac{1}{2}\left( {{e^{\dfrac{{A + B}}{2}}} - {e^{\dfrac{{ - A - B}}{2}}}} \right)\left( {{e^{\dfrac{{A - B}}{2}}} - {e^{\dfrac{{ - A + B}}{2}}}} \right)$
Now we multiply the brackets.
$\Rightarrow \left( {2\sinh (\dfrac{{A + B}}{2})\sinh (\dfrac{{A - B}}{2})} \right) = \dfrac{1}{2}\left[ {{e^{\dfrac{{A + B}}{2}}} \times {e^{\dfrac{{A - B}}{2}}} - {e^{\dfrac{{A + B}}{2}}} \times {e^{\dfrac{{ - A + B}}{2}}} - {e^{\dfrac{{ - A - B}}{2}}} \times {e^{\dfrac{{A - B}}{2}}} + {e^{\dfrac{{ - A - B}}{2}}} \times {e^{\dfrac{{ - A + B}}{2}}}} \right]$
Now using the law of indices, when the base is the same powers can be added. Here the base is same i.e. e, so we can add the powers of each term
$\Rightarrow \left( {2\sinh (\dfrac{{A + B}}{2})\sinh (\dfrac{{A - B}}{2})} \right) = \dfrac{1}{2}\left[ {{e^{\dfrac{{A + B}}{2} + \dfrac{{A - B}}{2}}} - {e^{\dfrac{{A + B}}{2} + \dfrac{{ - A + B}}{2}}} - {e^{\dfrac{{ - A - B}}{2} + \dfrac{{A - B}}{2}}} + {e^{\dfrac{{ - A - B}}{2} + \dfrac{{ - A + B}}{2}}}} \right]$
Taking LCM of terms in power
$\Rightarrow \left( {2\sinh (\dfrac{{A + B}}{2})\sinh (\dfrac{{A - B}}{2})} \right) = \dfrac{1}{2}\left[ {{e^{\dfrac{{A + B + A - B}}{2}}} - {e^{\dfrac{{A + B - A + B}}{2}}} - {e^{\dfrac{{ - A - B + A - B}}{2}}} + {e^{\dfrac{{ - A - B - A + B}}{2}}}} \right]$
$\Rightarrow \left( {2\sinh (\dfrac{{A + B}}{2})\sinh (\dfrac{{A - B}}{2})} \right) = \dfrac{1}{2}\left[ {{e^{\dfrac{{2A}}{2}}} - {e^{\dfrac{{2B}}{2}}} - {e^{\dfrac{{ - 2B}}{2}}} + {e^{\dfrac{{ - 2A}}{2}}}} \right]$
$\Rightarrow \left( {2\sinh (\dfrac{{A + B}}{2})\sinh (\dfrac{{A - B}}{2})} \right) = \dfrac{1}{2}\left[ {{e^A} - {e^B} - {e^{ - B}} + {e^{ - A}}} \right]$
Grouping the terms in the bracket
$\Rightarrow \left( {2\sinh (\dfrac{{A + B}}{2})\sinh (\dfrac{{A - B}}{2})} \right) = \dfrac{1}{2}\left[ {({e^A} - {e^B}) + ({e^{ - A}} - {e^{ - B}})} \right]$
$\Rightarrow \left( {2\sinh (\dfrac{{A + B}}{2})\sinh (\dfrac{{A - B}}{2})} \right) = \dfrac{{({e^A} - {e^B}) + ({e^{ - A}} - {e^{ - B}})}}{2}$ … (2)
From equation (1) and equation (2)
$\dfrac{{({e^A} - {e^B}) + ({e^{ - A}} - {e^{ - B}})}}{2} = \dfrac{{({e^A} - {e^B}) + ({e^{ - A}} - {e^{ - B}})}}{2}$
We can say LHS = RHS
Hence proved.

Note: Students are likely to make calculation mistakes while calculating the powers in the second step, they are advised to find the two values of hyperbolic functions of sin separately and then multiply.
Also, they should not try to solve the part with negative powers by converting it into reciprocal as it will complicate our solution, we will have to take reciprocal, then take LCM and multiply all the terms with the LCM which is a long process.