Question

Prove that \[\sinh \left( {a + b} \right) = \sinh a\cosh b + \sinh b\cosh a\].

Answer 1

Hint: We will take the term on the right-hand side and we will use the identity i.e., \[\cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2}\] and \[\sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2}\]. Then we will put these values in the right-hand side and simplify. At last, we will put \[\sinh \left( {a + b} \right) = \dfrac{{{e^{a + b}} - {e^{ - \left( {a + b} \right)}}}}{2}\] to prove \[\sinh \left( {a + b} \right) = \sinh a\cosh b + \sinh b\cosh a\].

Complete step by step answer:
As we know that \[\cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2}\] and \[\sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2}\].
Similarly, \[\cosh a = \dfrac{{{e^a} + {e^{ - a}}}}{2}\], \[\cosh b = \dfrac{{{e^b} + {e^{ - b}}}}{2}\], \[\sinh a = \dfrac{{{e^a} - {e^{ - a}}}}{2}\] and \[\sinh b = \dfrac{{{e^b} - {e^{ - b}}}}{2}\].
Here, \[{\text{RHS}} = \sinh a\cosh b + \sinh b\cosh a\]
Putting the values in the right-hand side, we get
\[ \Rightarrow {\text{RHS}} = \sinh a\cosh b + \sinh b\cosh a\]
\[ = \left( {\dfrac{{{e^a} - {e^{ - a}}}}{2}} \right)\left( {\dfrac{{{e^b} + {e^{ - b}}}}{2}} \right) + \left( {\dfrac{{{e^b} - {e^{ - b}}}}{2}} \right)\left( {\dfrac{{{e^a} + {e^{ - a}}}}{2}} \right)\]
On multiplying the terms, we get
\[ = \dfrac{{{e^{a + b}} + {e^{a - b}} - {e^{b - a}} - {e^{ - \left( {a + b} \right)}}}}{4} + \dfrac{{{e^{a + b}} + {e^{b - a}} - {e^{a - b}} - {e^{ - \left( {a + b} \right)}}}}{4}\]
On taking LCM and common, we get
\[ = \dfrac{1}{4}\left( {{e^{a + b}} + {e^{a - b}} - {e^{b - a}} - {e^{ - \left( {a + b} \right)}} + {e^{a + b}} + {e^{b - a}} - {e^{a - b}} - {e^{ - \left( {a + b} \right)}}} \right)\]
Cancelling the same terms with opposite sign, we get
\[ = \dfrac{1}{4}\left( {{e^{a + b}} - {e^{ - \left( {a + b} \right)}} + {e^{a + b}} - {e^{ - \left( {a + b} \right)}}} \right)\]
\[ = \dfrac{1}{4}\left( {2{e^{a + b}} - 2{e^{ - \left( {a + b} \right)}}} \right)\]
Taking two common, we get
\[ = \dfrac{2}{4}\left( {{e^{a + b}} - {e^{ - \left( {a + b} \right)}}} \right)\]
\[ = \dfrac{{{e^{a + b}} - {e^{ - \left( {a + b} \right)}}}}{2}\]
As \[\sinh \left( {a + b} \right) = \dfrac{{{e^{a + b}} - {e^{ - \left( {a + b} \right)}}}}{2}\]. Using this, we get right hand side as \[\sinh \left( {a + b} \right)\] i.e.,
\[ \Rightarrow {\text{RHS}} = \sinh \left( {a + b} \right)\], which is equal to the left-hand side.
Hence, proved that \[\sinh \left( {a + b} \right) = \sinh a\cosh b + \sinh b\cosh a\].

Note:
The hyperbolic trigonometric functions consider the area as their argument, known as ‘the hyperbolic angle’. Hyperbolic functions share almost similar properties to trigonometric functions like in the trigonometry, the derivative of \[\sin x\] is \[\cos x\] and \[\cos x\] is \[ - \sin x\] and the derivative of \[\sinh x\] is \[\cosh x\] and \[\cosh x\] is \[ \sinh x\].