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

Prove that sinh⁡(a+b)=sinh⁡acosh⁡b+sinh⁡bcosh⁡a\sinh \left( {a + b} \right) = \sinh a\cosh b + \sinh b\cosh a.

Explanation

Solution

We will take the term on the right-hand side and we will use the identity i.e., cosh⁡x=ex+e−x2\cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2} and sinh⁡x=ex−e−x2\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⁡(a+b)=ea+b−e−(a+b)2\sinh \left( {a + b} \right) = \dfrac{{{e^{a + b}} - {e^{ - \left( {a + b} \right)}}}}{2} to prove sinh⁡(a+b)=sinh⁡acosh⁡b+sinh⁡bcosh⁡a\sinh \left( {a + b} \right) = \sinh a\cosh b + \sinh b\cosh a.

Complete step by step answer:
As we know that cosh⁡x=ex+e−x2\cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2} and sinh⁡x=ex−e−x2\sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2}.
Similarly, cosh⁡a=ea+e−a2\cosh a = \dfrac{{{e^a} + {e^{ - a}}}}{2}, cosh⁡b=eb+e−b2\cosh b = \dfrac{{{e^b} + {e^{ - b}}}}{2}, sinh⁡a=ea−e−a2\sinh a = \dfrac{{{e^a} - {e^{ - a}}}}{2} and sinh⁡b=eb−e−b2\sinh b = \dfrac{{{e^b} - {e^{ - b}}}}{2}.
Here, RHS=sinh⁡acosh⁡b+sinh⁡bcosh⁡a{\text{RHS}} = \sinh a\cosh b + \sinh b\cosh a
Putting the values in the right-hand side, we get
⇒RHS=sinh⁡acosh⁡b+sinh⁡bcosh⁡a\Rightarrow {\text{RHS}} = \sinh a\cosh b + \sinh b\cosh a
=(ea−e−a2)(eb+e−b2)+(eb−e−b2)(ea+e−a2)= \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
=ea+b+ea−b−eb−a−e−(a+b)4+ea+b+eb−a−ea−b−e−(a+b)4= \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
=14(ea+b+ea−b−eb−a−e−(a+b)+ea+b+eb−a−ea−b−e−(a+b))= \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
=14(ea+b−e−(a+b)+ea+b−e−(a+b))= \dfrac{1}{4}\left( {{e^{a + b}} - {e^{ - \left( {a + b} \right)}} + {e^{a + b}} - {e^{ - \left( {a + b} \right)}}} \right)
=14(2ea+b−2e−(a+b))= \dfrac{1}{4}\left( {2{e^{a + b}} - 2{e^{ - \left( {a + b} \right)}}} \right)
Taking two common, we get
=24(ea+b−e−(a+b))= \dfrac{2}{4}\left( {{e^{a + b}} - {e^{ - \left( {a + b} \right)}}} \right)
=ea+b−e−(a+b)2= \dfrac{{{e^{a + b}} - {e^{ - \left( {a + b} \right)}}}}{2}
As sinh⁡(a+b)=ea+b−e−(a+b)2\sinh \left( {a + b} \right) = \dfrac{{{e^{a + b}} - {e^{ - \left( {a + b} \right)}}}}{2}. Using this, we get right hand side as sinh⁡(a+b)\sinh \left( {a + b} \right) i.e.,
⇒RHS=sinh⁡(a+b)\Rightarrow {\text{RHS}} = \sinh \left( {a + b} \right), which is equal to the left-hand side.
Hence, proved that sinh⁡(a+b)=sinh⁡acosh⁡b+sinh⁡bcosh⁡a\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\sin x is cos⁡x\cos x and cos⁡x\cos x is −sin⁡x - \sin x and the derivative of sinh⁡x\sinh x is cosh⁡x\cosh x and cosh⁡x\cosh x is sinh⁡x \sinh x.