$Sinhx = \dfrac{e^x-e^{-x} }{2}$ Expansion of $e^x = 1 +\dfrac{x }{ 1!} +\dfrac{x^2 }{2!}+\dfrac{x^3 }{3!} +\dfrac{x^4 }{ 4!}+\dfrac{x^5 }{ 5!} +\dfrac{x^6 }{ 6!}+\dfrac{x^7 }{7!} +\dfrac{x^8 }{ 8!} \cdots$ --- Equation 1 Expansion of $e^{-x} = 1 +\dfrac{-x }{ 1!} +\dfrac{x^2}{2!}+\dfrac{{-x}^3 }{ 3!} +\dfrac{x^4 }{ 4!}+\dfrac{-x^5 }{ 5!} +\dfrac{x^6 }{6!}+\dfrac{-x^7 }{ 7!} +\dfrac{x^8 }{ 8!} \cdots$--- Equation 2 On sutracting 2 from 1 $e^x-e^{-x}= \dfrac{2x}{1!}+\dfrac{2x^3}{3!}+\dfrac{2x^5}{ 5!}+\dfrac{2x^7}{7!} \cdots $ $\mathrm {Sinh}x = \dfrac{e^x-e^{-x} }{2}=\dfrac{\dfrac{2x}{ 1!}+\dfrac{2x^3}{3!}+\dfrac{2x^5}{ 5!}+\dfrac{2x^7}{7!}}{2}$$= \dfrac{x}{1!}+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\dfrac{x^7}{7!} \cdots$

Hence proved