# Relationship between trigonometric and hyperbolic sine

Why is the following identity true?

$$\sin(ix) = i\sinh(x)$$

When I do the calculation, I get this:$$\sin(ix) = \frac{{e^{i(ix)}}-e^{-i(ix)}}{2i}=\frac{e^{-x}-e^x}{2i}=-\frac{e^x-e^{-x}}{2i}=-\left(\frac{\sinh(x)}{i}\right)$$

• Remember $i^2=-1$ and thus $i=-1/i$ – marwalix Feb 1 '15 at 9:18
• oops, I was using p on a problem, fixed it. – user2809114 Feb 1 '15 at 9:20

$$\sin(ix) = \frac{{e^{i(ix)}}-e^{-i(ix)}}{2i}=\frac{e^{-x}-e^x}{2i}=-\frac{e^x-e^{-x}}{2i}=-\left(\frac{\operatorname{sinh}(x)}{i}\right)=i\operatorname{sinh}(x)$$ because $1/i=-i$

• all you did was put " = isinh(x)" at the end of my equation – user2809114 Feb 1 '15 at 9:30
• not at all ! marwalix used the fact that $-\frac 1i=i$ which finishes the proof. – Claude Leibovici Feb 1 '15 at 9:38
• I had hinted that in a comment to the question. By the way I also formatted the hyperbolic sine using \operatorname – marwalix Feb 1 '15 at 10:05
• @marwalix No need: the command \sinh already gives $\;\sinh x\;$ , and likewise \cosh x . +1 – Timbuc Feb 1 '15 at 10:07
• Wasn't sure that the LaTex add-on of SE has that. Not all have – marwalix Feb 1 '15 at 10:10

Another derivation:

$$\displaystyle \sin(ix) = ix - \frac {(ix)^3}{3!} + \frac {(ix)^5}{5!} - \frac {(ix)^7}{7!}+ \cdots$$

$$= ix + \frac {ix^3} {3!} + \frac {ix^5}{5!} + \frac {ix^7}{7!} + \cdots$$

$$= i \left({x + \frac {x^3}{3!} + \frac {x^5}{5!} + \frac {x^7}{7!} + \cdots }\right)= i \sinh x$$