1
$\begingroup$

Why is the following identity true?

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

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

$\endgroup$
  • 1
    $\begingroup$ Remember $i^2=-1$ and thus $i=-1/i$ $\endgroup$ – marwalix Feb 1 '15 at 9:18
  • $\begingroup$ oops, I was using p on a problem, fixed it. $\endgroup$ – user2809114 Feb 1 '15 at 9:20
8
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ all you did was put " = isinh(x)" at the end of my equation $\endgroup$ – user2809114 Feb 1 '15 at 9:30
  • 3
    $\begingroup$ not at all ! marwalix used the fact that $-\frac 1i=i$ which finishes the proof. $\endgroup$ – Claude Leibovici Feb 1 '15 at 9:38
  • $\begingroup$ I had hinted that in a comment to the question. By the way I also formatted the hyperbolic sine using \operatorname $\endgroup$ – marwalix Feb 1 '15 at 10:05
  • $\begingroup$ @marwalix No need: the command \sinh already gives $\;\sinh x\;$ , and likewise \cosh x . +1 $\endgroup$ – Timbuc Feb 1 '15 at 10:07
  • $\begingroup$ Wasn't sure that the LaTex add-on of SE has that. Not all have $\endgroup$ – marwalix Feb 1 '15 at 10:10
0
$\begingroup$

Another derivation:

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.