# How to show $\sin(-iy)=i\sinh(y)$?

How to show $\sin(-iy)=i \sinh(y)$?

I get: $\sin(-iy)=\frac{1}{2i}(e^{-iy}-e^{iy})=\frac{1}{2i}(\cos(y)-i\sin(y)-\cos(y)-i\sin(y))=...=-\sin(y)$.

I don't get it. $-sin(y) \neq i sinh(y)$ - look at here

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use Euler's identity $sin(x)= \frac{e^{ix}-e^{-ix}}{2i}$ and the definition of the hyperbolic sine $sin(x)= \frac{e^{x}-e^{-x}}{2}$ now just compare :) since $i(-i)=1$ –  Jose Garcia Mar 10 '13 at 10:49
You have the identity incorrect: it's $\sin(iy)=i\sinh y$. See here –  John Moeller Mar 10 '13 at 10:55
@JoseGarcia please use \sin instead of sin. The latter looks like the product of $s$, $i$ and $n$. –  kahen Mar 10 '13 at 11:00

You're missing the $i$'s already in $\sin$; $\sin x = \frac{1}{2i}(e^{ix} - e^{-ix})$. So:

$$\sin(iy) = \frac{1}{2i}(e^{i^2y} - e^{-i^2y}) = \frac{-i}{2}(e^{-y} - e^{y}) = i\sinh(y)$$

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$$\sin(x) = \frac{e^{i x} - e^{-i x}}{2 i}$$, then $$\sin(i x) = \frac{e^{i^2 x} - e^{-i^2 x}}{2 i}= \frac{e^{-x} - e^{ x}}{2 i}$$
Three upvotes for a wrong formula for $\sin(x)$. –  Did Mar 10 '13 at 11:06
Perhaps instead of complaining about the upvotes to a wrong answer let us try to help the answerer to check his answer: @ox2207, your first formula is wrong: it must be $$\sin x:=\frac{e^{ix}-e^{-ix}}{2i}$$ Of course, also the second formula is wrong... –  DonAntonio Mar 10 '13 at 14:51