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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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$$ \sin(x) = \frac{e^{i x} - e^{-i x}}{2 i}$$, then $$ \sin(i x) = \frac{e^{i^2 x} - e^{-i^2 x}}{2 i}= \frac{e^{-x} - e^{ x}}{2 i}$$

0x2207
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    Three upvotes for a wrong formula for $\sin(x)$. – Did Mar 10 '13 at 11:06
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    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
  • You both are right. However, it doesn't matter, because topic-starter's complication was in argument substitution, not in formulas, which he/she can easily look at wikipedia or somewhere else. – 0x2207 Mar 10 '13 at 18:46
  • @DonAntonio Perhaps you could calm down, don't you think? – Did Mar 10 '13 at 19:20
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    I'm pretty cool and calm down the last few weeks, @Did. Why? Do you see anything out of order somewhere? – DonAntonio Mar 10 '13 at 20:35
  • However, it doesn't matter... I disagree. – Did Mar 11 '13 at 06:59
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You're missing the $i$'s already in $\sin$; $\sin x = \frac{1}{2i}(e^{ix} - e^{-ix})$. So:

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

Josephine Moeller
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