I've been looking around the internet for a way to solve $\sin(i)$, and I found something about $\sinh(x)$
I furthered my search and found something about $\sinh(x)= \frac{e^x - e^{-x}}{2}$. When I tried plugging in $i\pi=x$, I got two very separate answers: $$\sinh(i\pi) = 1.2246468 \times 10^{-16}i,$$ but when I plug $i\pi$ into the other, I get $0$. So if there is some rule law or theorem I'm missing, I'd appreciate the knowledge.
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$\begingroup$$\sin$ and $\sinh$ are different functions,$$\sin(x)=\frac{\exp(ix)-\exp(-ix)}{2i},\quad\sinh(x)=\frac{\exp(x)-\exp(-x)}{2}.$$
They are connected by the relation $$\sinh(x)=-i\sin(ix).$$
$\endgroup$ 4 $\begingroup$Computer calculations are not exact, because a real number cannot be specified with infinite precision using only a finite number of bits. The common double precision floating point arithmetic will only represent numbers to within approximately $10^{-16}$, which is the magnitude of the difference between the two calculations.
In other words, this is a rounding error. The two expressions are exactly equal.
$\endgroup$ $\begingroup$A famous relation states$$e^{ix}=\cos x+i\sin x$$from which we conclude that$$\sin x={e^{ix}-e^{-ix}\over 2i}$$also$$\sinh x={e^{x}-e^{-x}\over 2}$$therefore$$\sinh x={e^{i(-ix)}-e^{i(ix)}\over2}=i\cdot {e^{i(-ix)}-e^{i(ix)}\over2}=-i\sin -ix=i\sin ix$$so we can write$$\sinh u=i\sin iu\\\sin u=i\sinh iu$$in this case$$\sin i=-i\sinh 1$$
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