I know the exponential form of $\sin(x)=\dfrac{e^{ix}-e^{-ix}}{2i}$. I wish to derive the exponential form of $\sinh(x)=\dfrac{e^x-e^{-x}}{2}$, and I am stuck.
I try to replace $\sin(x)$ with $\sin(i^4x)$, then multiply $i=\sqrt{-1}\implies i^4=1$ with $\sin(i^4x)$ and I have
$$i\sin(i^4x)=i\dfrac{e^{i^4x}-e^{-i^4x}}{2i}=\dfrac{e^{x}-e^{-x}}{2}\tag{1}$$
But this is a nonsensical derivation, because it implies that $i\sin(x)=\sinh(x).$ The correct identity must be $i\sin(x)=\sinh(ix)$
I think my result is clearly wrong. What is wrong with it? How should one derive the exponential form of $\sinh(x)$
$\endgroup$ 11 Answer
$\begingroup$You know the correct answer $i \sin(x)=\sinh (ix)$, but you want to get to the RHS via the LHS, so let $y=ix$, and the above becomes
$$i\sin(-iy)=\sinh(y)$$
Now calculate the LHS using the definition of $\sin$ (first take out the minus in the $\sin$)
$$-i\frac{e^{i(iy)}-e^{-i(iy)}}{2i}=\frac{e^y-e^{-y}}{2}=\sinh (y)$$
To answer “what’s wrong with your answer”: replacing 1 with $i^4$ is pointless —- you should see why we should substitute $x=-iy$ from the above.
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