Sin(i) using Euler's Equation

Problem: Find \(\sin i\) using Euler's Equation, which is:

\[e^{ix}=\cos x +i\sin x\] (a proof of this equation can be found here)


Substituting in \(\cos x =\pm \sqrt{1-\sin^2 x}\) and \(x=i\) gives:

\[e^{i^2}=\pm \sqrt{1-\sin^2 i} +i\sin i\]

\[e^{-1}=\pm \sqrt{1-\sin^2 i} +i\sin i\]

\[e^{-1} - i\sin i=\pm \sqrt{1-\sin^2 i}\]

Squaring both sides (this will not give extraneous solutions since we have \(\pm\) on the right hand side):

\[(e^{-1} - i\sin i)^2=(\pm \sqrt{1-\sin^2 i})^2\]

\[e^{-2} - 2ie^{-1} \sin i+(i\sin i)^2=1-\sin^2 i\]

\[e^{-2} - 2ie^{-1} \sin i - \sin^2 i=1-\sin^2 i\]

\[e^{-2} - 2ie^{-1} \sin i =1\]

Rearranging: \[  \sin i =\frac{1-e^{-2}}{- 2ie^{-1}}\]

\[  \sin i =\frac{e-e^{-1}}{- 2i}\]

Notice that \(\frac{1}{i}=\frac{1}{i} \times \frac{i}{i}=\frac{i}{i^2}=\frac{i}{-1}=-i\). Substituting this in:

\[  \sin i =\frac{-i(e-e^{-1})}{- 2}\]

\[\therefore  \sin i =i \frac{(e-e^{-1})}{2}\]

Notice that \(\frac{(e-e^{-1})}{2}=\frac{(e^1-e^{-1})}{2}=\sinh 1\)

\[\therefore  \sin i =i \sinh 1\]

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