## 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)

Solution:

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

$\therefore \sin i =i \sinh 1$