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LammettHash
Euler's formula tells us that

[tex]e^{i\theta}=\cos\theta+i\sin\theta[/tex]
[tex]e^{-i\theta}=\cos\theta-i\sin\theta[/tex]

Suppose we subtract the two. This eliminates the cosine terms.

[tex]e^{i\theta}-e^{-i\theta}=i\sin\theta-(-i\sin\theta)=2i\sin\theta[/tex]

Divide both sides by [tex]2i[/tex] and you're done.

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