Question

Prove the following with Euler's formula to write e^itheta and e^-itheta in terms of sine and cosine. Then subtract them.

Answer 1

Euler's formula tells us that

$e^{i\theta}=\cos\theta+i\sin\theta$
$e^{-i\theta}=\cos\theta-i\sin\theta$

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

$e^{i\theta}-e^{-i\theta}=i\sin\theta-(-i\sin\theta)=2i\sin\theta$

Divide both sides by $2i$ and you're done.