Simplify $|e^{2iwt} - 1|$

Question

So, I have been trying to solve how to simplify

$$|e^{2iwt} - 1|$$ to

$$2|\sin(wt)|.$$

However, I've been working on the problem for quite a while now without making any progress. I've tried to solve it by using Euler's formulas, but it's not helping much. If anyone could help me, by just giving me a tip, I would be really thankful.

Answer 1

Hint

$$e^{2i\omega t}-1=e^{i\omega t}\cdot 2i\cdot \frac{e^{i\omega t}-e^{-i\omega t}}{2i}=2ie^{i\omega t}\sin(\omega t).\tag{*}$$ Taking the modulus of the RHS of $(*)$ give the wished result.

Answer 2

$$|e^{2iwt} - 1|=\\ =|\cos{2iwt}+i\sin{2iwt} - 1|=\\|(\cos{2iwt}-1)+i\sin{2iwt}|\\ =\sqrt{()^2+()^2}\\ =\sqrt{(\cos{2iwt}-1)^2+(\sin{2iwt})^2}\\ =\sqrt{\cos^2{2iwt}+1-2\cos{2iwt}+\sin^2{2iwt}}\\ =\sqrt{\underbrace{\cos^2{2iwt}+\sin^2{2iwt}}_{1}+1-2\cos{2iwt}}\\ =\sqrt{2-2\cos{2iwt}}\\ =\sqrt{2(1-\cos{2iwt})}\\ =\sqrt{2.2\sin^2{(\dfrac{2iwt}{2})}}\\$$

Answer 3

$$|e^{i2a}-1|=\sqrt{(\cos2a-1)^2+\sin^22a}=\sqrt{4\sin^4a+4\cos^2a\sin^2a}=2|\sin a|$$ or

$$|e^{i2a}-1|=|e^{ia}||e^{ia}-e^{-ia}|=2|\sin a|$$

or, with $z=e^{ia}$,

$$|z^2-1|=\sqrt{(z-1)(z^*-1)(z+1)(z^*+1)}=\sqrt{(2-2\Re z)(2+2\Re z)}=2\sqrt{1-(\Re z)^2}$$

or

as hinted by @Neal, in an isoceles triangle of common side $1$ and angle $2a$, as defined by the points of affixes $e^{i2a}$ and $1$,

the length of the third side is

$$|e^{i2a}-1|=2|\sin a|,$$ or by the cosine law,

$$\sqrt{1^2+1^2-2\cdot1\cdot1\cos2a}=\sqrt{4\sin^2a}.$$