The expression $e^{2ix} - e^{-2ix}$ can be simplified using Euler's formula, which states:

$e^{ix} = \cos(x) + i\sin(x)$ and $e^{-ix} = \cos(x) - i\sin(x)$

Step-by-Step Simplification:

1. Apply Euler's formula: Using Euler's formula, we rewrite $e^{2ix}$ and $e^{-2ix}$:

   $e^{2ix} = \cos(2x) + i\sin(2x)$ $e^{-2ix} = \cos(2x) - i\sin(2x)$

2. Subtract the two expressions:

   $e^{2ix} - e^{-2ix} = (\cos(2x) + i\sin(2x)) - (\cos(2x) - i\sin(2x))$

3. Simplify: Group the real and imaginary terms:

   $e^{2ix} - e^{-2ix} = \cos(2x) - \cos(2x) + i\sin(2x) + i\sin(2x)$

   The real parts $\cos(2x) - \cos(2x)$ cancel out, and we are left with:

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

Final Simplified Form:

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

This is the simplified expression.

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Related Questions:

1. What is Euler's formula, and how is it derived?
2. How does the simplification change if the exponents were different, like $e^{ix} - e^{-ix}$?
3. Can $\sin(2x)$ be expanded further using trigonometric identities?
4. How can this expression be useful in solving differential equations involving complex exponentials?
5. What is the relationship between this result and the Fourier series?

Tip:

When dealing with complex exponentials, Euler's formula is your best friend for simplifying expressions into trigonometric terms.