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I am trying to go through and understand the derivation of

\[X_{c} = \frac{1}{j\omega C}\]

Where $X_{c}$ is capacitive reactance in ohms, $\omega$ is the angular velocity or $2\pi frequency$ and $C$ is capacitance in Farads

To start with we already have $I=C\frac{dv}{dt}$, $V=V_{pk}\sin\left (\omega \right )t$ and that $X_{c}=\frac{V}{I}$

Then

\[I=C\frac{d}{vt}V_{pk}\sin \left ( \omega \right )t = C\cdot V_{pk}\omega \cos \left ( \omega \right )t\]

Therefore

\[X_{c}=\frac{V_{pk}\sin\left (\omega \right )t}{C\cdot V_{pk}\omega \cos \left ( \omega \right )t}\]

$V_{pk}$ and $t$ cancel out and we can be left with

\[X_{c}=\frac{1}{\omega C}\cdot \frac{\sin \left (\omega \right )}{\cos \left (\omega \right )}\]

Now I am up to the part I don't understand. The only non-Euler derivation of this formula that I have been able to find was on youtube and unfortunately it gets a bit vague here. He says that because we know that for capacitance current leads the voltage that any real number multiplied by $90^{\circ}$, or $pi$ on the Argand plane, it swings up to the imaginary axis and thus it becomes $-j$.

I'd like to see how the second term can be mathematically turned into $-j$ or $\frac{1}{j}$ for that matter.

Any help or guidance would be greatly appreciated