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sin and cos in terms of complex numbers

We know how to get trig relations from complex numbers already. Recall that $e^{\theta i} = \cos\theta + i\sin\theta$. Then, we want to find $\sin\theta$ and $\cos\theta$. How do we do this? Well, conjugates!

Let $z=e^{\theta i}$. Then $\overline z = e^{-\theta i}$. Let's try adding these, as conjugates tend to make things cancel out. We have: $$\begin{align} z + \overline z &= e^{\theta i} + e^{-\theta i} \\ &= \cos\theta + i\sin\theta + \cos(-\theta) + i\sin(-\theta) \\ &= \cos\theta + i\sin\theta + \cos\theta - i\sin\theta \\ &= 2\cos\theta \end{align}$$ Aha! We have it! We can divide both sides by $2$ to get: $$\dfrac{e^{\theta i} + e^{-\theta i}}2 = \cos\theta$$

Now, finding $\sin \theta$ should be easy now. We need to cancel out the real terms now while keeping the imaginary parts. Think about what we should do, and press continue when you think you have it.

We subtract the conjugate of the complex number from the number instead of adding it. Then, we have: $$\begin{align} z + \overline z &= e^{\theta i} - e^{-\theta i} \\ &= \cos\theta + i\sin\theta - \cos(-\theta) - i\sin(-\theta) \\ &= \cos\theta + i\sin\theta - \cos\theta + i\sin\theta \\ &= 2i\sin\theta \end{align}$$ Then, we can divide both sides by $2i$ to get: $$\dfrac{e^{\theta i} - e^{-\theta i}}{2i} = \sin\theta$$

That's all there is for this explanation about expressing trig functions with complex numbers. Thank you for reading!