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Given a plane wave with the electric field component $$ \tilde{E}_{in} = E_0(-\hat{x} \cos \theta + \hat{z}\sin \theta)\exp [ i \frac{\omega}{c}(x \sin \theta + z \cos \theta) - i \omega t]$$

I've calculated the reflection on a metallic plane and it yielded

$$ \tilde{E}_{ref} = E_0(\hat{x} \cos \theta + \hat{z}\sin \theta)\exp [ i \frac{\omega}{c}(x \sin \theta - z \cos \theta) - i \omega t]. $$

I want to calculate the total complex electric field; i.e $ \tilde{E} = \tilde{E}_{in} + \tilde{E}_{ref} $.

The answer reads "With Eulers formulas we obtain $$ \tilde{E} = 2E_0 \exp[i \frac{\omega}{c} x \sin \theta - i \omega t](-i\hat{x} \cos \theta \sin (\frac{\omega}{c}z cos \theta) + \hat{z} \sin \theta \cos( \frac{\omega}{c} z \cos \theta))"$$

Q: Is there any smart way to conclude this?

Seems like a lot of work to obtain it through $ e^{i\omega t} = \cos \omega t + i \sin \omega t $

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I don't know if this is the most intelligent way to conclude this, but it does not require repeated use of $\exp(i\theta) = \cos\theta + i\sin\theta$. Recall that, \begin{equation} \cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2} \quad \text{and} \quad \sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i} . \end{equation}

Therefore, adding the two electric field contributions \begin{equation} \begin{split} \tilde{E} &= E_0\exp\left[i\frac{\omega}{c}(x\sin\theta) - i\omega t\right] \left[ \left(-e^{\frac{i\omega}{c}z\cos\theta} + e^{-\frac{i\omega}{c}z\cos\theta}\right)\cos\theta\ \hat{x} + \left(e^{\frac{i\omega}{c}z\cos\theta} + e^{-\frac{i\omega}{c}z\cos\theta}\right)\sin\theta\ \hat{z} \right]\\[0.25cm] &= 2E_0\exp\left[i\frac{\omega}{c}(x\sin\theta) - i\omega t\right] \left[ -\frac{i}{2i}\left(e^{\frac{i\omega}{c}z\cos\theta} - e^{-\frac{i\omega}{c}z\cos\theta}\right)z\cos\theta\ \hat{x} + \frac{1}{2}\left(e^{\frac{i\omega}{c}z\cos\theta} + e^{-\frac{i\omega}{c}z\cos\theta}\right)\sin\theta\ \hat{z} \right]\\[0.25cm] &= 2E_0\exp\left[i\frac{\omega}{c}(x\sin\theta) - i\omega t\right] \left[ -i\sin\left(\frac{\omega}{c}z\cos\theta\right)\cos\theta\ \hat{x} + \cos\left(\frac{\omega}{c}z\cos\theta\right)\sin\theta\ \hat{z} \right] \end{split} \end{equation} which is the expected result. Hope this helps!

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