The question:

Let $\displaystyle z = (1 + \sqrt{3} i)$ and w = (1 + i). Find Arg(z) and Arg(w) and hence Arg(zw). Evaluate zw and hence show that $\displaystyle cos\frac{7\pi}{12} = \frac{1 - \sqrt{3}}{2\sqrt{2}}$. Find a similar expression for $\displaystyle sin \frac{7\pi}{12}$

I found:

Arg(z) = $\displaystyle \frac{\pi}{3}$

Arg(w) = $\displaystyle \frac{\pi}{4}$

Arg(zw) = $\displaystyle \frac{\pi}{12}$

zw = $\displaystyle 2\sqrt{2}e^{i\frac{\pi}{12}}$

The last bit of the question is giving me trouble. I tried using de Moivre's theorem as follows:

$\displaystyle (2\sqrt{2}e^{i\frac{\pi}{12}})^7$

$\displaystyle (2\sqrt{2})^7e^{i\frac{7\pi}{12}}$

This gives me the argument I want. But I'm unsure of how to solve this. Changing it to sin/cos form (not sure of the proper name for it) seems to be the way to go, but I'm still stumped:

$\displaystyle (2\sqrt{2})^7(cos(\frac{7\pi}{12}) + isin(\frac{7\pi}{12}))$

Any assistance would be great!