Finding an angle $\theta$ in a complex number If we know that $z = \frac{1}{\sqrt2}(\cos\theta+i\cdot\sin\theta)$ and also that $z = \frac{(\sqrt3-1)+i(\sqrt3+1)}{4}$
How can I find $\cos\theta$ and $\sin\theta$? Using a calculator it gives me the angle $\frac{5\pi}{12}$. I tried the following:
$$\cos\theta = \frac{\Re z}{|z|} = \frac{\frac{1}{4}(\sqrt3-1)}{\frac{1}{\sqrt2}} = \frac{\sqrt3-1}{2\cdot\sqrt2}$$ How would I even proceed here, seems near-impossible without a calculator to me.

$$
\cos(\frac{5\pi}{12}) = \cos(\frac{3\pi}{4}-\frac{\pi}{3})=\cos(\frac{3\pi}{4})\cos(\frac{\pi}{3})+\sin(\frac{3\pi}{4})\sin(\frac{\pi}{3})=\frac{-\sqrt{2}}{2}\cdot\frac{1}{2}+\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2} = $$ $$\frac{\sqrt{6}-\sqrt{2}}{4}=\frac{\sqrt{3}-1}{2\sqrt{2}}
$$
Now work backwards, amigo. 
