Finding an angle $\theta$ in a complex number

Question

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.

Answer 1

Write this as $r(\cos t+i\sin t)$

Using the definition of atan2, $0<t<\dfrac\pi2$

and $t=\arctan\dfrac{\sqrt3+1}{\sqrt3-1}=\arctan\dfrac{1+1/\sqrt3}{1-1/\sqrt3}=\arctan\dfrac{\tan45^\circ+\tan30^\circ}{1-\tan45^\circ\tan30^\circ} $ $=45^\circ+30^\circ$

Hope you can find $r$ easily

Answer 2

see here http://en.wikipedia.org/wiki/Complex_number we have $\phi=\arctan\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)$

Answer 3

$$ \cos(\frac{5\pi}{12}) = \cos(\frac{3\pi}{4}-\frac{\pi}{3})=\cos(\frac{3\pi}{4})\cos(\frac{\pi}{3})+\sin(\frac{3\pi}{4})\sin(\frac{\pi}{3})=\frac{-\sqrt{2}}{2}\cdot\frac{1}{2}+\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2} = $$ $$\frac{\sqrt{6}-\sqrt{2}}{4}=\frac{\sqrt{3}-1}{2\sqrt{2}} $$

Now work backwards, amigo.