<s>[INST] How can I deduce exact values of $\cos\left(\frac{\pi}{12}\right)$ and $\sin\left(\frac{\pi}{12}\right)$? I have been struggling on the problem for some time; I would like a gentle tip on answering the question:

Given the equation $z^4 = 1 + \sqrt{3}i $ where $ z  \in \mathbb{C}$ , deduce the exact values of $\cos\left(\frac{\pi}{12}\right)$ and $\sin\left(\frac{\pi}{12}\right)$.

So far I am proceeding in the following manner: 
I pose $Z' = z^2$
therefore I have $Z'^2 = 1 + \sqrt{3}i $
I calculate that $Z'$ = $\frac{\sqrt{6}}{2} + \frac{\sqrt{2}}{2}i$ 
Right now I have to solve for $Z' = z^2 \iff z^2= \frac{\sqrt{6}}{2} + \frac{\sqrt{2}}{2}i$
After solving this equation I get that $ z = \frac{1}{2}\sqrt{\sqrt{6}+1} + \frac{1}{2}\sqrt{1-\sqrt{6}}i$
If I am taking that $\cos\left(\frac{\pi}{12}\right) = \Re(z)$ and $\sin\left(\frac{\pi}{12}\right) = \Im(z)$, certainly my solution is wrong. 
 [/INST] You start with $z^4=2(\cos\frac\pi3+i\sin\frac\pi3)$, so your final $z$ should be equal to $\sqrt[4]2(\cos\frac\pi{12}+i\sin\frac\pi{12})$. So, to get your cosine and sine divide by $\sqrt[4]2$ in the end. But your $z$ is wrong (square it); I get $\frac12\sqrt{2\sqrt2+\sqrt6}+\frac i2\sqrt{2\sqrt2-\sqrt6}$
</s>