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.

• use half-angle formulas knowing that $\cos {\pi/ 6}=0.5$ – Vasya Oct 31 '18 at 18:32
• Use $$\frac{\pi}{12}=\frac{\pi}{3}-\frac{\pi}{4}$$ – Crostul Oct 31 '18 at 18:34
• Please read the question fully. – AverageMarcin Oct 31 '18 at 18:36
• You have two errors in your expression for $z$. 1) $\sqrt{6} > 1$, so your expression for $z$ is a real number. 2) If we switch the quantity under the radical to $\sqrt{6}-1$, you have $z^2 = (1+i\sqrt{5})/2\ne (\sqrt{6}+i\sqrt{2})/2$. – eyeballfrog Oct 31 '18 at 18:38
• Another issue is that, given the defining equation for $z$, the real and imaginary parts of $z$, and hence of the roots of $z$, can't be cosines or sines since $\vert z\vert\ne1$. What must you do to relate $z$ to a complex number with unit modulus? – Will Orrick Oct 31 '18 at 18:41

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 $$\frac12\sqrt{2\sqrt2+\sqrt6}+\frac i2\sqrt{2\sqrt2-\sqrt6}$$
• This can be written without nested radicals using $\sqrt{2\sqrt{2}\pm\sqrt{6}} = (\sqrt{3}\pm1)/2^{1/4}$. – eyeballfrog Oct 31 '18 at 18:52