$\begingroup$Find the exact values of those two numbers but by using the two following complex numbers complexs numbers :
$$z = \frac{-\frac{\sqrt3}{2} -i\frac{1}{2}}{\sqrt8}$$ and $$w = \frac{\frac{\sqrt2}{2} +i\frac{\sqrt2}{2}}{\sqrt8}$$
I added them up and computed $z/w$ but it led me to nowhere I feel like I have to do some combined computation with those rwo complex numbers and then by identification I can know the exact values but I haven't found the right operations to do yet. help me please.
$\endgroup$23 Answers
$\begingroup$If you know $\cos$ and $\sin$ for $\pi/4$ and $\pi/3$, you can easily spot that $\sqrt{8}z=e^{-i\pi/3}$ and $\sqrt{8}w=e^{i\pi/4}$, so that: $$\exp\Big(\frac{11}{12}i\pi\Big)=\exp\Big(\frac{2}{3}i\pi\Big)\exp\Big(\frac{1}{4}i\pi\Big)=-8\overline{z}^2\sqrt{8}w.$$
$\endgroup$$\begingroup$$$\cos\frac{\pi}{12}+i\sin\frac{\pi}{12} = e^{i\pi/12} = e^{i\pi/3}\cdot e^{-i\pi/4} \tag{1}$$
$$ = \left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)\cdot\left(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\right) \tag{2}$$ $$ = \left(\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}\right)+i\left(\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}\right)\tag{3} $$ hence $$ \cos\frac{11\pi}{12} = -\frac{\sqrt{2}+\sqrt{6}}{4},\qquad \sin\frac{11\pi}{12}=\frac{\sqrt{6}-\sqrt{2}}{4}.\tag{4}$$
$\endgroup$$\begingroup$$\frac {z}{w} = \cos \frac {11\pi}{12} + i\sin \frac {11\pi}{12}$
I don't think it is easist way to get there, but it will work.
after the more obvious cancellations:
$\frac {-\sqrt 3 - i}{\sqrt2 + \sqrt 2 i}$
multiply the numerator and denominator by the conjugate of the denominator.
$\frac {(-\sqrt 6 - \sqrt 2) + i(\sqrt 6 - \sqrt 2)}{4}$
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