<s>[INST] Using the polar form of $1 + i$ and $\sqrt3 + i$ to deduce $\cos (\frac{\pi}{12}), \sin(\frac{\pi}{12})$ I have been beating my head against the following problem and would like a gentle nudge in the right direction.
The question states, by writing $1 + i$ and $\sqrt3 + i$ in polar form, deduce that
$$\cos (\frac{\pi}{12}) = \frac{\sqrt3 + 1}{2\sqrt2}, \sin(\frac{\pi}{12}) = \frac{\sqrt3 - 1}{2\sqrt2}$$
so I have written them in polar form
EDIT (as polar forms were incorrect):
$$1 + i = \sqrt2e^{i\pi/4}, \sqrt3 + i = 2e^{i\pi/6}$$
another part of the question also asks you to put $\frac{1 + i}{\sqrt3 + i}$ into form $x + yi$ which I figured is
$$\frac{\sqrt3 + 1}{4} + \frac{1 - \sqrt3}{4}i$$
I just can't seem to connect it all together unfortunately so any help would be greatfully received.
 [/INST] Notice 
$$1 + i = \sqrt2e^{i\frac{\pi}{4}}$$
$$\sqrt{3} - i = 2e^{-i\frac{\pi}{6}}$$
Multiply both and we get
$$(1 + i)(\sqrt{3} - i) = \sqrt2e^{i(\frac{\pi}{4} - \frac{\pi}{6})}$$
$$(\sqrt3 + 1) + (\sqrt3-1)i= 2\sqrt{2}e^{i\frac{\pi}{12}}$$
$$(\sqrt3 + 1) + (\sqrt3-1)i= 2\sqrt2\left(\sin{\frac{\pi}{12}} + i\cos\frac{\pi}{12}\right)$$
$$\frac{\sqrt3 + 1}{2\sqrt2} + \frac{\sqrt3 - 1}{2\sqrt2}i = \cos{\frac{\pi}{12}} + i\sin\frac{\pi}{12}$$
Now, compare both real and imaginary parts to deduce:
$$\sin{\frac{\pi}{12}} = \frac{\sqrt3 + 1}{2\sqrt2}$$
$$\cos{\frac{\pi}{12}} = \frac{\sqrt3 - 1}{2\sqrt2}$$
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