Using the polar form of $1 + i$ and $\sqrt3 + i$ to deduce $\cos (\frac{\pi}{12}), \sin(\frac{\pi}{12})$

Question

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.

Answer 1

Hint: $\frac14-\frac16=\frac{3}{12}-\frac{2}{12}=\frac{1}{12}$.

Answer 2

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:

$$\sin{\frac{\pi}{12}} = \frac{\sqrt3 + 1}{2\sqrt2}$$ $$\cos{\frac{\pi}{12}} = \frac{\sqrt3 - 1}{2\sqrt2}$$