Find the product of the complex numbers z1=cos⁡π6+isin⁡π6 and z2=cos⁡π4+isin⁡π4 Leave answer in polar form.

Step 1
Your input: simplify and calculate different forms of
$(\cos (\frac{\pi }{4})+i\sin (\frac{\pi }{4}))(\cos (\frac{\pi }{6})+i\sin (\frac{\pi }{6}))$
Rewrite:
$=(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}i}{2})(\frac{\sqrt{3}}{2}+\frac{i}{2})$
Use FOIL to multiply (for steps, see foil calculator), don't forget that $i^2=-1$
$((\frac{\sqrt{2}}{2}+\frac{\sqrt{2}i}{2})(\frac{\sqrt{3}}{2}+\frac{i}{2}))=(-\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}+i(\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}))$
Simplify:
$(-\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}+i(\frac{\sqrt{2}}{4}+\frac{6}{4}))=(-\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}+\frac{i(\sqrt{2}+\sqrt{6})}{4})$
Hence: $=-\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}+\frac{i(\sqrt{2}+\sqrt{6})}{4}$
Step 2
Polar form
For a complex number a+bi, polar form given by $r(\cos (\theta )+i\sin (\theta ))$, where $r=\sqrt{a^2+b^2}$ and $\theta =a\tan (\frac{b}{a})$
We have that $a=-\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}and=\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}$
Thus, $r=\sqrt{(-\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}{)}^2+(\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}{)}^2}=1$
Also, $\theta =a\tan (\frac{\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}}{-\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}})=\frac{5\pi }{12}$
Therefore: $-\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}+i(\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4})=\cos (\frac{5\pi }{12})+i\sin (\frac{5\pi }{12})$