Consider (frac{frac{1}{2}+frac{sqrt{3}}{2}i}{frac{sqrt{2}}{2}+frac{sqrt{2}}{2}i})

Algebraically, simplify it and write it in the form of (a+bi).

[frac{sqrt6+sqrt2}{4}+frac{sqrt6-sqrt2}{4} i]

We know the answer. But as a wise man once said, mathematics is all about seeing the same thing from different angles (or something like this).  So let us, calculate the same thing in some other way.

We write (frac{1}{2}+frac{sqrt{3}}{2}i) and (frac {sqrt{2}}{2}+frac{sqrt{2}}{2}i) in the form (r e^{ialpha}). And then evaluate the original fraction , that is going to be:

[frac{e^{i frac{pi}{3}}}{e^{i frac{pi}{4}}}]

That can be simplified as [e^{i frac{pi}{12}}] That, can be written as [cos frac{pi}{12}+i  sinfrac{pi}{12}] Compare this result with the one we had in the form of (a+bi) above, where (a) and (b) were in a radical form. They are the same things, represented in two different forms. Thus, their real parts should be equal with each other, and also their imaginary part should be equal with each other. So, we have: 
[cosfrac{pi}{12} = frac{sqrt6+sqrt2}{4}]
[sinfrac{pi}{12} = frac{sqrt6-sqrt2}{4}]

Design a problem to find (cos frac{pi}{8}) and (sin frac{pi}{8}) in the radical form. 

* We know that (a post will be written for this): 

[(cos alpha + i sin alpha)^2=cos (2alpha) + i sin(2alpha)]

Remember what the wise man once said and find ((cosalpha+i sinalpha)^{2}) by actually finding the power. Then we have:

[(cos alpha + i sin alpha)^2=(cos^2alpha -sin^2alpha)+ i (2 sinalpha cosalpha)] The right sides of the above two equality are the same represented in two different forms (both are equal to the same left side). So the real parts of the two should be equal, and the imaginary parts of the two should be also equal.

[cos 2alpha=cos^2 alpha -sin^2 alpha][sin 2alpha=2 sin alpha cosalpha]

Design a problem to find (cos 3alpha) and (sin 3alpha) .