One of the many interesting aspects of imaginary numbers is that we can use them to find out “real” facts (facts about real numbers).

Perhaps the most used examples are the derivation of the trigonometric identities for \(sin 2\theta\) and \(cos 2\theta\)*.

This post offers something more exciting and less-known: the radical forms of \(cos \frac{\pi}{12}\) and \(sin \frac{\pi}{12}\).

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 \(r e^{i\alpha}\). 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 \[cos \frac{\pi}{12}+i sin\frac{\pi}{12}\] 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:

\[cos\frac{\pi}{12} = \frac{\sqrt6+\sqrt2}{4}\]

\[sin\frac{\pi}{12} = \frac{\sqrt6-\sqrt2}{4}\]

\[cos\frac{\pi}{12} = \frac{\sqrt6+\sqrt2}{4}\]

\[sin\frac{\pi}{12} = \frac{\sqrt6-\sqrt2}{4}\]

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):

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

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

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

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

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

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