Prove that $45°$ angle can be trisected with straightedge and compass.

Question

I want to prove that $45°$ angle can be trisected for this i have to show that $\sin 15°$ or $\cos 15°$ is constructible.

How can i show that $\sin 15° \in \mathbb{Q}(\sqrt{2},\sqrt{3})$?

Answer 1

Use the fact that

$$\sin{(a - b)} = \sin{a} \cos{b} - \cos{a} \sin{b}$$

with $a = 45^{\circ}$ and $b = 30^{\circ}$.

Answer 2

Reference : Check my answer to the question http:// Find out the angle of <ABC

Answer 3

There is a similar problem solved here. What you have to do is to build a $30^\circ$ angle internal to the $45^\circ$, internal and sharing one arm. Then bisect the $30^\circ$ in order to get the $45^\circ$ trisected.