To show that an angle [math]\phi[/math] cannot be trisected, it suffices to show that [math]\theta=\frac{\phi}{3}[/math] is not a constructible angle.[br][br]Let's consider [math]\phi=60^\circ[/math]. Then [math]\theta=20^\circ[/math]. Now we use the following trigonometric identity:[br][br][center][math]\cos(3\theta)=4\cos^3(\theta)-3\cos(\theta)[/math][/center][br][br]Let [math]\beta=\cos(\theta)[/math], we have [br][br][center][math] \frac{1}{2}=\cos(\phi)=4\beta^3-3\beta [/math][/center][br][center][math] \Rightarrow 8\beta^3-6\beta-1=0[/math][/center][br][br]It can be shown that [math] 8\beta^3-6\beta-1[/math] is irreducible over [math]\mathbb{Q}[/math]. Therefore, the degree of [math]\beta[/math] over [math]\mathbb{Q}[/math] is [math]3[/math] and hence by the Main Theorem, [math]\cos(20^\circ)=\beta[/math] is not a constructible number.[br][br]In other words, [math]20^\circ[/math] is not a constructible angle, which in turn implies that [math]60^\circ[/math] cannot be trisected.