Rysunek poglądowy:

Niech:

$\left|AB\right|=a$  

 

Wiemy, że:

$\left|AC\right|=\left|AB\right|\cdot \sqrt{2}$  

$\left|AC\right|=a\sqrt{2}$  

Czyli:

$\left|AO\right|=\frac{1}{2}\cdot \left|AC\right|=\frac{a\sqrt{2}}{2}$  

Z twierdzenia Pitagorasa w trójkącie ACO₁:

${\left|AO\right|}^2+{\left|O{O}_1\right|}^2={\left|A{O}_1\right|}^2$  

$\frac{a^2}{2}+a^2={\left|A{O}_1\right|}^2$  

${\left|A{O}_1\right|}^2=\frac{3}{2}{a}^2$  

$\left|A{O}_1\right|=\frac{\sqrt{3}}{\sqrt{2}}a$  

$\left|A{O}_1\right|=\frac{a\sqrt{6}}{2}$  

Wiemy, że:

$\left|\angle O_{1}OA\right|=\frac{1}{2}\left|O_{1}CA\right|=\frac{1}{2}\cdot \alpha =\frac{\alpha }{2}$  

Zatem

$\cos \left(\frac{\alpha }{2}\right)=\frac{\left|O{O}_1\right|}{\left|A{O}_1\right|}=\frac{a}{\frac{a\sqrt{6}}{2}}=\frac{2}{\sqrt{6}}$  

${\cos }^2\left(\frac{\alpha }{2}\right)=\frac{4}{6}=\frac{2}{3}$  

Z jedynki trygonometrycznej:

${\sin }^2\left(\frac{\alpha }{2}\right)+{\cos }^2\left(\frac{\alpha }{2}\right)=1$  

${\sin }^2\left(\frac{\alpha }{2}\right)+\frac{2}{3}=1$  

${\sin }^2\left(\frac{\alpha }{2}\right)=\frac{1}{3}$  

Ze wzoru:

$\cos \left(2x\right)={\cos }^{2}x-{\sin }^{2}x$  

A więc:

$2x=\alpha$  

$x=\frac{\alpha }{2}$  

czyli

$\cos \alpha ={\cos }^2\left(\frac{\alpha }{2}\right)-{\sin }^2\left(\frac{\alpha }{2}\right)=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$