Respuesta :

[tex]\cos \frac{\theta}{2}=\sqrt{\frac{1+\cos \theta}{2}} \\ \\ \cos{\frac{\pi}{24}}=\cos{\frac{\frac{\pi}{12}}{2}}=\sqrt{\frac{1+\cos \frac{\pi}{12}}{2}}= \sqrt{ \frac{1}{2}+ \frac{\cos \frac{\pi}{12}}{2} } \\ \\ \cos{\frac{\pi}{12}}=\cos{\frac{\frac{\pi}{6}}{2}}=\sqrt{\frac{1+\cos \frac{\pi}{6}}{2}}= \sqrt{ \frac{1}{2}+ \frac{ \frac{ \sqrt{3} }{2} }{2} } =\sqrt{ \frac{2}{4}+ \frac{ \sqrt{3} }{4} } = \sqrt{\frac{ 2+\sqrt{3} }{4} } = \\ \\ =\frac{ \sqrt{2+\sqrt{3}} }{2} [/tex]

[tex]\cos{\frac{\pi}{24}}= \sqrt{ \frac{1}{2}+ \frac{\cos \frac{\pi}{12}}{2} } = \sqrt{ \frac{1}{2}+ \frac{\frac{ \sqrt{2+\sqrt{3}} }{2}}{2} } = \sqrt{ \frac{2}{4}+ \frac{ \sqrt{2+\sqrt{3}} }{4}} } =\sqrt{ \frac{ 2+\sqrt{2+\sqrt{3}} }{4}} } \\ \\\cos{\frac{\pi}{24}}=\frac{\sqrt{2+\sqrt{2+\sqrt{3}}}} {2}[/tex]