Respuesta :
[tex]\bf cos\left(\cfrac{{{ \theta}}}{2}\right)=\pm \sqrt{\cfrac{1+cos({{ \theta}})}{2}}\\\\
-------------------------------\\\\
\cfrac{\pi }{12}\cdot 2\implies \cfrac{\pi }{6}\qquad therefore\qquad \cfrac{\quad \frac{\pi }{6}\quad }{2}\implies \cfrac{\pi }{12}\qquad then
\\\\\\
cos\left( \frac{\pi }{12} \right)\implies cos\left( \cfrac{\frac{\pi }{6}}{2} \right)=\pm\sqrt{\cfrac{1+cos\left( \frac{\pi }{6} \right)}{2}}[/tex]
[tex]\bf cos\left( \cfrac{\frac{\pi }{6}}{2} \right)=\pm\sqrt{\cfrac{1+\frac{\sqrt{3}}{2}}{2}}\implies cos\left( \cfrac{\frac{\pi }{6}}{2} \right)=\pm\sqrt{\cfrac{\frac{2+\sqrt{3}}{2}}{2}} \\\\\\ cos\left( \cfrac{\frac{\pi }{6}}{2} \right)=\pm\sqrt{\cfrac{2+\sqrt{3}}{4}}\implies cos\left( \cfrac{\frac{\pi }{6}}{2} \right)=\pm\cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{4}} \\\\\\ cos\left( \cfrac{\frac{\pi }{6}}{2} \right)=\pm\cfrac{\sqrt{2+\sqrt{3}}}{2}[/tex]
[tex]\bf cos\left( \cfrac{\frac{\pi }{6}}{2} \right)=\pm\sqrt{\cfrac{1+\frac{\sqrt{3}}{2}}{2}}\implies cos\left( \cfrac{\frac{\pi }{6}}{2} \right)=\pm\sqrt{\cfrac{\frac{2+\sqrt{3}}{2}}{2}} \\\\\\ cos\left( \cfrac{\frac{\pi }{6}}{2} \right)=\pm\sqrt{\cfrac{2+\sqrt{3}}{4}}\implies cos\left( \cfrac{\frac{\pi }{6}}{2} \right)=\pm\cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{4}} \\\\\\ cos\left( \cfrac{\frac{\pi }{6}}{2} \right)=\pm\cfrac{\sqrt{2+\sqrt{3}}}{2}[/tex]
Answer:
[tex]\cos \left(\frac{\pi }{12}\right)=\frac{\sqrt{2+\sqrt{3}}}{2}[/tex]
Step-by-step explanation:
To find the exact value of [tex]\cos \left(\frac{\pi }{12}\right)[/tex] using half angle identities you must:
Write [tex]\cos \left(\frac{\pi }{12}\right)[/tex] as [tex]\cos \left(\frac{\frac{\pi }{6}}{2}\right)[/tex]
Using the half angle identity [tex]\cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos \left(x\right)}{2}}[/tex]
[tex]\cos \left(\frac{\frac{\pi }{6}}{2}\right)=\sqrt{\frac{1+\cos \left(\frac{\pi }{6}\right)}{2}}[/tex]
Use the following identity: [tex]\cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}[/tex]
[tex]\sqrt{\frac{1+\cos \left(\frac{\pi }{6}\right)}{2}}=\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}[/tex]
Join [tex]1+\frac{\sqrt{3}}{2}[/tex]
[tex]1+\frac{\sqrt{3}}{2}=\frac{1\cdot \:2}{2}+\frac{\sqrt{3}}{2}=\frac{2+\sqrt{3}}{2}[/tex]
[tex]\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}=\sqrt{\frac{\frac{2+\sqrt{3}}{2}}{2} } =\sqrt{\frac{2+\sqrt{3}}{4}} =\frac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}=\frac{\sqrt{2+\sqrt{3}}}{2}[/tex]
Therefore,
[tex]\cos \left(\frac{\pi }{12}\right)=\frac{\sqrt{2+\sqrt{3}}}{2}[/tex]
