Respuesta :
Answer;
[tex]B\text{. }\frac{\sqrt[]{2}+\sqrt[]{6}}{4}[/tex]Explanation;
Given that;
[tex](60^0-45^0)=15^0[/tex]Hence;
[tex]\text{Cos 15}^0=Cos(60^0-45^0)[/tex]According to trigonometry identity;
[tex]\begin{gathered} Cos(60^0-45^0\text{) = Cos60 Cos45 + Sin60Sin45} \\ Cos(60^0-45^0\text{) }=\frac{1}{2}(\frac{1}{\sqrt[]{2}})+\frac{\sqrt[]{3}}{2}(\frac{1}{\sqrt[]{2}}) \end{gathered}[/tex]Evaluate the result by finding the LCM
[tex]Cos(60^0-45^0\text{) }=\frac{1+\sqrt[]{3}}{2\sqrt[]{2}}[/tex]Rationalize;
[tex]\begin{gathered} Cos(60^0-45^0\text{) }=\frac{1+\sqrt[]{3}}{2\sqrt[]{2}}\times\frac{\sqrt[]{2}}{\sqrt[]{2}} \\ Cos(60^0-45^0\text{) }=\frac{\sqrt[]{2}(1+\sqrt[]{3})}{2\cdot2} \\ Cos(60^0-45^0\text{) }=\frac{\sqrt[]{2}+\sqrt[]{6}}{4} \end{gathered}[/tex]Hence the required reusult is;
[tex]\frac{\sqrt[]{2}+\sqrt[]{6}}{4}[/tex]