Answer:
[tex]\frac{2+\sqrt{3} }{4}[/tex]
Step-by-step explanation:
We can use the identity [tex]Cos^2(x)=\frac{1}{2}+\frac{1}{2}Cos(2x)[/tex] to write it as:
[tex]Cos^2(\frac{\pi}{12})=\frac{1}{2}+\frac{1}{2}Cos(2(\frac{\pi}{12}))\\=\frac{1}{2}+\frac{1}{2}Cos(\frac{\pi}{6})\\=\frac{1}{2}+\frac{1}{2}(\frac{\sqrt{3} }{2})\\=\frac{1}{2}+\frac{\sqrt{3} }{4}\\=\frac{2+\sqrt{3} }{4}[/tex]
Note: The value of [tex]Cos(\frac{\pi}{6})[/tex] is [tex]\frac{\sqrt{3} }{2}[/tex]
The answer is the 2nd one, [tex]\frac{2+\sqrt{3} }{4}[/tex]
