use a half - angle formula to find the exact value of \\( \\cos \\frac{\\pi}{12} \\).\n\\( \\cos…

Explanation:

Step1: Recall the half - angle formula

The half - angle formula for cosine is (\cos\frac{\alpha}{2}=\pm\sqrt{\frac{1 + \cos\alpha}{2}}). Since (\frac{\pi}{12}=\frac{\frac{\pi}{6}}{2}) and (\frac{\pi}{12}) is in the first quadrant ((\cos x>0) when (x\in(0,\frac{\pi}{2}))), we take the positive sign. Here (\alpha=\frac{\pi}{6}).

Step2: Substitute (\alpha = \frac{\pi}{6}) into the formula

We know that (\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}). Substitute into (\cos\frac{\alpha}{2}=\sqrt{\frac{1+\cos\alpha}{2}}), we get (\cos\frac{\pi}{12}=\sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}}).

Step3: Simplify the expression

First, simplify the fraction inside the square - root: (\frac{1+\frac{\sqrt{3}}{2}}{2}=\frac{\frac{2 + \sqrt{3}}{2}}{2}=\frac{2+\sqrt{3}}{4}). Then (\cos\frac{\pi}{12}=\sqrt{\frac{2+\sqrt{3}}{4}}=\frac{\sqrt{2+\sqrt{3}}}{2}). Another way is to rationalize further. We know that (\cos\frac{\pi}{12}=\cos(15^{\circ})), and (\cos(A - B)=\cos A\cos B+\sin A\sin B), (\cos15^{\circ}=\cos(45^{\circ}-30^{\circ})=\cos45^{\circ}\cos30^{\circ}+\sin45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}).

Answer:

(\frac{\sqrt{6}+\sqrt{2}}{4})