Question

use a half - angle formula to find the exact value of $\cos\frac{\pi}{12}$. $\cos\frac{\pi}{12}=\square$

Explanation:

Step1: Identify 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 (\(\cos x>0\) for \(x\in(0,\frac{\pi}{2})\)), we use the positive form. 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:

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