To find the exact value of \(\cos \frac{\pi}{12}\), we can use the half-angle formula: \[ \cos \frac{\theta}{2} = \sqrt{\frac{1 + \cos \theta}{2}} \] Here, we can let \(\theta = \frac{\pi}{6}\), so that: \[ \frac{\pi}{12} = \frac{1}{2} \times \frac{\pi}{6} \] We know the cosine of \(\frac{\pi}{6}\): \[ \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \] Now we apply the half-angle formula: \[ \cos \frac{\pi}{12} = \cos \left(\frac{\pi}{6} \cdot \frac{1}{2}\right) = \sqrt{\frac{1 + \cos \frac{\pi}{6}}{2}} \] Substituting \(\cos \frac{\pi}{6}\): \[ \cos \frac{\pi}{12} = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} \] Now, simplifying the expression in the numerator: \[ 1 + \frac{\sqrt{3}}{2} = \frac{2}{2} + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2} \] Thus, we have: \[ \cos \frac{\pi}{12} = \sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{3}}{4}} = \frac{\sqrt{2 + \sqrt{3}}}{2} \] Therefore, the exact value of \(\cos \frac{\pi}{12}\) is: \[ \boxed{\frac{\sqrt{2 + \sqrt{3}}}{2}} \]
Find the exact value of cos π/12
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