given that \\( \\cos \\alpha = \\frac { 1 } { 3 } \\) and \\( 0 < \\alpha < \\frac { \\pi } { 2 } \\)…

Explanation:

Step1: Determine the quadrant of (\frac{\alpha}{2})

Since (0 < \alpha<\frac{\pi}{2}), then (0 < \frac{\alpha}{2}<\frac{\pi}{4}). So (\frac{\alpha}{2}) is in the first - quadrant, and (\cos\frac{\alpha}{2}>0).

Step2: Use the half - angle formula

The half - angle formula for cosine is (\cos\frac{\theta}{2}=\sqrt{\frac{1 + \cos\theta}{2}}). Here (\theta=\alpha) and (\cos\alpha=\frac{1}{3}). Substitute (\cos\alpha) into the formula: (\cos\frac{\alpha}{2}=\sqrt{\frac{1+\frac{1}{3}}{2}}).

Step3: Simplify the expression

First, simplify the numerator of the fraction inside the square root: (1+\frac{1}{3}=\frac{3 + 1}{3}=\frac{4}{3}). Then (\cos\frac{\alpha}{2}=\sqrt{\frac{\frac{4}{3}}{2}}=\sqrt{\frac{4}{3}\times\frac{1}{2}}=\sqrt{\frac{2}{3}}). Rationalize the denominator: (\sqrt{\frac{2}{3}}=\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{2}\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}=\frac{\sqrt{6}}{3}).

Answer:

(\frac{\sqrt{6}}{3})