Question

given that \\( \cos \alpha = \frac { 1 } { 3 } \\) and \\( 0 < \alpha < \frac { \pi } { 2 } \\), determine the exact value of \\( \cos \frac { \alpha } { 2 } \\)

\\( \cos \frac { \alpha } { 2 } = \\)
(simplify your answer, including any radicals. use integers or fractions for any nu

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:

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