given that \\( \\cos 2 \\alpha = \\frac { 1 } { 3 } \\) and \\( \\alpha \\) terminates in quadrant i, find…

Explanation:

Step1: Use the double - angle formula

The double - angle formula for cosine is (\cos2\alpha = 1 - 2\sin^{2}\alpha). Given (\cos2\alpha=\frac{1}{3}), we substitute it into the formula: (\frac{1}{3}=1 - 2\sin^{2}\alpha).

Step2: Solve for (\sin^{2}\alpha)

Rearrange the equation (\frac{1}{3}=1 - 2\sin^{2}\alpha) to get (2\sin^{2}\alpha=1-\frac{1}{3}). Calculate (1-\frac{1}{3}=\frac{3 - 1}{3}=\frac{2}{3}), so (2\sin^{2}\alpha=\frac{2}{3}). Then (\sin^{2}\alpha=\frac{1}{3}).

Step3: Determine the sign of (\sin\alpha)

Since (\alpha) terminates in quadrant I, (\sin\alpha>0). Take the square root of (\sin^{2}\alpha=\frac{1}{3}), we get (\sin\alpha=\sqrt{\frac{1}{3}}). Simplify (\sqrt{\frac{1}{3}}=\frac{\sqrt{1}}{\sqrt{3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}).

Answer:

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