Question

given that \\( \cos 2 \alpha = \frac { 1 } { 3 } \\) and \\( \alpha \\) terminates in quadrant i, find the exact value of \\( \sin \alpha \\).
\\( \sin \alpha = \square \\)
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)

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:

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