watch the video and then solve the problem given below.\nclick here to watch the video.\ngiven that \\(…
watch the video and then solve the problem given below.\nclick here to watch the video.\ngiven that \\( \\cos \\alpha = - \\frac { 1 } { 3 } \\) and \\( 0 < \\alpha < \\frac { \\pi } { 2 } \\), determine the exact value of \\( \\cos \\frac { \\alpha } { 2 } \\).\n\\( \\cos \\frac { \\alpha } { 2 } = \\square \\)\n(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)
Answer
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=\frac{1}{3}) into the formula: [ \begin{align*} \cos\frac{\alpha}{2}&=\sqrt{\frac{1+\frac{1}{3}}{2}}\ &=\sqrt{\frac{\frac{3 + 1}{3}}{2}}\ &=\sqrt{\frac{4}{6}}\ &=\sqrt{\frac{2}{3}}\ &=\frac{\sqrt{2}}{\sqrt{3}}\ &=\frac{\sqrt{2}\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}\ &=\frac{\sqrt{6}}{3} \end{align*} ]
Answer:
(\frac{\sqrt{6}}{3})