Question

given that \\( \cos \theta = - \frac { \sqrt { 7 } } { 3 }, \frac { \pi } { 2 } < \theta < \pi \\), find the exact value of each of the following.
(a) \\( \sin ( 2 \theta ) \\) (b) \\( \cos ( 2 \theta ) \\) (c) \\( \sin \frac { \theta } { 2 } \\) (d) \\( \cos \frac { \theta } { 2 } \\)
(a) \\( \sin ( 2 \theta ) = - \frac { 2 \sqrt { 14 } } { 9 } \\)
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expres
(b) \\( \cos ( 2 \theta ) = \frac { 5 } { 9 } \\)
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expres
(c) \\( \sin \frac { \theta } { 2 } = \square \\)
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expres

Explanation:

Step1: Determine the value of \(\sin\theta\)

Since \(\sin^{2}\theta+\cos^{2}\theta = 1\) and \(\frac{\pi}{2}<\theta<\pi\) (so \(\sin\theta> 0\)), we have \(\sin\theta=\sqrt{1-\cos^{2}\theta}\).
Substitute \(\cos\theta=-\frac{\sqrt{7}}{3}\) into the formula: \(\sin\theta=\sqrt{1 - (-\frac{\sqrt{7}}{3})^{2}}=\sqrt{1-\frac{7}{9}}=\sqrt{\frac{2}{9}}=\frac{\sqrt{2}}{3}\)

Step2: Use the half - angle formula for \(\sin\frac{\theta}{2}\)

The half - angle formula is \(\sin\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{2}}\)
Substitute \(\cos\theta =-\frac{\sqrt{7}}{3}\) into the formula:
\(\sin\frac{\theta}{2}=\sqrt{\frac{1-(-\frac{\sqrt{7}}{3})}{2}}=\sqrt{\frac{3 + \sqrt{7}}{6}}=\frac{\sqrt{6 + 2\sqrt{7}}}{6}=\frac{\sqrt{(\sqrt{7}+1)^{2}}}{6}=\frac{\sqrt{7}+1}{2\sqrt{3}}=\frac{\sqrt{21}+\sqrt{3}}{6}\)

Answer:

\(\frac{\sqrt{21}+\sqrt{3}}{6}\)