Use the information given about the angle , to find the exact values of the foliowing.

Question

Use the information given about the angle $$ 	heta, \cos 	heta=-\frac{\sqrt{3}}{3}, \frac{\pi}{2}<	heta<\pi $$, to find the exact values of the foliowing. $$ \begin{array}{llll}	ext { (a) } \sin (2 	heta) & 	ext { (b) } \cos (2 	heta) & 	ext { (c) } \sin \frac{	heta}{2} & 	ext { (d) } \cos \frac{	heta}{2} \ 	ext { (a) } \sin (2 	heta)=-\frac{2 \sqrt{2}}{3} & 	ext { (Type an exact answer, using radicals as needed.) } \ 	ext { (b) } \cos (2 	heta)=-\frac{1}{3} 	ext { (Type an exact answer, using radicals as needed.) } \ 	ext { (c) } \sin \frac{	heta}{2}=\square 	ext { (Type an exact answer, using radicals as needed.) }\end{array} $$

UpStudy AI · Accepted Answer

To solve the problem, we start with the known condition $$ \cos \theta = -\frac{\sqrt{3}}{3} $$ and the range $$ \frac{\pi}{2} < \theta < \pi $$. This indicates that $$ \theta $$ is in the second quadrant, where sine is positive and cosine is negative.

### Step 1: Find $$ \sin \theta $$

Using the Pythagorean identity:
$$
\sin^2 \theta + \cos^2 \theta = 1
$$
Substituting the value of $$ \cos \theta $$:
$$
\sin^2 \theta + \left(-\frac{\sqrt{3}}{3}\right)^2 = 1
$$
$$
\sin^2 \theta + \frac{3}{9} = 1
$$
$$
\sin^2 \theta + \frac{1}{3} = 1
$$
$$
\sin^2 \theta = 1 - \frac{1}{3} = \frac{2}{3}
$$
Taking the square root, since $$ \sin \theta $$ is positive in the second quadrant:
$$
\sin \theta = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}
$$

### Step 2: Calculate $$ \sin(2\theta) $$ and $$ \cos(2\theta) $$

Using the double angle formulas:
$$
\sin(2\theta) = 2 \sin \theta \cos \theta
$$
Substituting the values:
$$
\sin(2\theta) = 2 \left(\frac{\sqrt{6}}{3}\right) \left(-\frac{\sqrt{3}}{3}\right) = -\frac{2\sqrt{18}}{9} = -\frac{6\sqrt{2}}{9} = -\frac{2\sqrt{2}}{3}
$$

Now for $$ \cos(2\theta) $$:
$$
\cos(2\theta) = \cos^2 \theta - \sin^2 \theta
$$
Substituting the values:
$$
\cos(2\theta) = \left(-\frac{\sqrt{3}}{3}\right)^2 - \left(\frac{\sqrt{6}}{3}\right)^2
$$
$$
= \frac{3}{9} - \frac{6}{9} = -\frac{3}{9} = -\frac{1}{3}
$$

### Step 3: Calculate $$ \sin\left(\frac{\theta}{2}\right) $$ and $$ \cos\left(\frac{\theta}{2}\right) $$

Using the half-angle formulas:
$$
\sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos \theta}{2}}
$$
Substituting the value of $$ \cos \theta $$:
$$
\sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \left(-\frac{\sqrt{3}}{3}\right)}{2}} = \sqrt{\frac{1 + \frac{\sqrt{3}}{3}}{2}} = \sqrt{\frac{\frac{3}{3} + \frac{\sqrt{3}}{3}}{2}} = \sqrt{\frac{3 + \sqrt{3}}{6}}
$$

For $$ \cos\left(\frac{\theta}{2}\right) $$:
$$
\cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos \theta}{2}}
$$
Substituting the value of $$ \cos \theta $$:
$$
\cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \left(-\frac{\sqrt{3}}{3}\right)}{2}} = \sqrt{\frac{1 - \frac{\sqrt{3}}{3}}{2}} = \sqrt{\frac{\frac{3}{3} - \frac{\sqrt{3}}{3}}{2}} = \sqrt{\frac{3 - \sqrt{3}}{6}}
$$

### Final Answers

Thus, the exact values are:
- (a) $$ \sin(2\theta) = -\frac{2\sqrt{2}}{3} $$
- (b) $$ \cos(2\theta) = -\frac{1}{3} $$
- (c) $$ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{3 + \sqrt{3}}{6}} $$
- (d) $$ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{3 - \sqrt{3}}{6}} $$ (not requested but calculated for completeness)
$$ \sin(2\theta) = -\frac{2\sqrt{2}}{3} $$, $$ \cos(2\theta) = -\frac{1}{3} $$, $$ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{3 + \sqrt{3}}{6}} $$, $$ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{3 - \sqrt{3}}{6}} $$

Use the information given about the angle , to find the exact values of the foliowing.

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