Answer :
Let's solve this problem step by step:
1. Given: [tex]\(\cos \theta = -\frac{8}{17}\)[/tex]
2. Theta is in quadrant III. In quadrant III, both sine and cosine functions are negative.
3. Find [tex]\(\sin \theta\)[/tex]:
Using the Pythagorean identity:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
Substitute [tex]\(\cos \theta = -\frac{8}{17}\)[/tex]:
[tex]\[ \sin^2 \theta + \left(-\frac{8}{17}\right)^2 = 1 \][/tex]
[tex]\[ \sin^2 \theta + \frac{64}{289} = 1 \][/tex]
[tex]\[ \sin^2 \theta = 1 - \frac{64}{289} \][/tex]
[tex]\[ \sin^2 \theta = \frac{289}{289} - \frac{64}{289} \][/tex]
[tex]\[ \sin^2 \theta = \frac{225}{289} \][/tex]
Since [tex]\(\theta\)[/tex] is in quadrant III, [tex]\(\sin \theta\)[/tex] is negative:
[tex]\[ \sin \theta = -\sqrt{\frac{225}{289}} = -\frac{15}{17} \][/tex]
4. Find [tex]\(\cos 2\theta\)[/tex]:
Using the double-angle formula for cosine:
[tex]\[ \cos 2\theta = 2\cos^2 \theta - 1 \][/tex]
Substitute [tex]\(\cos \theta = -\frac{8}{17}\)[/tex]:
[tex]\[ \cos 2\theta = 2\left(-\frac{8}{17}\right)^2 - 1 \][/tex]
[tex]\[ \cos 2\theta = 2\left(\frac{64}{289}\right) - 1 \][/tex]
[tex]\[ \cos 2\theta = \frac{128}{289} - 1 \][/tex]
[tex]\[ \cos 2\theta = \frac{128}{289} - \frac{289}{289} \][/tex]
[tex]\[ \cos 2\theta = \frac{128 - 289}{289} \][/tex]
[tex]\[ \cos 2\theta = \frac{-161}{289} \][/tex]
5. Find [tex]\(\tan 2\theta\)[/tex]:
Using the relationship between [tex]\(\tan \theta\)[/tex] and [tex]\(\sin \theta\)[/tex], [tex]\(\cos \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]
Substitute [tex]\(\sin \theta = -\frac{15}{17}\)[/tex] and [tex]\(\cos \theta = -\frac{8}{17}\)[/tex]:
[tex]\[ \tan \theta = \frac{-\frac{15}{17}}{-\frac{8}{17}} = \frac{15}{8} \][/tex]
Using the double-angle formula for tangent:
[tex]\[ \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} \][/tex]
Substitute [tex]\(\tan \theta = \frac{15}{8}\)[/tex]:
[tex]\[ \tan 2\theta = \frac{2 \cdot \frac{15}{8}}{1 - \left(\frac{15}{8}\right)^2} \][/tex]
[tex]\[ \tan 2\theta = \frac{\frac{30}{8}}{1 - \frac{225}{64}} \][/tex]
[tex]\[ \tan 2\theta = \frac{\frac{30}{8}}{\frac{64}{64} - \frac{225}{64}} \][/tex]
[tex]\[ \tan 2\theta = \frac{\frac{30}{8}}{\frac{64 - 225}{64}} \][/tex]
[tex]\[ \tan 2\theta = \frac{\frac{30}{8}}{\frac{-161}{64}} \][/tex]
[tex]\[ \tan 2\theta = \frac{30}{8} \cdot \frac{64}{-161} \][/tex]
[tex]\[ \tan 2\theta = \frac{1920}{-1288} \][/tex]
[tex]\[ \tan 2\theta = -\frac{1920}{1288} = -\frac{480}{322} \approx -\frac{240}{161} \approx -1.49068 \][/tex]
In conclusion:
[tex]\[ \cos 2\theta = -\frac{161}{289} \][/tex]
[tex]\[ \tan 2\theta \approx -1.49068 \][/tex]
1. Given: [tex]\(\cos \theta = -\frac{8}{17}\)[/tex]
2. Theta is in quadrant III. In quadrant III, both sine and cosine functions are negative.
3. Find [tex]\(\sin \theta\)[/tex]:
Using the Pythagorean identity:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
Substitute [tex]\(\cos \theta = -\frac{8}{17}\)[/tex]:
[tex]\[ \sin^2 \theta + \left(-\frac{8}{17}\right)^2 = 1 \][/tex]
[tex]\[ \sin^2 \theta + \frac{64}{289} = 1 \][/tex]
[tex]\[ \sin^2 \theta = 1 - \frac{64}{289} \][/tex]
[tex]\[ \sin^2 \theta = \frac{289}{289} - \frac{64}{289} \][/tex]
[tex]\[ \sin^2 \theta = \frac{225}{289} \][/tex]
Since [tex]\(\theta\)[/tex] is in quadrant III, [tex]\(\sin \theta\)[/tex] is negative:
[tex]\[ \sin \theta = -\sqrt{\frac{225}{289}} = -\frac{15}{17} \][/tex]
4. Find [tex]\(\cos 2\theta\)[/tex]:
Using the double-angle formula for cosine:
[tex]\[ \cos 2\theta = 2\cos^2 \theta - 1 \][/tex]
Substitute [tex]\(\cos \theta = -\frac{8}{17}\)[/tex]:
[tex]\[ \cos 2\theta = 2\left(-\frac{8}{17}\right)^2 - 1 \][/tex]
[tex]\[ \cos 2\theta = 2\left(\frac{64}{289}\right) - 1 \][/tex]
[tex]\[ \cos 2\theta = \frac{128}{289} - 1 \][/tex]
[tex]\[ \cos 2\theta = \frac{128}{289} - \frac{289}{289} \][/tex]
[tex]\[ \cos 2\theta = \frac{128 - 289}{289} \][/tex]
[tex]\[ \cos 2\theta = \frac{-161}{289} \][/tex]
5. Find [tex]\(\tan 2\theta\)[/tex]:
Using the relationship between [tex]\(\tan \theta\)[/tex] and [tex]\(\sin \theta\)[/tex], [tex]\(\cos \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]
Substitute [tex]\(\sin \theta = -\frac{15}{17}\)[/tex] and [tex]\(\cos \theta = -\frac{8}{17}\)[/tex]:
[tex]\[ \tan \theta = \frac{-\frac{15}{17}}{-\frac{8}{17}} = \frac{15}{8} \][/tex]
Using the double-angle formula for tangent:
[tex]\[ \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} \][/tex]
Substitute [tex]\(\tan \theta = \frac{15}{8}\)[/tex]:
[tex]\[ \tan 2\theta = \frac{2 \cdot \frac{15}{8}}{1 - \left(\frac{15}{8}\right)^2} \][/tex]
[tex]\[ \tan 2\theta = \frac{\frac{30}{8}}{1 - \frac{225}{64}} \][/tex]
[tex]\[ \tan 2\theta = \frac{\frac{30}{8}}{\frac{64}{64} - \frac{225}{64}} \][/tex]
[tex]\[ \tan 2\theta = \frac{\frac{30}{8}}{\frac{64 - 225}{64}} \][/tex]
[tex]\[ \tan 2\theta = \frac{\frac{30}{8}}{\frac{-161}{64}} \][/tex]
[tex]\[ \tan 2\theta = \frac{30}{8} \cdot \frac{64}{-161} \][/tex]
[tex]\[ \tan 2\theta = \frac{1920}{-1288} \][/tex]
[tex]\[ \tan 2\theta = -\frac{1920}{1288} = -\frac{480}{322} \approx -\frac{240}{161} \approx -1.49068 \][/tex]
In conclusion:
[tex]\[ \cos 2\theta = -\frac{161}{289} \][/tex]
[tex]\[ \tan 2\theta \approx -1.49068 \][/tex]
