Answer :
To find the values of [tex]\(\cos 2\theta\)[/tex] and [tex]\(\tan 2\theta\)[/tex] given that [tex]\(\cos \theta = -\frac{8}{17}\)[/tex] and [tex]\(\theta\)[/tex] is in quadrant III, we need to use some trigonometric identities and properties of the functions in different quadrants:
1. Finding [tex]\(\sin \theta\)[/tex]:
Since [tex]\(\cos \theta = -\frac{8}{17}\)[/tex] and [tex]\(\theta\)[/tex] is in quadrant III where sine is negative as well, we use the Pythagorean identity:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
Plugging [tex]\(\cos \theta\)[/tex] into the identity, we get:
[tex]\[ \sin^2 \theta + \left(-\frac{8}{17}\right)^2 = 1 \][/tex]
Simplifying:
[tex]\[ \sin^2 \theta + \frac{64}{289} = 1 \][/tex]
[tex]\[ \sin^2 \theta = 1 - \frac{64}{289} \][/tex]
[tex]\[ \sin^2 \theta = \frac{225}{289} \][/tex]
Since [tex]\(\theta\)[/tex] is in quadrant III where sine is negative, we have:
[tex]\[ \sin \theta = -\frac{15}{17} \][/tex]
2. Finding [tex]\(\cos 2\theta\)[/tex]:
Using the double angle formula for cosine:
[tex]\[ \cos 2\theta = 2 \cos^2 \theta - 1 \][/tex]
Plugging in the value of [tex]\(\cos \theta\)[/tex]:
[tex]\[ \cos 2\theta = 2 \left(-\frac{8}{17}\right)^2 - 1 \][/tex]
Simplifying:
[tex]\[ \cos 2\theta = 2 \cdot \frac{64}{289} - 1 \][/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]
So, [tex]\(\cos 2\theta\)[/tex] is approximately [tex]\(-0.5571\)[/tex].
3. Finding [tex]\(\tan 2\theta\)[/tex]:
Using the double angle formula for tangent:
[tex]\[ \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} \][/tex]
First, we need to find [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{15}{17}}{-\frac{8}{17}} = \frac{15}{8} \][/tex]
Now, plug this into the double angle formula:
[tex]\[ \tan 2\theta = \frac{2 \cdot \frac{15}{8}}{1 - \left(\frac{15}{8}\right)^2} \][/tex]
Simplifying:
[tex]\[ \tan 2\theta = \frac{\frac{30}{8}}{1 - \frac{225}{64}} \][/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]\[ \tan 2\theta = \frac{30 \cdot 64}{8 \cdot -161} \][/tex]
[tex]\[ \tan 2\theta = \frac{1920}{-1288} \][/tex]
[tex]\[ \tan 2\theta = -\frac{1920}{1288} \][/tex]
So, [tex]\(\tan 2\theta\)[/tex] is approximately [tex]\(-1.4907\)[/tex].
Based on these calculations:
[tex]\[ \cos 2\theta \approx -0.5571 \quad \text{and} \quad \tan 2\theta \approx -1.4907 \][/tex]
Therefore, the completed statement is:
If [tex]\(\cos \theta = -\frac{8}{17}\)[/tex] and [tex]\(\theta\)[/tex] is in quadrant III, [tex]\(\cos 2\theta = -0.5571\)[/tex] and [tex]\(\tan 2\theta = -1.4907\)[/tex].
1. Finding [tex]\(\sin \theta\)[/tex]:
Since [tex]\(\cos \theta = -\frac{8}{17}\)[/tex] and [tex]\(\theta\)[/tex] is in quadrant III where sine is negative as well, we use the Pythagorean identity:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
Plugging [tex]\(\cos \theta\)[/tex] into the identity, we get:
[tex]\[ \sin^2 \theta + \left(-\frac{8}{17}\right)^2 = 1 \][/tex]
Simplifying:
[tex]\[ \sin^2 \theta + \frac{64}{289} = 1 \][/tex]
[tex]\[ \sin^2 \theta = 1 - \frac{64}{289} \][/tex]
[tex]\[ \sin^2 \theta = \frac{225}{289} \][/tex]
Since [tex]\(\theta\)[/tex] is in quadrant III where sine is negative, we have:
[tex]\[ \sin \theta = -\frac{15}{17} \][/tex]
2. Finding [tex]\(\cos 2\theta\)[/tex]:
Using the double angle formula for cosine:
[tex]\[ \cos 2\theta = 2 \cos^2 \theta - 1 \][/tex]
Plugging in the value of [tex]\(\cos \theta\)[/tex]:
[tex]\[ \cos 2\theta = 2 \left(-\frac{8}{17}\right)^2 - 1 \][/tex]
Simplifying:
[tex]\[ \cos 2\theta = 2 \cdot \frac{64}{289} - 1 \][/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]
So, [tex]\(\cos 2\theta\)[/tex] is approximately [tex]\(-0.5571\)[/tex].
3. Finding [tex]\(\tan 2\theta\)[/tex]:
Using the double angle formula for tangent:
[tex]\[ \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} \][/tex]
First, we need to find [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{15}{17}}{-\frac{8}{17}} = \frac{15}{8} \][/tex]
Now, plug this into the double angle formula:
[tex]\[ \tan 2\theta = \frac{2 \cdot \frac{15}{8}}{1 - \left(\frac{15}{8}\right)^2} \][/tex]
Simplifying:
[tex]\[ \tan 2\theta = \frac{\frac{30}{8}}{1 - \frac{225}{64}} \][/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]\[ \tan 2\theta = \frac{30 \cdot 64}{8 \cdot -161} \][/tex]
[tex]\[ \tan 2\theta = \frac{1920}{-1288} \][/tex]
[tex]\[ \tan 2\theta = -\frac{1920}{1288} \][/tex]
So, [tex]\(\tan 2\theta\)[/tex] is approximately [tex]\(-1.4907\)[/tex].
Based on these calculations:
[tex]\[ \cos 2\theta \approx -0.5571 \quad \text{and} \quad \tan 2\theta \approx -1.4907 \][/tex]
Therefore, the completed statement is:
If [tex]\(\cos \theta = -\frac{8}{17}\)[/tex] and [tex]\(\theta\)[/tex] is in quadrant III, [tex]\(\cos 2\theta = -0.5571\)[/tex] and [tex]\(\tan 2\theta = -1.4907\)[/tex].
