Given that \(\tan \theta = -\frac{7}{6}\) and \(\frac{\pi}{2} < \theta < \pi\), we can determine the values of the trigonometric functions using the properties of the triangle and identities.
Step 1: Determine values of \(\sin \theta\) and \(\cos \theta\)
Since \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), we can establish the following:
\[ \sin \theta = -7k \quad \text{and} \quad \cos \theta = 6k \]
where \(k\) is some positive constant. To maintain consistency with the given interval \(\frac{\pi}{2} < \theta < \pi\) which corresponds to Quadrant II, \(\sin \theta\) is positive and \(\cos \theta\) is negative.
Thus, let:
\[ \sin \theta = 7k \quad \text{and} \quad \cos \theta = -6k \]
To find \(k\), we must apply the Pythagorean identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Substituting the values of \(\sin \theta\) and \(\cos \theta\):
\[ (7k)^2 + (-6k)^2 = 1 \]
Calculating this gives:
\[ 49k^2 + 36k^2 = 1 \]
\[ 85k^2 = 1 \implies k^2 = \frac{1}{85} \implies k = \frac{1}{\sqrt{85}} \]
Now plug \(k\) back into our expressions for \(\sin \theta\) and \(\cos \theta\):
\[ \sin \theta = 7 \cdot \frac{1}{\sqrt{85}} = \frac{7}{\sqrt{85}} \]
\[ \cos \theta = -6 \cdot \frac{1}{\sqrt{85}} = -\frac{6}{\sqrt{85}} \]
Step 2: Calculate \(\sin 2\theta\), \(\cos 2\theta\), and \(\tan 2\theta\)
Now, we can use the double angle formulas:
- For \(\sin 2\theta\):
\[ \sin(2\theta) = 2 \sin \theta \cos \theta \]
Substituting the values:
\[ \sin(2\theta) = 2 \left(\frac{7}{\sqrt{85}}\right) \left(-\frac{6}{\sqrt{85}}\right) = -\frac{84}{85} \]
So, \(\sin 2\theta = -\frac{84}{85}\).
- For \(\cos 2\theta\):
\[ \cos(2\theta) = \cos^2 \theta - \sin^2 \theta \]
Calculating \( \cos^2 \theta \) and \( \sin^2 \theta \):
\[ \cos^2 \theta = \left(-\frac{6}{\sqrt{85}}\right)^2 = \frac{36}{85} \]
\[ \sin^2 \theta = \left(\frac{7}{\sqrt{85}}\right)^2 = \frac{49}{85} \]
Thus,
\[ \cos(2\theta) = \frac{36}{85} - \frac{49}{85} = -\frac{13}{85} \]
So, \(\cos 2\theta = -\frac{13}{85}\).
- For \(\tan 2\theta\):
Using the identity \(\tan(2\theta) = \frac{\sin(2\theta)}{\cos(2\theta)}\):
\[ \tan(2\theta) = \frac{-\frac{84}{85}}{-\frac{13}{85}} = \frac{84}{13} \]
Step 3: Calculate \( \csc 2\theta\), \( \sec 2\theta\), and \( \cot 2\theta\)
Now we find the other trigonometric functions:
- For \(\csc 2\theta\):
\(\csc 2\theta = \frac{1}{\sin 2\theta} = \frac{1}{-\frac{84}{85}} = -\frac{85}{84}\).
- For \(\sec 2\theta\):
\(\sec 2\theta = \frac{1}{\cos 2\theta} = \frac{1}{-\frac{13}{85}} = -\frac{85}{13}\).
- For \(\cot 2\theta\):
\(\cot 2\theta = \frac{1}{\tan 2\theta} = \frac{13}{84}\).
Summary of results:
- \(\sin 2\theta = -\frac{84}{85}\)
- \(\csc 2\theta = -\frac{85}{84}\)
- \(\cos 2\theta = -\frac{13}{85}\)
- \(\sec 2\theta = -\frac{85}{13}\)
- \(\tan 2\theta = \frac{84}{13}\)
- \(\cot 2\theta = \frac{13}{84}\)