To find the exact values of the remaining trigonometric functions given that $\sin \theta = \frac{\sqrt{2}}{6}$ and $\cos \theta > 0$, we proceed as follows:

1. Find $\cos \theta$:

Using the Pythagorean identity: $\sin^2 \theta + \cos^2 \theta = 1$

Substitute $\sin \theta = \frac{\sqrt{2}}{6}$: $\left( \frac{\sqrt{2}}{6} \right)^2 + \cos^2 \theta = 1$ $\frac{2}{36} + \cos^2 \theta = 1$ $\frac{1}{18} + \cos^2 \theta = 1$ $\cos^2 \theta = 1 - \frac{1}{18} = \frac{17}{18}$ Since $\cos \theta > 0$: $\cos \theta = \sqrt{\frac{17}{18}} = \frac{\sqrt{17}}{3\sqrt{2}} = \frac{\sqrt{34}}{6}$

2. Find $\tan \theta$:

$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{\sqrt{2}}{6}}{\frac{\sqrt{34}}{6}} = \frac{\sqrt{2}}{\sqrt{34}} = \frac{\sqrt{2}}{\sqrt{34}} \cdot \frac{\sqrt{34}}{\sqrt{34}} = \frac{\sqrt{68}}{34} = \frac{2\sqrt{17}}{34} = \frac{\sqrt{17}}{17}$

3. Find $\csc \theta$:

$\csc \theta = \frac{1}{\sin \theta} = \frac{6}{\sqrt{2}} = 3\sqrt{2}$

4. Find $\sec \theta$:

$\sec \theta = \frac{1}{\cos \theta} = \frac{6}{\sqrt{34}} = \frac{6\sqrt{34}}{34} = \frac{3\sqrt{34}}{17}$

5. Find $\cot \theta$:

$\cot \theta = \frac{1}{\tan \theta} = \frac{1}{\frac{\sqrt{17}}{17}} = \frac{17}{\sqrt{17}} = \sqrt{17}$

Would you like any further details on these calculations, or do you have any specific questions?

Related Questions:

1. How would the values change if $\cos \theta < 0$?
2. Can you verify these trigonometric identities using a unit circle approach?
3. What are the values of the trigonometric functions in other quadrants?
4. How do these calculations change for other trigonometric identities, such as $\cos 2\theta$?
5. Can you show the geometric interpretation of these values on a coordinate plane?

Tip: For trigonometric functions, always use identities like the Pythagorean identity and reciprocal identities to simplify your calculations and ensure the values are correct.